Differentiation — Paper 1 · A Level Mathematics 9709

8 10 The equation of a curve is y = 2x + x2. dy d2y (i) Obtain expressions for and . [3] dx dx2 (ii) Find the coordinates of the stationary point on the curve and determine the nature of the stationary point. [3] (iii) Show that the normal to the curve at the point (−2, −2) intersects the x-axis at the point (−10, 0). [3] (iv) Find the area of the region enclosed by the curve, the x-axis and the lines x = 1 and x = 2. [3]

12 marks

Mark scheme: dy 16 −16/x3.10 (i) = 2 − 3 B1 For dx x d 2 y 48 B1 For “2” and for “0”. 2 = 4 B1√ For d/dx of his −16/x3 providing −ve dx x [3] power differentiated. dy (ii) =0 → x = 2, y = 6. M1 Sets dy/dx to 0 + attempt at x. dx A1 Needs both coordinates. d 2 y 2 is +ve Minimum. A1√ Looks at sign. Correct conclusion for dx [3] his x and his 2nd differential. (iii) x = −2 m = 4 Perp gradient = −¼ M1 Uses m1m2 = −1 with dy/dx. y + 2 = − 14 ( x + 2) DM1 Correct form of equation (not for tan) Sets y to 0 → x = −10 A1 Co nb answer given. [3]  2 8  (iv) Area = x −x  B1 B1 For each term  Evaluated from 1 to 2 → 7 B1 Co. (−7 ⇒ 7 gets B0) [3] 2

This question in 9709/11 May/June 2007

11 6 The diagram shows the graph of y = f(x), where f : x → for x ≥0. 2x + 3 (i) Find an expression, in terms of x, for f′(x) and explain how your answer shows that f is a decreasing function. [3] (ii) Find an expression, in terms of x, for f−1(x) and find the domain of f−1. [4] (iii) Copy the diagram and, on your copy, sketch the graph of y = f−1(x), making clear the relationship between the graphs. [2] The function g is defined by g : x →12x for x ≥0. (iv) Solve the equation fg(x) = 32. [3]

12 marks

Mark scheme: 11 (i) f´(x) = −6(2x+3)-2 × 2 B1 B1 co.(−ve power ok) B1 for ×2 Always −ve → Decreasing B1√ Answer given. Correct explanation. [3] 6 (ii) y = M1 Reasonable attempt in making x the 2 x + 3 subject (ok to interchange x, y first) M1 Order of operations must be correct ie 1  6  → f -1(x) =  −3  ÷ y, −3 then ÷ 2. 2  x  A1 Correct expression as f–1(x). Gets 2/3 for correct expression with y. Domain of f –1: 0 < x ≤ 2 B1 Could be independent of answer for f -1. [4] Condone < or ≤ (iii) B1 Correct graph for f -1(curve, stops on axis) B1 Makes clear on graph, or in words or [2] by the line y=x marked, the symmetry. 6 (iv) fg(x) = M1 x + 3 = 1.5 → x = 1 M1 g first, then f. Reverse (3÷(2x+3) M0 [or , using f-1, → g(x) = ½, ⇒ x = 1.] A1 Not DM – so can get this if attempt ok [ M1 M1 A1] co [3] DM1 for quadratic. Quadratic must be set to 0. Factors. Attempt at two brackets. Each bracket set to 0 and solved. Formula. Correct formula. Correct use, but allow for numerical slips in b² and −4ac.

This question in 9709/11 May/June 2007

11 y A C D y = x 3 – 6x 2 + 9x x O B The diagram shows the curve y = x3 −6x2 + 9x for x ≥0. The curve has a maximum point at A and a minimum point on the x-axis at B. The normal to the curve at C (2, 2) meets the normal to the curve at B at the point D. (i) Find the coordinates of A and B. [3] (ii) Find the equation of the normal to the curve at C. [3] (iii) Find the area of the shaded region. [5]

11 marks

Mark scheme: dy 11 (i) = 3x2 – 12x + 9 B1 co (can be given in part (ii)) dx dy Solves = 0 M1 Attempt to solve dy/dx = 0. dx → A (1, 4), B (3, 0). A1 Both needed. [3] (ii) If x = 2, m = −3 Normal has m = 1 M1 Use of m1m2 = −1. needs calculus. 3 Eqn y −2 = 1 (x − 2) or 3y = x + 4. M1 A1 Correct form of equation – needs calculus. 3 [3] A1 any form. (iii) area under curve – integrate y. x2 → 1 x4 − 2x3 + 9 B2,1 For the 3 terms. −1 for each error. 4 2 Limits 2 to “his 3” → ¾ (0.75) M1 Using 2 to “his 3” with integration. Area of trapezium = ½ × 1 × (2 + 2⅓) M1 Any correct method for trapezium. = 2 1 6 Subtract → shaded area of 1 5 A1 co 12 [5]

This question in 9709/11 May/June 2009

7 P r cm q rad O Q A piece of wire of length 50 cm is bent to form the perimeter of a sector POQ of a circle. The radius of the circle is r cm and the angle POQ is θ radians (see diagram). (i) Express θ in terms of r and show that the area, A cm2, of the sector is given by A = 25r −r2. [4] (ii) Given that r can vary, find the stationary value of A and determine its nature. [4]

8 marks

Mark scheme: 7 (i) 2r + rθ = 50 M1 Must use s = rθ and link with perimeter 1 θ = (50 – 2r) A1 co r A = 1 r2θ M1 Used with θ as f(r) 2 → A = 25r – r2 A1 co (answer given) [4] d A (ii) = 25 − 2 r B1 co dr = 0 when r = 12.5 M1 sets differential to 0 + solution A = 156¼ A1 co 2nd differential negative → Maximum B1 Could be quoted directly from quadratic. [4] GCE A/AS LEVEL – October/November 2009 9709 12 3

This question in 9709/12 Oct/Nov 2009

3 8 The function f is such that f(x) = 2x 5 for x ∈>, x ≠−2.5. + (i) Obtain an expression for f ′(x) and explain why f is a decreasing function. [3] (ii) Obtain an expression for f −1(x). [2] (iii) A curve has the equation y = f(x). Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = 2 is rotated through 360◦about the x-axis. [4]

9 marks

Mark scheme: 3 8 x a 2 x + 5 (i) fV(x) = –3(2x + 5)–2 × 2 B1 B1 B1 for –3(2x + 5)–2. B1 for ×2 fV(x) is negative → decreasing B1√ √ providing bracket is squared. [3] (using value or values only B0) 3 3 (ii) y = → 2 x + 5 = M1 Attempt at making x the subject. 2 x + 5 y –1 1  3  3− 5 x → f (x) =  −5  or A1 co including f(x) not f(y) 2  x  2 x [2] 9 2 (iii) ∫ π ( 2 x + 5) dx B1 For –9(2x + 5)–1 = (–9π(2x + 5)–1 ÷ 2) B1 For ÷ 2 in ∫ of y2 Limits 0 to 2 → π (−½ − −0.9) M1 Use of correct limits with ∫ of y2. → = 0.4π (or 1.26) A1 co [4]

This question in 9709/12 Oct/Nov 2009

10 y y = x 2 – 4x + 7 2y = x + 5 B A x O (i) The diagram shows the line 2y x 5 and the curve y x2 7, which intersect at the points = + = −4x + A and B. Find (a) the x-coordinates of A and B, [3] (b) the equation of the tangent to the curve at B, [3] (c) the acute angle, in degrees correct to 1 decimal place, between this tangent and the line 2y = x + 5. [3] (ii) Determine the set of values of k for which the line 2y = x + k does not intersect the curve y x2 7. [4] = −4x +

13 marks

Mark scheme: 10 (i) (a) 2y = x + 5, y = x2 – 4x + 7 Sim equations → 2x2 – 9x + 9 = 0 M1 Complete elimination of x or y → x = 3 or x = 1½. DM1 A1 Correct method for quadratic. co. [3] dy (b) = 2 x − 4 B1 co dx → y – 4 = 2(x – 3) M1 A1 Correct form of eqn with m numeric. co [3] nb use of y + 4 or x, y interchanged M1 A0 (c) m = 2 → angle of 63.4º m = ½ → angle of 26.6º M1 Finds angle with x-axis once. → angle between = 37º M1A1 Subtracts two angles. co. [3] (i+2j).(2i+j) → 4=√5√5cosθ M1M1A1 or use of tan(A–B) M2A1 or Cosine rule with 3 sides found. (ii) y = x2 – 4x + 7 2y = x + k Sim eqns → 2x2 – 9x + 14 – k = 0 M1 A1 Eliminates y or x completely. Co (= 0) Uses b2 – 4ac, 81 − 8(14 − k) M1 Uses b2 – 4ac = 0, or < 0 or > 0 Key value is k = 3.875 or 31/8. k < 3.875 A1 Co condone Y. [4]

This question in 9709/12 Oct/Nov 2009

dy 6 5 The equation of a curve is such that dx = √(3x −2). Given that the curve passes through the point P (2, 11), find (i) the equation of the normal to the curve at P, [3] (ii) the equation of the curve. [4]

7 marks

Mark scheme: dy 6 5 = dx 3 x − 2 (i) x = 2, tangent has gradient 3 M1 Use of mlm2 = –1 with dy/dx 1 → normal has gradient − M1 A1 Correct form of line eqn. for normal 3 1 → y − 11 = − ( x − 2 ) [3] 3 3 x − 2 B1 Without the ÷3 ÷ 3 (ii) Integrate → 6 B1 For ÷3, even if B0 above 1 2 → y = 4 3 x − 2 + c through (2,11) M1 Using (2, 11) for c A1 co → y = 4 3 x − 2 + 3 [4]

This question in 9709/13 May/June 2010

9 y 4 y = x + x y = 5 A B M x O 4 The diagram shows part of the curve y = x + x which has a minimum point at M. The line y = 5 intersects the curve at the points A and B. (i) Find the coordinates of A, B and M. [5] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis. [6]

11 marks

Mark scheme: 4 9 y = x + x 4 (i) x + = 5 → A (1, 5), B(4, 5) B1 B1 co. co. x dy 4 = 1 − 2 M1 Differentiates. dx x = 0 when x = 2, M (2, 4). DM1 A1 Setting to 0. co. [5] (ii) Vol of cylinder = π52.3 B1 Any valid method. Vol under curve = π y 2 dx M1 Attempt at integrating y2 ∫ x 3 16 Integral = − + 8 x A2, 1, 0 Allow if no π present. 3 x Uses his limits “1 to 4” DM1 Using his limits. → 75π − 57π = 18π A1 co. [6] 2

This question in 9709/13 May/June 2010

8 x cm y cm x cm The diagram shows a metal plate consisting of a rectangle with sides x cm and y cm and a quarter-circle of radius x cm. The perimeter of the plate is 60 cm. (i) Express y in terms of x. [2] (ii) Show that the area of the plate, A cm2, is given by A 30x [2] = −x2. Given that x can vary, (iii) find the value of x at which A is stationary, [2] (iv) find this stationary value of A, and determine whether it is a maximum or a minimum value. [2] [Questions 9, 10 and 11 are printed on the next page.]

8 marks

Mark scheme: πx 8 (i) 2 x + 2 y + = 60 M1 Linking 60 with sum of at least 4 sides 2 and use of radians πx → y = 30 − x − A1 co 4 [2] πx 2 (ii) A = xy + 4 πx πx 2 1 2 = x (30 − x − ) + M1 Subs “y” into area eqn and use r θ 4 4 2 = 30x – x2 A1 co. [2] dA (iii) = 30 − 2 x Knowing to differentiate dx = 0 when x = 15 cm M1 A1 Sets differential to 0 + solution. co. [2] (iv) Max. M1 A1 Any valid method. co. [2] GCE AS/A LEVEL – October/November 2010 9709 11

This question in 9709/11 Oct/Nov 2010

10 The equation of a curve is y 3 4x = + −x2. (i) Show that the equation of the normal to the curve at the point is 2y x 9. [4] (3, 6) = + (ii) Given that the normal meets the coordinate axes at points A and B, find the coordinates of the mid-point of AB. [2] (iii) Find the coordinates of the point at which the normal meets the curve again. [4]

10 marks

Mark scheme: 10 y = 4x – x2 + 3 dy (i) = 4 − 2 x B1 co dx At x = 3, m = − 2 1 Gradient of normal = M1 Use of m1m2 = −1 2 Eqn of normal y − 6 = 12 ( x − 3) M1 A1 Use of y – k = m(x – h) or y = mx + c → 2y = x + 9 (where m is gradient of normal) [4] 9 (ii) Meets axes at (0, ) and (−9, 0) M1 Sets x and y to 0 + midpoint formula. 2 − 9 9  Mid-point is  ,  A1 co.  2 4  [2] (iii) 2y = x + 9, y = 4x – x2 + 3 → 2x2 – 7x + 3 = 0 oe M1 A1 Eliminates x completely. Correct eqn. → (½, 4¾) M1 A1 Solution of quadratic. co [4] GCE AS/A LEVEL – October/November 2010 9709 11 9 11 y = 2 − x dy 2

This question in 9709/11 Oct/Nov 2010

3 The length, x metres, of a Green Anaconda snake which is t years old is given approximately by the formula x = 0.7 √(2t −1), where 1 ≤t ≤10. Using this formula, find dx (i) , [2] dt (ii) the rate of growth of a Green Anaconda snake which is 5 years old. [2]

4 marks

Mark scheme: 3 (i) (k(2t – 1)–1/2 M1 k ≠ 1 0.7(2t – 1)–1/2 A1 oe [2] (ii) Sub t = 5 into their deriv M1 0.23(3) A1 Ignore units [2]

This question in 9709/12 Oct/Nov 2010

10 5 x 4 4 x 5 h 1 x 2 x The diagram shows an open rectangular tank of height h metres covered with a lid. The base of the tank has sides of length x metres and 2x1 metres and the lid is a rectangle with sides of length 4x5 metres and 5x4 metres. When full the tank holds 4 m3 of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is A m2. 3 24 (i) Express h in terms of x and hence show that A = 2x2 + x . [5] (ii) Given that x can vary, find the value of x for which A is a minimum, showing clearly that A is a minimum and not a maximum. [5]

10 marks

Mark scheme: 8 10 (i) h = 2 M1 Uses lbh = 4 x A1 co 1 2 1 5 4 A = x + 2 × xh + 2 xh + x × x M1 Allow 1 error but needs the lid 2 2 4 5 A = (3 / 2 ) x 2 + 3 xh 3 2 8 A = x + 3 x × 2 M1 For substitution of h as f(x) 2 x 3 2 24 A = x + A1 AG 2 x [5] dA 24 (ii) = 3 x − = 0 B1 Correct derivative. 2 dx x M1 Sets to 0 and attempts to solve. x = 2 A1 co d 2 A 48 = 3 + M1 Reasonable attempt – allow 1 error dx 2 x 3 > 0 when x = 2 hence minimum A1 co [5] AG (Result consistent with their f'')

This question in 9709/12 Oct/Nov 2010

1 5 A curve has equation y = + x. x −3 dy d2y (i) Find and . [2] dx dx2 (ii) Find the coordinates of the maximum point A and the minimum point B on the curve. [5]

7 marks

Mark scheme: dy 1 =5 (i) + 1 B1 oe dx ( x − 3) 2 d 2 y 2 = B1 oe d x 2 ( x − 3)3 [2] (ii) (x – 3)2 = 1 ⇒ x – 3 = ±1 M1 dy Set = 0 & reasonable attempt to dx solve x = 4, 2 A1 y = 5, 1 A1 d 2 y When x = 4 > 0 (= 2) ⇒ min M1 Investigate signs of f″ at a point or 2 d x other method d 2 y When x = 2 < 0 (= –2) ⇒ max A1 2 d x [5] GCE AS/A LEVEL – October/November 2010 9709 13

This question in 9709/13 Oct/Nov 2010

6 A curve has equation y = f(x). It is given that f ′(x) = 3x2 + 2x −5. (i) Find the set of values of x for which f is an increasing function. [3] (ii) Given that the curve passes through (1, 3), find f(x). [4]

7 marks

Mark scheme: 6 (i) (3x + 5)(x – 1)(> 0) M1 Attempt at factorisation –5/3, 1 A1 Both required x < –5/3, x > 1 A1 Ignore any words between answers Condone < > [3] (ii) f(x) = x3 + x2 – 5x (+ c) M1 Attempt at integration A1 Any unsimplified expression ok 3 = 1 + 1 – 5 + c M1 Sub. (1, 3) f(x) = x3 + x2 – 5x + 6 A1 Accept c = 6 [4]

This question in 9709/13 Oct/Nov 2010

11 y P y = 9 – x3 8 Q y = x3 x O a b 8 The diagram shows parts of the curves y = 9 −x3 and y = and their points of intersection P and Q. x3 The x-coordinates of P and Q are a and b respectively. (i) Show that x = a and x = b are roots of the equation x6 −9x3 + 8 = 0. Solve this equation and hence state the value of a and the value of b. [4] (ii) Find the area of the shaded region between the two curves. [5] (iii) The tangents to the two curves at x = c (where a < c < b) are parallel to each other. Find the value of c. [4]

13 marks

Mark scheme: 3 8 11 (i) 9 − x = 3 M1 Together with attempt to mult by x3 x x6 – 9x3 + 8 = 0 A1 AG completely correct working (X – 1)(X – 8) = 0 → X = 1 or 8 M1 Attempt to solve quadratic in X or x3 a = 1, b = 2 A1 [4] 2  3 8  dx M1 Intention to integrate the difference ( 9 − x ) − 3 (ii) ∫1  x  y1 – y2 not π(y1 – y2)  x 4  − 4  B1 9 x −  ⋅ 2 B1  4   x  1 18 − 4 + 1 − (9 − + 4 ) M1 Correct use of their limits once 4 1 2 A1 4 [5] dy − 24 dy (iii) = , = –3x2 B1, B1 cao dx dx x 4 − 24 = –3c2 c 4 c6 = 8 M1 Equating and solution c = 2 or 81/6 or 1.41(4...) A1 Accept x or c [4]

This question in 9709/13 Oct/Nov 2010

2 The volume of a spherical balloon is increasing at a constant rate of 50 cm3 per second. Find the rate of increase of the radius when the radius is 10 cm. [Volume of a sphere = 43πr3.] [4]

4 marks

Mark scheme:  dv  22  =  4πr M1 dr   2 A1 SOI at any point = 4π × 10 dv dr dt M1 Correct link between differentials with = OE used dt dv dr dr finally as subject dt 50 1 50 = or 0.0398 2 A1 Allow . 4π × 10 8π 400π [4] 0 Non-calculus methods 4

This question in 9709/11 May/June 2011

6 The variables x, y and ß can take only positive values and are such that ß = 3x + 2y and xy = 600. 1200 (i) Show that ß = 3x + x . [1] (ii) Find the stationary value of ß and determine its nature. [6]

7 marks

Mark scheme:  600  ( z 3 x )6 (i) z = 3 x + 2  or x = 600 OE B1  x  2 [1] → AG d z 1200 dz 1800 (ii) = 3 − or = 2 − B1 d x x 2 dy y 2 = 0 → x = 20 or = 0 → y = 30 M1A1 Set to 0 & attempt to solve. Allow ±20 Ft from their x provided positive 120 z = 60 + = 120 A1√ Or other valid method 20 d 2 z 2400 d 2 z k = B1√ Dep. on (k > 0) or other = dx 2 x 3 dx 2 x 3 > 0 ⇒ minimum B1 valid method. [6] 31( + 2 x ) −1 B1

This question in 9709/11 May/June 2011

dy 3 17 A curve is such that dx = and the point (1, 2) lies on the curve. (1 + 2x)2 (i) Find the equation of the curve. [4] (ii) Find the set of values of x for which the gradient of the curve is less than 3.1 [3]

7 marks

Mark scheme: 31( + 2 x ) 7 (i) + ( c ) B1 − 1 31( + 2 x ) −1 y = + ( c ) B1(indep) Division by 2 y = necessary − 2 Sub (1, (1/2)) M1 Dependent on c present 1 3 = + c ⇒ c = 1 A1 Use of y = mx + c etc. gets 0/4 2 − 6 [4] (ii) (1 + 2x)2(>)9 or 4x2 + 4x – 8(>)0 OE M1 1, ‒2 A1 x > 1, x < –2 ISW A1 [3]

This question in 9709/11 May/June 2011

2 A curve has equation y = 3x3 −6x2 + 4x + 2. Show that the gradient of the curve is never negative. [3]

3 marks

Mark scheme: 2 1 Allow +√ or √. Dep on final ans as (v) ( x − 2 ) = ( y − 2 ) M1 n 2 f of x 1 x = 2 ± ( y − 2 ) M1 2 1 ( ) ( )

This question in 9709/11 Oct/Nov 2011

7 x 2y 3y 3x y 4x The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of the garden is 48 m. (i) Find an expression for y in terms of x. [1] (ii) Given that the area of the garden is A m2, show that A = 48x −8x2. [2] (iii) Given that x can vary, find the maximum area of the garden, showing that this is a maximum value rather than a minimum value. [4]

7 marks

Mark scheme: 1 7 (i) y = oe B1 [1] 6(48 − 8 x ) (ii) A = 4 xy + 2 xy or 3 xy + 3 xy = 6 xy M1 A = x (48 − 8 x ) = 48 x − 8 x 2 A1 [2] AG δA (iii) = 48 − 16 x B1 δx Attempt to solve derivative = 0 A = 72 cao M1A1 Expect x = 3 δ 2 A = − 16 (< 0 ) ⇒ Maximum B1 [4] www Accept other complete methods 2 δx x x + y y + z z

This question in 9709/11 Oct/Nov 2011

10 y y = Ö(1 + 2x ) C B x A O meeting the x-axis at A and the y-axis at B. The The diagram shows the curve y = √(1 + 2x) y-coordinate of the point C on the curve is 3. (i) Find the coordinates of B and C. [2] (ii) Find the equation of the normal to the curve at C. [4] (iii) Find the volume obtained when the shaded region is rotated through 360◦about the y-axis. [5]

11 marks

Mark scheme: If B0B0 then SCB1 for both y 1 & 10 (i) B = ()1,0 C = (3,4) B1, B1 [2] x = 4 1 δy 1 − 1 − 2 required & at least one of 1 × 2 (ii) = × 2(1 + 2 x ) 2 M1A1 2 δx 2 for M1 Grad. of normal = −3 B1 y − 3 = −3( x − 4 ) or y = −3 x + 15 oe B1√ [4] Ft only from their C 2 1 2 2 1 x δy , square ( y − )1 & attempt ∫ (iii) y = 1 + 2 x ⇒ x = SOI B1 2 2 2 ( y − )1 n int 1 4 2 (π ) × × ( y − 2 y + 1)δy M1 ∫ 4 1  y 5 2 y 3  Apply limits 0 → their 1 (from their (π ) × − + y A1 B) 4  5 3  2 π 2 1  1  − + 1 (π ) × DM1 cao SCB1 for ∫ y δx →4 (scores 4  5 3  1/5) 2 π A1 [5] 15 ( ) 2 B 1 B1

This question in 9709/11 Oct/Nov 2011

dy 7 A curve is such that . The line 3y + x = 17 is the normal to the curve at the point P on the dx = 5 −8x2 curve. Given that the x-coordinate of P is positive, find (i) the coordinates of P, [4] (ii) the equation of the curve. [4]

8 marks

Mark scheme: dy 8 7 = 5 − 2 , Normal 3 y + x = 17 dx x (i) Gradient of line = −⅓ B1 co dy M1 Use of m1m2 = − 1 = 3 → x = 2, y = 5 DM1 DM1 solution. A1 co. dx A1 [4] (ii) y = 5 x + 8 x −1 (+ c ) B1 B1 co.co. doesn’t need +c. Uses (2, 5) → c = −9 M1 A1 Use of +c following integration. co. [4] GCE AS/A LEVEL – October/November 2011 9709 12 2

This question in 9709/12 Oct/Nov 2011

Find8 The equation of a curve is y = √(8x −x2). dy (i) an expression for dx, and the coordinates of the stationary point on the curve, [4] (ii) the volume obtained when the region bounded by the curve and the x-axis is rotated through 360◦about the x-axis. [4] [Questions 9 and 10 are printed on the next page.]

8 marks

Mark scheme: 8 y = 8 x − x 2 dy (i) 2 × (8 − 2 x ) B1 B1 for everything but ×(8-2x) = 12 (8 x − x 2 ) − 1 dx B1 B1 for × (8−2x), even if B0 = 0 when x = 4. M1 Sets to 0 + attempt at solution. → (4, 4) A1 Co – A0 if fortuitous because of B0 [4] earlier. (ii) y = 0 when x = 0 or 8 B1 Vol = π ∫ (8 x − x 2 d)x Anywhere  2 x 3  B2,1 = π 4 x −1 for each error (not including π)  −3  256π B1 → 3 [4] co

This question in 9709/12 Oct/Nov 2011

4 A watermelon is assumed to be spherical in shape while it is growing. Its mass, M kg, and radius, r cm, are related by the formula M kr3, where k is a constant. It is also assumed that the radius is increasing at a constant rate of 0.1 centimetres= per day. On a particular day the radius is 10 cm and the mass is 3.2 kg. Find the value of k and the rate at which the mass is increasing on this day. [5]

5 marks

Mark scheme: 2.3 2 4 1000k = 3.2 ⇒k = or or 0.0032 oe M1A1 1000 625  dM  2   = 3kr B1 dr   dM dM dr 2 = × used e.g. 3 × k × 10 × 1.0 M1 Must eventually make dM/dt subject dt dr dt cao. Non-calculus methods (e.g. → A1 0.096 0.09696) can score only 1st 2 marks [5] ( )2

This question in 9709/11 May/June 2012

9 y B (0, 3) 9 y = 2 x + 3 C A (3, 1) x O 9 The diagram shows part of the curve y = crossing the y-axis at the point B (0, 3). The point 2x + 3, A on the curve has coordinates (3, 1) and the tangent to the curve at A crosses the y-axis at C. (i) Find the equation of the tangent to the curve at A. [4] (ii) Determine, showing all necessary working, whether C is nearer to B or to O. [1] (iii) Find, showing all necessary working, the exact volume obtained when the shaded region is rotated through 360◦about the x-axis. [4]

9 marks

Mark scheme: → y 1 = 9 ( x )3 [4] (normal →max 2/4, no calculus 0/4) (ii) Meets the y-axis when x = 0, y = 1⅔ B1 Sets x to 0 in his tangent. This is nearer to B than to O. [1] The 1⅔ and part (i) must be correct.

This question in 9709/12 Oct/Nov 2012

6 The non-zero variables x, y and u are such that u = x2y. Given that y + 3x = 9, find the stationary value of u and determine whether this is a maximum or a minimum value. [7]

7 marks

Mark scheme: 6 u = x 2 y y + 3 x = 9 M1 Expressing u in terms of 1 2 variable 2  9 − y  u = x (9 − 3 x ) or   y  3  du du DM1A1 Knowing to differentiate. =18x − 9x² or =27 − 12y + y² dx dy = 0 when x = 2 or y = 3 → u = 12 DM1 Setting differential to 0. A1 d 2 u DM1 Any valid method 2 =18−18x −ve A1 dx [7]

This question in 9709/13 May/June 2013

10 y y = (3 – 2 x )3 2(1 , 8) x O The diagram shows the curve y = 3 −2x 3 and the tangent to the curve at the point 12, 8 . (i) Find the equation of this tangent, giving your answer in the form y = mx + c. [5] (ii) Find the area of the shaded region. [6]

11 marks

Mark scheme: dy 2 2 2 ] B1B1 OR − 54 + 72 x − 24 x B2,1,010 (i) = [ 3(3 − 2 x ) ]× [− dx 1 dy At x = , = −24 M1 2 dx  1  y − 8 = −24 x −  DM1  2  y = −24 x + 20 A1 [5]  (3 − 2 x )4  1  2 3 4 (ii) Area under curve =   × − B1B1 OR 27 x − 27 x + 12 x − 2 x B2,1,0 2   4   81  − 2 − −  M1 Limits 0→ ½ applied to integral with  8  intention of subtraction shown − 24x + 20 ) M1 or area trap =½(20 + 8) × ½ Area under tangent = ∫ ( = − 12 x 2 + 20 x or 7 (from trap) A1 Could be implied 9 or 1.125 A1 Dep on both M marks 8 [6]

This question in 9709/11 Oct/Nov 2013

k2 9 A curve has equation y = + x, where k is a positive constant. Find, in terms of k, the values of x + 2 x for which the curve has stationary points and determine the nature of each stationary point. [8]

8 marks

Mark scheme: dy 9 = − k 2 ( x + 2 )− 2 + 1 = 0 M1A1 Attempt differentiation & set to zero dx x + 2 = ± k DM1 Attempt to solve x = –2 ± k A1 cao d 2 y 2 − 3 = 2 k ( x + 2 ) M1 Attempt to differentiate again 2 d x d 2 y M1 Sub their x value with k in it into 2 d x d 2 y  2  When x = –2 = k, 2 =   which is (> 0) min A1 Only 1 of bracketed items needed for each d x  k  d d 2 2 y 2 y   = When x = –2 – k,   which is (< 0) A1 but 2 and x need to be correct. 2 d x k d x   − max [8] GCE A LEVEL – October/November 2013 9709 13

This question in 9709/13 Oct/Nov 2013

9 The base of a cuboid has sides of length x cm and 3x cm. The volume of the cuboid is 288 cm3. (i) Show that the total surface area of the cuboid, A cm2, is given by 768 A = 6x2 + . x (ii) Given that x can vary, find the stationary value of A and determine its nature. [5]

5 marks

Mark scheme: 9 (i) 3x²y = 288 y is the height B1 co A = 2(3x² + xy + 3xy) M1 Considers at least 5 faces (y ≠ x) Sub for y → A = 6x2 + 768 A1 co answer given x [3] d 768 A = 12 x − (ii) B1 co d x x 2 = 0 when x = 4 → A = 288. Allow (4 , 288) M1 A1 Sets differential to 0 + solution. co 2 d 1536 A = 12 + M1 Any valid method 2 d x x 3 (= 36) > 0 Minimum A1 co www dep on correct f″ and x = 4 [5] GCE AS/A LEVEL – May/June 2014 9709 13

This question in 9709/13 May/June 2014

9 The function f is defined for x > 0 and is such that f ′ x = 2x −2 . The curve y = f x passes through x2 the point P 2, 6 . (i) Find the equation of the normal to the curve at P. [3] (ii) Find the equation of the curve. [4] (iii) Find the x-coordinate of the stationary point and state with a reason whether this point is a maximum or a minimum. [4]

11 marks

Mark scheme: 9 (i) f ′( 2) = 4 − 12 = 72 → gradient of normal = − 72 B1M1 y − 6 = − 2 ( x − 2) AEF A1 Ft from their f ′(2 ) 7 [3] (ii) f ( x ) = x 2 + 2 ( + c ) B1B1 x 6 = 4 + 1 + c ⇒ c = 1 M1A1 Sub (2, 6) – dependent on c being present [4] (iii) 2 x − 2 = 0 ⇒ 2 x 3 − 2 = 0 M1 Put f ′( x ) = 0 and attempt to solve x 2 x = 1 A1 Not necessary for last A mark as x > 0 given 4 f ′′ ( x ) = 2 + or any valid method M1 x 3 f ′′(1) = 6 OR > 0 hence minimum A1 Dependent on everything correct [4] 2

This question in 9709/11 Oct/Nov 2014

12 4 A curve has equation y = 3 −2x. dy (i) Find dx. [2] A point moves along this curve. As the point passes through A, the x-coordinate is increasing at a rate of 0.15 units per second and the y-coordinate is increasing at a rate of 0.4 units per second. (ii) Find the possible x-coordinates of A. [4]

6 marks

Mark scheme: 12 4 y = 3 − 2 x (i) Differential = −12(3 – 2x)−2 × −2 B1 B1 co co (even if 1st B mark lost) [2] d y d y d x (ii) = ÷ = 0.4 ÷ 0.15 M1 Chain rule used correctly (AEF) d x d t d t 24 8 dy 8 3 → = M1 Equates their with their or 2 (3 − 2 x ) 3 dx 3 8 → x = 0 or 3 A1 A1 co co [4]

This question in 9709/12 Oct/Nov 2014

6 The equation of a curve is y = x3 + ax2 + bx, where a and b are constants. (i) In the case where the curve has no stationary point, show that a2 < 3b. [3] (ii) In the case where a = −6 and b = 9, find the set of values of x for which y is a decreasing function of x. [3]

6 marks

Mark scheme: 6 y = x3 + ax2 + bx dy (i) = 3x² + 2ax + b B1 co dx dy (ii) b² − 4ac = 4a² − 12b (I 0) M1 Use of discriminant on their quadratic dx or other valid method → a² I= 3b A1 co – answer given [3] (iii) y = x³ − 6x² + 9x dy = 3x² − 12x + 9 I 0 M1 Attempt at differentiation dx = 0 when x = 1 and 3 A1 co → 1 I x I 3 A1 condone < [3]

This question in 9709/12 Oct/Nov 2014

3 (i) Express 9x2 −12x + 5 in the form ax + b 2 + c. [3] (ii) Determine whether 3x3 −6x2 + 5x −12 is an increasing function, a decreasing function or neither. [3]

6 marks

Mark scheme: 3 (i) (3 x − 2 ) 2 + 1 B1B1B1 For either of 1st 2 marks bracket must be in the form (ax + b )2 except for 2  2  SCB2 for 9 x −  + 1  3  [3] (ii) f ′( x ) = 9 x 2 − 12 x + 5 B1 = their (3 x − 2 ) 2 + 1 M1 Ft from (i). Some > 0 (or > 1) hence an increasing function A1 reference/recognition [3] Allow > 1. Allow their 1 provided positive. Allow a complete alt method (2/2 or 0/2) 2 3

This question in 9709/13 Oct/Nov 2014

2 y y = 2x2 Q x X −2, 0 O P p, 0 The diagram shows the curve y = 2x2 and the points X −2, 0 and P p, 0 . The point Q lies on the curve and PQ is parallel to the y-axis. (i) Express the area, A, of triangle XPQ in terms of p. [2] The point P moves along the x-axis at a constant rate of 0.02 units per second and Q moves along the curve so that PQ remains parallel to the y-axis. (ii) Find the rate at which A is increasing when p = 2. [3]

5 marks

Mark scheme:  1  2   cos  x  = B1  2  3  2  3 1 x = 0.84 x = 1.68 only, aef M1A1 Looks up cos−1 first, then ×2 2 [3] (in given range) (ii) B1 y always +ve, m always –ve. B1 from (0, 8) to (2π, 2) (may be [2] implied) 2 (iii) No turning point on graph or 1:1 B1 cao, independent of graph in (ii) [1] 1   M1 Tries to make x subject. x (iv) y = 5 + 3cos    2  Order; −5, ÷3, cos−1, ×2 M1 Correct order of operations  x − 5  A1 cao x = 2cos−1    3  [3] 9 y = x 3 + px 2 dy (i) = 3x² + 2px B1 cao dx 2p Sets to 0 → x = 0 or − M1 Sets differential to 0 3 2 p 4 p 3  → (0, 0) or − , A1 A1 cao cao, first A1 for any correct    3 27  [4] turning point or any correct pair of x values. 2nd A1 for 2 complete TPs d 2 y (ii) 2 = 6x + 2p M1 Other methods include; clear dx demonstration of sign change of gradient, clear reference to the shape of the curve At (0, 0) → 2p +ve Minimum A1 www  

This question in 9709/11 May/June 2015

10 y A 2, 9 y = 9 + 6x −3x2 x O B C 3, 0 Points A 2, 9 and B 3, 0 lie on the curve y = 9 + 6x −3x2, as shown in the diagram. The tangent at A intersects the x-axis at C. Showing all necessary working, (i) find the equation of the tangent AC and hence find the x-coordinate of C, [4] (ii) find the area of the shaded region ABC. [5] [Question 11 is printed on the next page.]

9 marks

Mark scheme: d y10 (i) = 6 − 6 x B1 d x At x = 2 , gradient = −6 soi B1 y − 9 = −6 ( x − 2 ) oe Expect y = −6 x + 21 M1 Line through (2, 9) and with gradient their −6 When y = ,0 x = 3 12 cao A1 [4] (ii) Area under curve: ∫ 9 + 6 x − 3 x 2 dx = 9 x + 3 x 2 − x 3 B2,1,0 Allow unsimplified terms ( 27 + 27 − 27 ) − (18 + 12 − 8) M1 Apply limits 2,3. Expect 5 27 3 27 1 × × 9 ( = Area under tangent: 2 ( −6 x + 21) dx ( → ). Ft on their ) B1 OR ∫ 2 7 2 2 4 4 −x6 + 21 and/or their 7/2. 27 7 Area required − 5 = A1 4 4 [5]

This question in 9709/13 May/June 2015

8 5 A curve has equation y = + 2x. x dy d2y (i) Find and . [3] dx dx2 (ii) Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point. [5]

8 marks

Mark scheme: d y 8 5 (i) = − + 2 cao B1B1 d x x 2 d 2 y 16 = cao B1 dx 2 x 3 [3] 8 2 (ii) − + 2 = 0 → 2 x − 8 = 0 M1 Set = 0 and rearrange to quadratic form 2 x x = ± 2 A1 y = ± 8 A1 If A0A0 scored, SCA1 for just (2, 8) d 2 y  Ft for " correct" conclusion if  > 0 when x = 2 hence MINIMUM B1 2 2 dx  d y  2  2 incorrect or  d y < 0 when x = − 2 hence MAXIMUM B1  dx  dx 2 [5] any valid method inc. a good sketch  2 2

This question in 9709/11 Oct/Nov 2015

6 A vacuum flask (for keeping drinks hot) is modelled as a closed cylinder in which the internal radius is r cm and the internal height is h cm. The volume of the flask is 1000 cm3. A flask is most efficient when the total internal surface area, A cm2, is a minimum. 2000 (i) Show that A = 20r2 + . [3] r (ii) Given that r can vary, find the value of r, correct to 1 decimal place, for which A has a stationary value and verify that the flask is most efficient when r takes this value. [5]

8 marks

Mark scheme: 6 (i) A = 2π r 2 + 2π rh B1 2 1000 π r h = 1000 → h = 2 M1 π r 2 2000 Sub for h into A → A = 2π r + AG A1 r [3] d A 2000 (ii) = 0 ⇒ 4π r − 2 = 0 M1A1 Attempt differentiation & set = 0 d r r r = = 5.4 DM1 A1 Reasonable attempt to solve to 3r = d 2 A 4000 = 4π + dr 2 r 3 > 0 hence MIN hence MOST EFFICIENT AG B1 Or convincing alternative method [5] 3

This question in 9709/12 Feb/March 2016

10 y Q 3, 4 y = 161 3x −1 2 x O P R The diagram shows part of the curve y = 1 3x −1 2, which touches the x-axis at the point P. The 16 point Q 3, 4 lies on the curve and the tangent to the curve at Q crosses the x-axis at R. (i) State the x-coordinate of P. [1] Showing all necessary working, find by calculation (ii) the x-coordinate of R, [5] (iii) the area of the shaded region PQR. [6]

12 marks

Mark scheme: 10 (i) x = 1/ 3 B1 [1] dy  2  = (ii) [ 3] B1B1  ( 3 x − 1)  16 dx   dy When x = 3 = 3 soi M1 dx Equation of QR is y − 4 = 3 ( x − 3 ) M1 When y = 0 x = 5 / 3 A1 [5]  1 3  1  (iii) Area under curve =  ( 3 x − 1) ×  B1B1  16 × 3  3  1 3 32 1 8 − 0  = M1A1 Apply limits: their and 3   16 × 9 9 3 Area of ∆= 8 / 3 B1 32 8 8 Shaded area = − = (or 0.889) A1 9 3 9 [6]

This question in 9709/12 Feb/March 2016

10 y 8 y = + 2x x M x O 8 The diagram shows the part of the curve y 2x for x 0, and the minimum point M. = x + > dy d2y (i) Find expressions for dx, dx2 and Ó y2 dx. [5] (ii) Find the coordinates of M and determine the coordinates and nature of the stationary point on the part of the curve for which x 0. [5] < (iii) Find the volume obtained when the region bounded by the curve, the x-axis and the lines x 1 = and x 2 is rotated through about the x-axis. [2] = 360Å

12 marks

Mark scheme: ff(x) = 10 − 3 (10 − 3x ) B1 Correct unsimplified expression 10 gf(2) = (= −2) B1 Correct unsimplified expression ( ( ) )

This question in 9709/12 May/June 2016

dy k 3 A curve is such that = 6x2 + and passes through the point P 1, 9 . The gradient of the curve dx x3 at P is 2. (i) Find the value of the constant k. [1] (ii) Find the equation of the curve. [4]

5 marks

Mark scheme: 3 (i) 6 + k = 2 → k = −4 B1 [1] 6 x 3 4 2 k 2 (ii) ( y ) = x− (+c) B1B1 ft on their k. Accept + x− 3 −−2 − 2 9 = 2 + 2 + c c must be present M1 Sub (1,9) with numerical k. Dep on attempt ∫ ( y ) = 2 x 3 + 2 x−2 + 5 A1 Equation needs to be seen [4] Sub (2, 3) →c = –13½ scores M1A0 3 + 2 d 3 + 12 d 2 3 + 12 d

This question in 9709/13 May/June 2016

1 9 11 The point P 3, 5 lies on the curve y = − x −1 x −5. (i) Find the x-coordinate of the point where the normal to the curve at P intersects the x-axis. [5] (ii) Find the x-coordinate of each of the stationary points on the curve and determine the nature of each stationary point, justifying your answers. [6]

11 marks

Mark scheme: dy −2 −211 (i) = − ( x − 1) + 9 ( x − 5 ) M1A1 May be seen in part (ii) dx 1 9 mtangent =− + = 2 B1 4 4 Equation of normal is y − 5 = −½ ( x − 3 ) M1 Through (3, 5) and with m = −1/ mtangent x = 13 A1 [5] (ii) 2 2 dy ( x − 5 ) = 9 ( x − 1) B1 Set = 0 and simplify dx x 2 − x − 2 = 0 M1 Simplify further and attempt x − 5 = ( ± ) 3 ( x − 1) or ( 8 )( ) solution x = − 1 or 2 A1 d 2 y − 3 − 3 = 2 ( x − 1) − 18 ( x − 5 ) B1 If change of sign used, x values 2 d x close to the roots must be used and all must be correct d 2 y 1 When x = − 1, =− < 0 MAX B1 2 d x 6 d 2 y 8 When x = 2, = > 0 MIN B1 2 d x 3 [6]

This question in 9709/11 Oct/Nov 2016

3 7 The equation of a curve is y = 2 + 2x −1. dy (i) Obtain an expression for dx. [2] (ii) Explain why the curve has no stationary points. [1] At the point P on the curve, x = 2. (iii) Show that the normal to the curve at P passes through the origin. [4] (iv) A point moves along the curve in such a way that its x-coordinate is decreasing at a constant rate of 0.06 units per second. Find the rate of change of the y-coordinate as the point passes through P. [2]

9 marks

Mark scheme: dy −3 7 (i) = × 2 B1 B1for a single correct term (unsimplified) dx ( 2 x − 1) 2 without ×2. B1 [2] dy (ii) e.g. Solve for = 0 is impossible. B1 Satisfactory explanation. dx [1] dy −6 (iii) If x = 2, = and y = 3 M1* Attempt at both needed. dx 9 9 Perpendicular has m = M1* Use of m1m2 = −1 numerically. 6 3 → y − 3 = ( x − 2 ) DM1 Line equation using (2, their 3) and their m. 2 Shows when x=0 then y=0 AG A1 [4] dx (iv) = −0.06 dt dy dy dx 2 = × → − × −0.06 = 0.04 M1 A1 dt dx dt 3 [2]

This question in 9709/12 Oct/Nov 2016

3 12h h The diagram shows a water container in the form of an inverted pyramid, which is such that when the height of the water level is h cm the surface of the water is a square of side 12h cm. (i) Express the volume of water in the container in terms of h. [1] [The volume of a pyramid having a base area A and vertical height h is 3Ah.]1 … … … … … … … … … … … … … … … … Water is steadily dripping into the container at a constant rate of 20 cm3 per minute. (ii) Find the rate, in cm per minute, at which the water level is rising when the height of the water level is 10 cm. [4] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3(i) 3 1 12 V h = oe B1 Total: 1 3(ii) 2 d 1 d 4 V h h = or ( ) 2/3 d 4 12 d h v V − = M1A1 Attempt differentiation. Allow incorrect notation for M. For A mark accept their letter for volume - but otherwise correct notation. Allow V ′ d d d d d d h h V t V t = × 2 4 20 = × h soi DM1 Use chain rule correctly with ( ) d 20. d V t = Any equivalent formulation. Accept non-explicit chain rule (or nothing at all) d d h t       = 2 4 20 10 × = 0.8 or equivalent fraction A1 Total: 4

This question in 9709/12 Feb/March 2017

9 The point A 2, 2 lies on the curve y = x2 −2x + 2. (i) Find the equation of the tangent to the curve at A. [3] … … … … … … … … … … … … … … … The normal to the curve at A intersects the curve again at B. (ii) Find the coordinates of B. [4] … … … … … … … … … … … … … … … The tangents at A and B intersect each other at C. (iii) Find the coordinates of C. [4] … … … … … … … … … … … … … … …

11 marks

Mark scheme: 9(i) d 2 2 d y x x = − . At x = 2, m = 2 B1B1 Numerical m Equation of tangent is ( ) 2 2 2 y x −= − B1 Expect y = 2x ‒ 2 Total: 3 9(ii) Equation of normal ( ) 2 ½ 2 y x −= − − M1 Through (2, 2) with gradient = ‒1/m . Expect ½ 3 y x = − + 2 2 2 2 ½ 3 2 3 2 0 x x x x x − + = − + → − − = M1 Equate and simplify to 3-term quadratic ½, 3¼ x y = − = A1A1 Ignore answer of (2, 2) Total: 4 Question Answer Marks Guidance 9(iii) At ( ) ½, grad 2 ½ 2 3 x = − = − − =− B1 Ft their ‒½. Equation of tangent is ( ) 3¼ 3 ½ y x − = − + *M1 Through their B with grad their ‒3 (not m1 or m2). Expect 3 7 / 4 y x = − + 2 2 3 7 / 4 x x − = − + DM1 Equate their tangents or attempt to solve simultaneous equations 3 / 4, ½ x y = = − A1 Both required. Total: 4

This question in 9709/12 Feb/March 2017

6 The horizontal base of a solid prism is an equilateral triangle of side x cm. The sides of the prism are vertical. The height of the prism is h cm and the volume of the prism is 2000 cm3. (i) Express h in terms of x and show that the total surface area of the prism, A cm2, is given by ï3 24 000 A = x2 + x−1. [3] 2 ï3 … … … … … … … … … … … … … … … … … … … … … … (ii) Given that x can vary, find the value of x for which A has a stationary value. [3] … … … … … … … … … … … … … … … (iii) Determine, showing all necessary working, the nature of this stationary value. [2] … … … … … … … … …

8 marks

Mark scheme: 6(i) Volume = 1 3 ² 2 2 x h       = 2000 → h = 8000 3 ²x √ M1 Use of (area of triangle, with attempt at ht) × h =2000, f h x = A = ( ) 2 1 3 3 2 2 2 xh x   + × × ×     M1 Uses 3 rectangles and at least one triangle Sub for h → 2 1 3 24 000 2 3 A x x− √ = + A1 AG Total: 3 6(ii) 2 d 3 24000 2 d 2 3 A x x x − = − B1 CAO, allow decimal equivalent = 0 when x³ = 8000 → x = 20 M1 A1 Sets their d d A x to 0 and attempt to solve for x Total: 3 Question Answer Marks Guidance 6(iii) 3 d² 3 48000 2 d ² 2 3 A x x − = + > 0 M1 Any valid method, ignore value of d² d ² A x providing it is positive → Minimum A1 FT FT on their x providing it is positive Total: 2

This question in 9709/11 May/June 2017

10 y 4 y = 5 −3x x O 1 4 The diagram shows part of the curve y = 5 −3x. (i) Find the equation of the normal to the curve at the point where x = 1 in the form y = mx + c, where m and c are constants. [5] … … … … … … … … … … … … … … … … … The shaded region is bounded by the curve, the coordinate axes and the line x = 1. (ii) Find, showing all necessary working, the volume obtained when this shaded region is rotated through 360Å about the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) ( ) d 4 d 5 3 ² y x x = − × (−3) B1 B1 B1 without ×(−3) B1 For ×(−3) Gradient of tangent = 3, Gradient of normal – ⅓ *M1 Use of m1m2 = −1 after calculus → eqn: ( ) 1 2 1 3 y x − = − − DM1 Correct form of equation, with (1, their y), not (1,0) → 1 7 3 3 y x = − + A1 This mark needs to have come from y = 2, y must be subject Total: 5 10(ii) Vol = π ( ) 1 0 16 d 5 3 ² x x − ∫ M1 Use of ²d V y x π = ∫ with an attempt at integration π ( ) 16 3 5 3x   − ÷ −   −     A1 A1 A1 without( ÷ −3), A1 for (÷ −3) = ( π 16 16 6 15   −     ) = 8 5 π (if limits switched must show – to +) M1 A1 Use of both correct limits M1 Total: 5

This question in 9709/11 May/June 2017

12 5 A curve has equation y = 3 + 2 −x. (i) Find the equation of the tangent to the curve at the point where the curve crosses the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) A point moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.04 units per second. Find the rate of change of the y-coordinate when x = 4. [2] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) Crosses x-axis at (6, 0) B1 6 x is sufficient. d d y x = (0 +) −12 (2 – x)−2 × (−1) B2,1,0 −1 for each incorrect term of the three or addition of + C. Tangent ( ) 6 y x = − ¾ or 4 3 18 y x = − M1 A1 Must use dy/dx, x= their 6 but not x = 0 (which gives m = 3), and correct form of line equation. Using = + y mx c gets A1 as soon as c is evaluated. Total: 5 5(ii) If x = 4, dy/dx = 3 d 3 0.04 0.12 d y t = × = M1 A1FT M1 for (“their m” from d d y x and x = 4) × 0.04. Be aware: use of x = 0 gives the correct answer but gets M0. Total: 2

This question in 9709/12 May/June 2017

9 The equation of a curve is y = 8 x −2x. (i) Find the coordinates of the stationary point of the curve. [3] … … … … … … … … … … … … … … d2y (ii) Find an expression for and hence, or otherwise, determine the nature of the stationary point. dx2 [2] … … … … … … … … (iii) Find the values of x at which the line y = 6 meets the curve. [3] … … … … … … … … … … … … … … … … (iv) State the set of values of k for which the line y = k does not meet the curve. [1] … … … … … … … …

9 marks

Mark scheme: 9(i) ½ d 4 2 d − = − y x x B1 Accept unsimplified. = 0 when x = 2 x = 4, y = 8 B1B1 Total: 3 9(ii) 3 2 d² 2 d ² − = − y x x B1FT FT providing –ve power of x d² 1 d ² 4   = −     y x → Maximum B1 Correct d² d ² y x and x=4 in (i) are required. Followed by“< 0 or negative” is sufficient” but d² d ² y x must be correct if evaluated. Total: 2 9(iii) EITHER: Recognises a quadratic in x (M1 Eg x =u → 2 ² 8 6 0 − + = u u 1 and 3 as solutions to this equation A1 → x = 9, x = 1. A1) Question Answer Marks Guidance OR: Rearranges then squares (M1 x needs to be isolated before squaring both sides. → 2 10 9 0 − + = x x oe A1 → x = 9, x = 1. A1) Both correct by trial and improvement gets 3/3 Total: 3 9(iv) k > 8 B1 Total: 1

This question in 9709/12 May/June 2017

6 The line 3y + x = 25 is a normal to the curve y = x2 −5x + k. Find the value of the constant k. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 6 Gradient of normal is – 1/3 → gradient of tangent is 3 SOI B1 B1 FT FT from their gradient of normal. dy/dx = 2x – 5 = 3 M1 Differentiate and set = their 3 (numerical). x = 4 *A1 Sub x = 4 into line → y = 7 & sub their (4, 7) into curve DM1 OR sub x = 4 into curve → y = k ‒ 4 and sub their(4, k ‒ 4) into line OR other valid methods deriving a linear equation in k (e.g. equating curve with either normal or tangent and sub x = 4). k = 11 A1 Total: 6

This question in 9709/13 May/June 2017

10 (a) y y = h y = x2 −1 x O Fig. 1 Fig. 1 shows part of the curve y = x2 −1 and the line y = h, where h is a constant. (i) The shaded region is rotated through 360Å about the y-axis. Show that the volume of revolution, V, is given by V = 0 12h2 + h . [3] … … … … … … … … … (ii) Find, showing all necessary working, the area of the shaded region when h = 3. [4] … … … … … … … … … … … … … … … (b) h Fig. 2 Fig. 2 shows a cross-section of a bowl containing water. When the height of the water level is h cm, the volume, V cm3, of water is given by V = 0 12h2 + h . Water is poured into the bowl at a constant rate of 2 cm3 s−1. Find the rate, in cm s−1, at which the height of the water level is increasing when the height of the water level is 3 cm. [4] … … … … … … … … … …

11 marks

Mark scheme: 10(a)(i) Attempt to integrate 1 d V y y π = ∫ + M1 2 1 ½ h h h ∫ + = + is M0. Use of 2d y x ∫ is M0 ( ) 2 2 y y π   = +     A1 2 2 h h π   = +     A1 AG. Must be from clear use of limits 0→ h somewhere. Total: 3 10(ii) ( ) 1/2 1 d y y ∫ + ALT 6 ‒ ( ) 2 1 d x x ∫ − M1 Correct variable and attempt to integrate ( ) 3/2 1 y + ⅔ oe ALT 6 ‒ ( 3x x − ⅓ ) CAO *A1 Result of integration must be shown [ ] 8 1 − ⅔ ALT 8 1 6 [ 1 1 3 3     − − − −         ] DM1 Calculation seen with limits 0→3 for y. For ALT, limits are 1→2 and rectangle. 14/3 ALT 6 ‒ 4/3 = 14/3 A1 16/3 from 8 × ⅔ gets DM1A0 provided work is correct up to applying limits. Total: 4 Question Answer Marks Guidance 10(b) Clear attempt to differentiate wrt h M1 Expect ( ) d 1 d V h h π = + . Allow h + 1. Allow h. Derivative = 4π SOI *A1 2 derivative their . Can be in terms of h DM1 2 1 or or 0.159 4 2 π π A1 Total: 4

This question in 9709/13 May/June 2017

2 A function f is defined by f : x →x3 −x2 −8x + 5 for x < a. It is given that f is an increasing function. Find the largest possible value of the constant a. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 f′(x) 2 3 2 8 = − − x x M1 4 3 − , 2 SOI A1 f′(x) > 0 4 3 x ⇒ < − SOI M1 Accept x > 2 in addition. FT their solutions Largest value of a is 4 3 − A1 Statement in terms of a. Accept a ⩽ 4 3 − or 4 3 a < − . Penalise extra solutions 4

This question in 9709/11 Oct/Nov 2017

4 Machines in a factory make cardboard cones of base radius r cm and vertical height h cm. The volume, V cm3, of such a cone is given by V = 130r2h. The machines produce cones for which h + r = 18. (i) Show that V = 60r2 −130r3. [1] … … … … … … … (ii) Given that r can vary, find the non-zero value of r for which V has a stationary value and show that the stationary value is a maximum. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … … (iii) Find the maximum volume of a cone that can be made by these machines. [1] … … … … … … … … … … …

6 marks

Mark scheme: 4(i) ( ) 2 2 3 1 1 18 6 3 3 V r r r r π π π = − = − B1 1 4(ii) 2 d 12 0 d V r r r π π = − = M1 Differentiate and set = 0 ( ) 12 0 12 π − = → = r r r A1 2 2 d 12 2 d V r r π π = − M1 Sub r = 12 → 12 24 12 MAX π π π − = − → A1 AG 4 4(iii) Sub 12, 6 Max 288 or 905 π = = → = r h V B1 1

This question in 9709/11 Oct/Nov 2017

10 y y = 5x −1 P 2, 3 Q x O The diagram shows part of the curve y = 5x −1 and the normal to the curve at the point P 2, 3 . This normal meets the x-axis at Q. (i) Find the equation of the normal at P. [4] … … … … … … … … … … … … … … … … … (ii) Find, showing all necessary working, the area of the shaded region. [7] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) ( ) 1 2 d 1 5 1 d 2 − = × − y x x × 5 ( = 5 6 ) of normal = m 6 5 − M1 Uses m1m2 = −1 with their numeric value from their dy/dx Equation of normal ( ) 6 3 2 5 − = − − y x OE or 5y + 6x = 27 or 6 27 5 5 − = + y x A1 Unsimplified. Can use = + y mx c to get 5.4 = c ISW Question Answer Marks Guidance 10(ii) EITHER: For the curve ( ) 5 1d ∫ − x x = ( ) 3 2 5 1 3 2 − x ÷ 5 (B1 Correct expression without ÷5 B1 For dividing an attempt at integration of y by 5 Limits from 1 5 to 2 used → 3.6 or 18 5 OE M1 A1 Using 1 5 and 2 to evaluate an integrand ( ) 2 may be ∫y Normal crosses x-axis when y = 0, → x= (4½) M1 Uses their equation of normal, NOT tangent Area of triangle = 3.75 or 15 4 OE A1 This can be obtained by integration Total area=3.6 + 3.75 = 7.35, 147 20 OE A1) OR: For the curve: ( ) ( ) 2 1 1 d 5 ∫ + y y = 3 1 5 3   +     y y (B2, 1, 0 –1 each error or omission. Limits from 0 to 3 used → 2.4 or 12 5 OE M1 A1 Using 0 and 3 to evaluate an integrand Uses their equation of normal, NOT tangent. M1 Either to find side length for trapezium or attempt at integrating between 0 and 3 Area of trapezium = ( ) 1 39 3 2 4½ 3 9 2 4 4 + × = or A1 This can be obtained by integration Shaded area = 39 12 147 7.35, 4 5 20 − = OE A1) Question Answer Marks Guidance 7

This question in 9709/12 Oct/Nov 2017

11 y 1 y = x −1 2 B 5, 2 x O A 1, 0 1 The diagram shows the curve y = x −1 2 and points A 1, 0 and B 5, 2 lying on the curve. (i) Find the equation of the line AB, giving your answer in the form y = mx + c. [2] … … … … … … (ii) Find, showing all necessary working, the equation of the tangent to the curve which is parallel to AB. [5] … … … … … … … … … … … … … … … … … … … (iii) Find the perpendicular distance between the line AB and the tangent parallel to AB. Give your answer correct to 2 decimal places. [3] … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(i) Gradient of AB = 1 2 B1 Equation of AB is y = 1 2 x – 1 2 B1 2 11(ii) d d y x ( ) 1 2 ½ 1 x − = − B1 ( ) 1 2 ½ 1 ½ x − − = . Equate their d d y x to their ½ *M1 2, 1 = = x y A1 y ‒ 1 = ½(x ‒ 2) (thro' their(2,1) & their ½) → ½ = y x DM1 A1 5 Question Answer Marks Guidance 11(iii) EITHER: sin sin 1 d d θ θ = → = (M1 Where θ is angle between AB and the x-axis gradient of ( ) ½ tan ½ 26.5 7 AB θ θ = ⇒ = ⇒ = ° B1 ( ) sin26.5 7 0.45 = ° = d (or 1 5 ) A1) OR1: Perpendicular through O has equation 2 = − y x (M1 Intersection with AB: 1 2 2 ½ ½ , 5 5 −   − = − →    x x A1 2 2 1 2 0.45 5 5     = + =         d (or 1 5 ) A1) OR2: Perpendicular through (2, 1) has equation 2 5 = − + y x (M1 Intersection with AB: 11 3 2 5 ½ ½ , 5 5   − + = − →    x x A1 2 2 1 2 5 5     = +         d = 0.45 (or 1/√5) A1) Question Answer Marks Guidance 11(iii) OR3: OAC ∆ has area 1 4 [where C = (0, 1 2 − )] (B1 1 2 × 5 2 × d = 1 4 → d = 1 5 M1 A1) 3

This question in 9709/13 Oct/Nov 2017

3 28 A curve has equation y 8x. = 12x2 −4x + (i) Find the x-coordinates of the stationary points. [5] … … … … … … … … … … … … … … … … … … … … … … … … d2y (ii) Find . [1] dx2 … … … … … … (iii) Find, showing all necessary working, the nature of each stationary point. [2] … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(i) ½ d / d 6 8 y x x x = − + Set to zero and attempt to solve a quadratic for ½ x M1 Could use a substitution for ½ x or rearrange and square correctly* ½ ½ 4 or 2 x x = = [ 2 4 x and x = = gets M1 A0] A1 Implies M1. ‘Correct’ roots for their d / d y x also implies M1 16or 4 = x A1FT Squares of their solutions *Then A1,A1 for each answer 5 Question Answer Marks Guidance 8(ii) 2 2 d / d 1 3 − = − y x x ½ B1FT FT on their dy/dx, providing a fractional power of x is present 1 8(iii) (When x = 16) 2 2 d / d y x = 1/4 > 0 hence MIN M1 Checking both of their values in their 2 2 d / d y x (When x = 4) 2 2 d / d y x = ‒1/2 < 0 hence MAX A1 All correct Alternative methods ok but must be explicit about values of x being considered 2

This question in 9709/12 Feb/March 2018

1 9 A curve has equation y c and a line has equation y cx where c is a constant. = x + = −3, (i) Find the set of values of c for which the curve and the line meet. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) The line is a tangent to the curve for two particular values of c. For each of these values find the x-coordinate of the point at which the tangent touches the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(i) 2 2 1 3 3 1 0 cx cx x cx x c + = − → − + − = equality Use ( ) ( ) 2 2 2 2 4 3 4 10 9 or 5 16 b ac c c c c c   − = + + = + + + −   M1 Select their correct coefficients which must contain ‘c’ twice Ignore = 0, < 0, >0 etc. at this stage (Critical values) ‒1, ‒9 A1 SOI 9, 1 − − c c - . A1 4 Question Answer Marks Guidance 9(ii) Sub their c to obtain a quadratic ( ) 2 1 2 1 0 c x x   = −→− − − =   M1 1 = − x A1 Sub their c to obtain a quadratic ( ) ( 2 [ 9 9 6 1 0 c x x  = − →− + − =  M1 1/ 3 = x A1 [Alt 1: 2 / 1/ dy dx x c = − = , when 1 1, 1, 9, 3 c x c x = − = ± = − = ± Give M1 for equating the gradients, A1 for all four answers and M1A1 for checking and eliminating] [Alt 2: 2 / 1/ dy dx x c = − = leading to ( ) 2 2 1/ 1/ x ( 1/ x ) x 3 x − = − − Give M1 A1 at this stage and M1A1 for solving] 4

This question in 9709/12 Feb/March 2018

5 2 A point is moving along the curve y = 2x + in such a way that the x-coordinate is increasing at a x constant rate of 0.02 units per second. Find the rate of change of the y-coordinate when x = 1. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 5 2 y x x = + → d 5 2 d ² y x x = − = −3 (may be implied) when x = 1. M1 A1 d d y t = d d d d y x x t × → −0.06 M1 A1 Ignore notation, but needs to multiply d d y x by 0.02. 4

This question in 9709/11 May/June 2018

10 The curve with equation y = x3 −2x2 + 5x passes through the origin. (i) Show that the curve has no stationary points. [3] … … … … … … … … … … … … … … (ii) Denoting the gradient of the curve by m, find the stationary value of m and determine its nature. [5] … … … … … … … … … … … … … … (iii) Showing all necessary working, find the area of the region enclosed by the curve, the x-axis and the line x = 6. [4] … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10 y = x³ − 2x² + 5x 10(i) d d y x = 3x² − 4x + 5 B1 CAO Using ² 4 b ac − → 16 – 60 → negative → some explanation or completed square and explanation M1 A1 Uses discriminant on equation (set to 0). CAO 3 10(ii) m = 3x² − 4x + 5 d d m x = 6x – 4 (= 0) (must identify as d d m x ) B1FT FT providing differentiation is equivalent → x = 2 3 , m = 11 3 or 11 3 dy dx = Alt1: 2 2 11 3 3 3 m x   = − +     , 11 3 m = Alt2: 2 2 11 3 4 5 0, 4 0, 3 x x m b ac m − + − = − = = M1 A1 Sets to 0 and solves. A1 for correct m. Alt1: B1 for completing square, M1A1 for ans Alt2: B1 for coefficients, M1A1 for ans d²m d ²x = 6 +ve → Minimum value or refer to sketch of curve or check values of m either side of x = 2 3 , M1 A1 M1 correct method. A1 (no errors anywhere) 5 Question Answer Marks Guidance 10(iii) Integrate → 4 2 ³ 5 ² 4 3 2 x x x − + B2,1 Loses a mark for each incorrect term Uses limits 0 to 6 → 270 (may not see use of lower limit) M1 A1 Use of limits on an integral. CAO Answer only 0/4 4

This question in 9709/11 May/June 2018

3 y y = 5x Q R P y = x 9 −x2 x O The diagram shows part of the curve y x 9 and the line y 5x, intersecting at the origin O and the point R. Point P lies on the line y = 5x between−x2 O and R and= the x-coordinate of P is t. Point Q lies on the curve and PQ is parallel to the= y-axis. (i) Express the length of PQ in terms of t, simplifying your answer. [2] … … … … (ii) Given that t can vary, find the maximum value of the length of PQ. [3] … … … … … … … … … … …

5 marks

Mark scheme: 3(i) B1 B1 subsequent working. B1 for PQ allow 4 – ³ t t or ³ – 4 t t . Note: 4x – x3 from equating line and curve 0/2 even if x then replaced by t. [2] Question Answer Marks Guidance 3(ii) ( ) d d PQ t = 4 – 3t² B1FT B1FT for differentiation of their PQ, which MUST be a cubic expression, but can be ( ) d f x dx from (i) but not the equation of the curve. = 0 → t = + 2 3 √ M1 Setting their differential of PQ to 0 and attempt to solve for t or x. → Maximum PQ = 16 3 3 √ or 16 3 9 A1 Allow 3.08 awrt. If answer comes from wrong method in (i) award A0. Correct answer from correct expression by T&I scores 3/3. 3

This question in 9709/12 Oct/Nov 2018

11 y M y = 3 4x + 1 −2x A x O The diagram shows part of the curve y 3 4x 1 The curve crosses the y-axis at A and the stationary point on the curve is M. = + −2x. dy (i) Obtain expressions for and y dx. [5] dx Ó … … … … … … … … … … … … … … … (ii) Find the coordinates of M. [3] … … … … … … … … … (iii) Find, showing all necessary working, the area of the shaded region. [4] … … … … … … … … … … … … … … …

12 marks

Mark scheme: 11(i) d d y x = ( ) 1 2 3 4 1 2 x −   × +     [×4] [− 2] 6 2 4 1 x   −   +   B2,1,0 d y x ∫ = ( ) 3 2 3 3 4 1 2 x   + ÷     [ ÷ 4 ] [ − 2 ² 2 x ] (+ C) ( ) 3 2 2 4 1 2 x x   +   = −       B1 B1 B1 B1 for ( ) 3 2 3 3 4 1 2 x + ÷ B1 for ‘÷4’. B1 for ‘− 2 2 2 x ’. Ignore omission of + C. If included isw any attempt at evaluating. 5 11(ii) At M, d d y x = 0 → 6 4 1 x + = 2 M1 Sets their 2 term d d y x to 0 and attempts to solve (as far as x = k) x = 2, y = 5 A1 A1 3 Question Answer Marks Guidance 11(iii) Area under the curve = ( ) 2 3 2 0 1 4 1 ² 2 x x   + −     M1 Uses their integral and their ‘2’ and 0 correctly (13.5 – 4) – 0.5 or 9.5 – 0.5 = 9 A1 No working implies use of integration function on calculator M0A0. Area under the chord = trapezium = ½ × 2 × (3 + 5) = 8 Or 2 2 0 3 8 2 x x   + =     M1 Either using the area of a trapezium with their 2, 3 and 5 or ( ) 3 their x dx ∫ + using their ‘2’ and 0 correctly. (Shaded area = 9 – 8) = 1 A1 Dependent on both method marks, OR Area between the chord and the curve is: ( ) 2 0 3 4 1 2 3 x x x dx + − − + ∫ 2 0 3 4 1 3 3 x x dx = + − − ∫ M1 Subtracts their line from given curve and uses their ‘2’ and 0 correctly. ( ) 2 2 3 2 0 1 3 4 1 6 2 x x x   = + − −     A1 All integration correct and limits 2 and 0. 27 1 3 2 2 6 6       = − − −             M1 Evidence of substituting their ‘2’ and 0 into their integral. 1 1 1 3 3 1 2 6 3    = − = =       A1 No working implies use of a calculator M0A0. [4]

This question in 9709/12 Oct/Nov 2018

10 y x = 32 A 3 x = 3 y = 2 3x −1 −1 x O 1 2 3 −1 The diagram shows part of the curve y = 2 3x −1 3 and the lines x = 2 and x = 3. The curve and the 3 line x = 2 intersect at the point A. 3 (i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. [5] … … … … … … … … … … … … … … … (ii) Find the equation of the normal to the curve at A, giving your answer in the form y = mx + c. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) ( ) ( ) ( ) ( ) [ ] 1/3 2/3 3 1 4 3 1 d 4 3 1/ 3 x V x x π π −   − = ∫ − =  ÷     for [ ] [ ] ( )[ ] 4 2 1 π − DM1 Expect ( )( ) 13 4 3 1 x π − 4π or 12.6 A1 Apply limits ⅔ → 3. Some working must be shown. 5 Question Answer Marks Guidance 10(ii) ( ) 4/3 d / d ( 2 / 3) 3 1 3 y x x − = − − × B1 Expect ( ) 4/3 2 3 1 x − − − When 2 / 3, 2 x y = = soi d / d 2 y x = − B1B1 2nd B1 dep. on correct expression for dy//dx Equation of normal is ( ) 23 2 ½ y x − = − M1 Line through (⅔, their 2) and with grad ‒1/m. Dep on m from diffn 1 5 2 3 y x = + A1 5

This question in 9709/13 Oct/Nov 2018

9 y y = x3 + x2 P x O 3 The diagram shows part of the curve with equation y = x3 + x2 . The shaded region is bounded by the curve, the x-axis and the line x = 3. (i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. [4] … … … … … … … … … … … … … … … … (ii) P is the point on the curve with x-coordinate 3. Find the y-coordinate of the point where the normal to the curve at P crosses the y-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) 3 2 d π = ∫ + V x x x M1 Attempt 2d ∫y x ( ) 3 4 3 0 4 3 x x π   +     A1 ( ) ( ) 81 9 0 4 π   + −     DM1 May be implied by a correct answer 117 4 π oe A1 Accept 91.9 If additional areas rotated about x-axis, maximum of M1A0DM1A0 4 Question Answer Marks Guidance 9(ii) ( ) ( ) 1/2 3 2 2 d 1 3 2 d 2 − = + × + y x x x x x B2,1,0 Omission of 2 3 2 + x x is one error (At x = 3,) y = 6 B1 At x = 3, 1 1 11 33 2 6 4 = × × = m soi DB1ft Ft on their dy / dx providing differentiation attempted Equation of normal is ( ) 4 6 3 11 − = − − y x DM1 Equation through (3, their 6) and with gradient ‒1/their m When x = 0, y = 7 1 11 oe A1 6

This question in 9709/12 Feb/March 2019

10 y 1 2 y = 4x x O 1 The diagram shows the curve with equation y = 4x 2. (i) The straight line with equation y = x + 3 intersects the curve at points A and B. Find the length of AB. [6] … … … … … … … … … … … … … … … … (ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T. [3] … … … … … … … … … … … (iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB. [3] … … … … … … … … … … … …

12 marks

Mark scheme: 10(i) 1/2 4 3 = + → x x ( ) 1/2 2 1/2 ( ) 4 3 0 − + = x x OR 2 16 6 9 = + + x x x M1 Either treat as quad in 1/2 x OR square both sides and RHS is 3-term 1/2 1 or 3 = x ( ) 2 10 9 0 − + = x x A1 If in 1st method 1/2 x becomes x, allow only M1 unless subsequently squared x = 1 or 9 A1 4 or1 2 = y A1ft Ft from their x values If the 2 solutions are found by trial substitution B1 for the first coordinate and B3 for the second coordinate ( ) ( ) 2 2 2 9 1 12 4 = − + − AB M1 128 or 8 2 = AB oe or 11.3 A1 6 10(ii) dy/dx = 2 1/2 − x B1 2 1/2 − x = 1 M1 Set their derivative = their gradient of AB and attempt to solve (4, 8) A1 Alternative method without calculus: MAB = 1, tangent is y = mx + c where m = 1 and meets y = 4x1/2 when 4x1/2 = x + c. This is a quadratic with b2 = 4ac, so 16 – 4 × 1 × ܿ= 0 so c = 4 B1 Solving 4x1/2 = x + 4 gives x = 4 and y = 8 M1A1 3 Question Answer Marks Guidance 10(iii) Equation of normal is ( ) 8 1 4 − = − − y x M1 Equation through their T and with gradient ‒1/their gradient of AB. Expect 12 = −+ y x , Eliminate y (or x) → 12 3 or 3 12 −+ = + − = − x x y y M1 May use their equation of AB (4½, 7½) A1 3

This question in 9709/12 Feb/March 2019

2 The line 4y = x + c, where c is a constant, is a tangent to the curve y2 = x + 3 at the point P on the curve. (i) Find the value of c. [3] … … … … … … … … … … … … … (ii) Find the coordinates of P. [2] … … … … … … … … …

5 marks

Mark scheme: 2(i) Eliminates x or y → ² 4 3 0 y y c or ( ) ² 2 16 ² 48 0 x c x c + − + − = M1 Uses ² 4 b ac = → 4c – 28 = 0 M1 Uses discriminant = 0. (c the only variable) Any valid method (may be seen in part (i)) c = 7 A1 Alternative method for question 2(i) 1 1 4 2 ( 3) dy dx x = = + M1 Solving M1 c = 7 A1 3 2(ii) Uses c = 7, y² − 4y + 4 = 0 M1 Ignore (1,–2), c=-9 (1, 2) A1 2

This question in 9709/11 May/June 2019

d2y 10 A curve for which = 2x −5 has a stationary point at 3, 6 . dx2 (i) Find the equation of the curve. [6] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the x-coordinate of the other stationary point on the curve. [1] … … … … … … … … (iii) Determine the nature of each of the stationary points. [2] … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(i) integrating → d d y x = x² − 5x (+c) B1 = 0 when x = 3 M1 Uses the point to find c after ∫ = 0. c = 6 A1 integrating again → ( ) ³ 5 ² 6 3 2 x x y x d = − + + B1 FT Integration again FT if a numerical constant term is present. use of (3, 6) M1 Uses the point to find d after ∫ = 0. d = 1½ A1 6 Question Answer Marks Guidance 10(ii) d d y x = x² − 5x + 6 = 0 → x = 2 B1 1 10(iii) x = 3, d²y 1 d ²x = and/or +ve Minimum. x = 2, d²y 1 d ²x =− and/or −ve Maximum B1 www May use shape of ‘ 3 x + ’ curve or change in sign of dy dx B1 www SC: 3 x = , minimum, 2 x = , maximum, B1 2

This question in 9709/11 May/June 2019

dy 3 A curve is such that x3 . The point P 2, 9 lies on the curve. dx = −4x2 (i) A point moves on the curve in such a way that the x-coordinate is decreasing at a constant rate of 0.05 units per second. Find the rate of change of the y-coordinate when the point is at P. [2] … … … … … … … … (ii) Find the equation of the curve. [3] … … … … … … … … … … … … … …

5 marks

Mark scheme: 3(i) d d d d d d = × y y x t x t = 7 × – 0.05 M1 −0.35 (units/s) or Decreasing at a rate of (+) 0.35 A1 Ignore notation and omission of units 2 3(ii) ( ) 4 4 4 = + x y x (+c) oe B1 Accept unsimplified Uses (2, 9) in an integral to find c. M1 The power of at least one term increase by 1. c = 3 or ( ) 4 4 4 y x x = + + 3 oe A1 A0 if candidate continues to a final equation that is a straight line. 3

This question in 9709/12 May/June 2019

9 The curve C1 has equation y x2 7. The curve C2 has equation y2 4x k, where k is a = −4x + = + constant. The tangent to C1 at the point where x 3 is also the tangent to C2 at the point P. Find the = value of k and the coordinates of P. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. 2x – 2= 4 + x k (→ 4 ² 12 4 0 − + − = x x k ) *M1 Forms an equation in one variable using tangent & C2 Use of ² 4 − b ac = 0 on a 3 term quadratic set to 0. *DM1 Uses ‘discriminant = 0’ 144 = 16(4 – k) → k = − 5 A1 4 ² 12 4 0 − + − = x x k → 4 ² 12 9 0 − + = x x DM1 Uses k to form a 3 term quadratic in x x = 3 2 1 2       or , y = 1(or – 1). A1 Condone ‘correct’ extra solution. Alternative method for question 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. For C2: 1 2 (4 ) − = + dy A x k dx *M1 Finds dy dx for C2 in the form 1 2 (4 ) − + A x k Question Answer Marks Guidance 9 At P: ‘their 2’ = 1 2 (4 ) " − + A x k → ( 1 4 1 4 − = + = k x or x k ) *DM1 Equating ‘their 2’ to ‘their dy dx ’ and simplify to form a linear equation linking 4x + k and a constant. ( ) 2 2 2 4 − = + x x k → ( ) ( ) 2 2 2 2 1 4 8 3 0 − = → − + = x x x DM1 Using their 2 2 = − y x , y2 = 4x + k and their 4 1 + = x k (but not =0) to form a 3 term quadratic in x. 3 1 2 2   =     x or and from ( ) 5 1 k or = − − A1 Needs correct values for x and k. from y2 = 4x + k, y = 1(or – 1). A1 Condone ‘correct’ extra solution. Alternative method for question 9 For C1: d d y x = 2x – 4 → m = 2 B1 y – ‘their 4’ = ‘their m’ (x – 3) or using y = mx + c M1 Use of : d d y x and (3, their 4) to find the tangent equation. y – 4 = 2( x – 3) or 2 2 = − y x A1 If using = + mx c , getting 2 = − c is enough. For C2: 1 2 (4 ) − = + dy A x k dx *M1 Finds dy dx for C2 in the form 1 2 (4 ) − + A x k At P: ‘their 2’ = 1 2 (4 ) " − + A x k → ( 1 4 1 4 − = + = k x or x k ) *DM1 Equating ‘their 2’ to ‘their dy dx ’ and simplify to form a linear equation linking 4x + k and a constant. From 4x + k = 1 and y2 = 4x + k → y2 = 1 DM1 Using their 4x + k = 1 (but not =0) and C2 to form y2 = a constant Question Answer Marks Guidance 9 y = 1(or – 1) and 3 1 2 2   =     x or A1 Needs correct values for y and x. From 4 1 x k + = , k = –5 ( or – 1) A1 Condone ‘correct’ extra solution 8

This question in 9709/12 May/June 2019

10 y A 1 2 y = 3x + 4 x O 4 1 The diagram shows part of the curve with equation y = 3x + 4 2 and the tangent to the curve at the point A. The x-coordinate of A is 4. (i) Find the equation of the tangent to the curve at A. [5] … … … … … … … … … … … … … … … … (ii) Find, showing all necessary working, the area of the shaded region. [5] … … … … … … … … … … … … … … … … … … … … … … … … [Question 10 (iii) is printed on the next page.] (iii) A point is moving along the curve. At the point P the y-coordinate is increasing at half the rate at which the x-coordinate is increasing. Find the x-coordinate of P. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 10(i) ( ) 1 2 1 3 4 2 x −   +     ( ) 1 2 d 1 3 4 3 d 2 y x x −   = + ×     B1 Must have ‘ 3 × ’ At x = 4, d 3 d 8 y x = soi B1 Line through (4, their4) with gradient their 3 8 M1 If y ≠ 4 is used then clear evidence of substitution of x = 4 is needed Equation of tangent is ( ) 3 4 4 8 y x − = − or 3 5 8 2 y x = + A1 oe 5 Question Answer Marks Guidance 10(ii) Area under line 1 5 4 4 13 2 2   = + × =     B1 OR [ ] 4 2 0 3 5 3 5 3 10 13 8 2 16 2 x x x   + = + = + =     ∫ Area under curve: ( ) ( ) [ ] 3/2 1 2 3 4 3 4 3 3 / 2 x x   + ∫ + =  ÷     B1B1 Allow if seen as part of the difference of 2 integrals First B1 for integral without [ ] 3 ÷ Second B1 must have [ ] 3 ÷ 128 16 112 4 12 9 9 9 9 − = = M1 Apply limits 0 → 4 to an integrated expression Area = 13 ‒ 4 12 9 = 5 9 (or 0.556) A1 Alternative method for question 10(ii) Area for line = 1/2 × 4 × 3/2 = 3 B1 OR ( ) [ ] 4 2 5/2 1 1 1 8 20 4 20 16 25 3 3 3 3 y y   − = − = − + =   ∫ Area for curve = 3 2 4 ( 4) 9 3 y y y     ∫ − = −         ⅓ B1B1 64 16 8 8 32 9 3 9 3 9     − − − =         M1 Apply limits 2 → 4 to an integrated expression for curve Area = 32 3 9 − = 5 9 (or 0.556) A1 5 Question Answer Marks Guidance 10(iii) d 1 d 2 y x = B1 ( ) 1 2 3 3 4 2 x − + = 1 2 M1 Allow M1 for ( ) 1 2 3 3 4 2 x − + = 2. ( ) 1 2 3 4 3 x + = →3 4 9 x x + = → 5 = 3 oe A1 3

This question in 9709/13 May/June 2019

dy 1 9 A curve for which = 5x −1 2 −2 passes through the point 2, 3 . dx (i) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … d2y (ii) Find . [2] dx2 … … … … (iii) Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point. [4] … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) 1/2 3 2 [ 5 1 5 2 ] x x − ÷ ÷ − B1 B1 ( ) 27 3 4 3 / 2 5 = − + × c M1 Substitute x = 2, y = 3 ( ) 3 2 2 5 1 18 17 17 7 2 5 5 15 5   −   = − = → = − +       x c y x A1 9(ii) ( ) [ ] 1/2 2 2 d / d ½ 5 1 5 −   = − ×   y x x B1 B1 9(iii) ( ) 1/2 5 1 2 0 5 1 4 1 x x x − − = → −= = M1A1 Set d 0 d y x = and attempt solution (M1) 16 17 37 2 25 5 15 y = − + = A1 Or 2.47 or 37 1, 15       2 d 5 1 5 2 2 4 d x y x = × = (> 0) hence minimum A1 OE

This question in 9709/11 Oct/Nov 2019

10 y 4 y = 1 − 2 2x + 1 B x O A 4 The diagram shows part of the curve y 1 . The curve intersects the x-axis at A. The 2 = − 2x 1 + normal to the curve at A intersects the y-axis at B. dy (i) Obtain expressions for and y dx. [4] dx Ó … … … … … … … … … … … … (ii) Find the coordinates of B. [4] … … … … … … … … … … … … (iii) Find, showing all necessary working, the area of the shaded region. [4] … … … … … … … … … … … …

12 marks

Mark scheme: 10(i) [ ] 3 d 0 (2 1) d y x x −   = + +  × [+ 16] B2,1,0 OE. Full marks for 3 correct components. Withhold one mark for each error or omission. ∫ydx = [ ] [ ] 1 (2 1) 2 −   + + × +   x x (+c) B2,1,0 OE. Full marks for 3 correct components. Withhold one mark for each error or omission. 4 10(ii) At A, x = ½. B1 Ignore extra answer x = −1.5 d d y x = 2 → Gradient of normal ( ) ½ =− *M1 With their positive value of x at A and their dy dx , uses m₁m₂ = −1 Equation of normal: ( ) 0 ½ ½ − = − − y x or y − 0 = −½ (0 – ½) or 0 = −½×½ + c DM1 Use of their x at A and their normal gradient. B (0, ¼) A1 4 Question Answer Marks Guidance 10(iii) ( ) ( ) 1 2 2 0 4 1 d 2 1 − + ∫ x x *M1 d y x ∫ SOI with 0 and their positive x coordinate of A. [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. Area of triangle above x-axis = ½ × ½ × ¼ 1 16   =     B1 Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) Alternative method for question 10(iii) ( ) 0 1 3 2 1 1 d 2 (1 ) − − − ∫ y y *M1 d ∫x y SOI. Where x is of the form 1 2 1 ) −   − +       k y c with 0 and their negative y intercept of curve. [ ] 3 2 4 2   − −−+     = (½) DM1 Substitutes both 0 and their –3 into their ∫xdy and subtracts. Area of triangle above x-axis = ½ × ½ × ¼ 1 16   =     B1 Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) Question Answer Marks Guidance Alternative method for question 10(iii) 1 2 0 1 1 d 2 4 − + − ∫ x y x *M1 ∫(their normal curve) with 0 and their positive x coordinate of A. Curve [½ + 1] – [0 + 2] = (−½) DM1 Substitutes both 0 and their ½ into their ∫ydx and subtracts. 1 2 0 1 1 d 2 4 − + ∫ x x = 2 4 4 − + x x = [ ] 1 1 – 0 16 8 −   +     1 16   =     B1 Substitutes both 0 and ½ into the correct integral and subtracts. Total area of shaded region = 9 16 A1 OE (including AWRT 0.563) 4

This question in 9709/12 Oct/Nov 2019

3 The equation of a curve is y = x3 + x2 −8x + 7. The curve has no stationary points in the interval a < x < b. Find the least possible value of a and the greatest possible value of b. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 d d y x = 3x2 + 2x ‒ 8 B1 Set to zero (SOI) and solve M1 (Min) a = ‒2, (Max) b = 4/3. – in terms of a and b. A1 A1 Accept 4 3 2, a b − . - SC: A1 for 4 3 2, a b > − < or for 4 3 2 x −< < 4

This question in 9709/13 Oct/Nov 2019

5 x cm 4x cm 2x cm The dimensions of a cuboid are x cm, 2x cm and 4x cm, as shown in the diagram. (i) Show that the surface area S cm2 and the volume V cm3 are connected by the relation 2 S = 7V 3. [3] … … … … … … … … … … … … … … … … … (ii) When the volume of the cuboid is 1000 cm3 the surface area is increasing at 2 cm2 s−1. Find the rate of increase of the volume at this instant. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) B1B1 SOI 2 2 3 7 7 4 V x S = × = B1 AG, WWW 3 5(ii) 1 3 d 14 14 d 3 30 S V V −  = =     SOI when V = 1000 *M1 A1 Attempt to differentiate For M mark d d S V       to be of form 1 3 kV − d d d d d d V S V t t S   = ×     OE used with d 2 d S t = and 14 30 1 their DM1 30 7 or 4.29 A1 OE Alternative method for question 5(ii) 3 2 1 2 d 3 1 30 d 2 14 7 7 7 7 S V V S S   = → = × × =     SOI when S = 700 *M1 A1 Attempt to differentiate For M mark 1 2 d to be of form d V kS S       d d d d d d V S V t t S   = ×     OE used with d 2 d S t = and 14 30 1 their DM1 30 7 or 4.29 A1 OE Question Answer Marks Guidance 5(ii) Alternative method for question 5(ii) Attempt to find either d d V x or d d and d d S V x S       together with either d d x t or x *M1 d d V x = 24x2 or d d 3 56 and d d 7 S V x x x S   = =     , d d x t = 1 140 or x = 5 (A1) A1 Correct method for d d V t DM1 30 7 or 4.29 A1 OE 4

This question in 9709/13 Oct/Nov 2019

11 y A 2, 3 B y = x −1 −2 + 2 x O 1 3 The diagram shows part of the curve y = x −1 −2 + 2, and the lines x = 1 and x = 3. The point A on the curve has coordinates 2, 3 . The normal to the curve at A crosses the line x = 1 at B. (i) Show that the normal AB has equation y = 12x + 2. [3] … … … … … … … … … … … … … (ii) Find, showing all necessary working, the volume of revolution obtained when the shaded region is rotated through 360Å about the x-axis. [8] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(i) ( ) 3 d 2 1 d y x x − = − − B1 When x = 2, m = ‒2 → gradient of normal = 1 m − M1 m must come from differentiation Equation of normal is ( ) 3 ½ 2 ½ 2 y x y x − = − → = + A1 AG Through (2, 3) with gradient 1 m − . Simplify to AG 3 Question Answer Marks Guidance 11(ii) ( ( ) ( ) ( ) 2 2 1 2 π) d , π d y x y x ∫ ∫ *M1 Attempt to integrate 2 y for at least one of the functions ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 2 4 4 2 π 2 or 2 4 π 1 4 1 4 x x x x x − − + + + − + − + ∫ ∫ A1A1 A1 for ( ) 2 1 2 2 x + depends on an attempt to integrate this form later ( ) ( ) ( ) ( ) ( ) 3 3 2 2 1 1 3 2 12 3 1 π 2 or 4 1 4 1 π 4 3 1 x x x x x x x − − + + + − − + + − −             A1A1 Must have at least 2 terms correct for each integral (π) 125 2 1 18 4 8 1 4 12 3 12 or − + + − + +             1 1 2 12 4 8 24 3 − − − + − − +             DM1 Apply limits to at least 1 integrated expansion Attempt to add 2 volume integrals (or 1 volume integral + frustum) π{ } 7 7 7 6 12 24 + DM1 13 7 8 π or 111π 8 or 13.9π or 43.6 A1 2 1 4 8 1 4 3 12 + + − + +       1 1 2 12 4 8 24 3 − − − + − − +       8

This question in 9709/13 Oct/Nov 2019

4 A curve has equation y = x2 −2x −3. A point is moving along the curve in such a way that at P the y-coordinate is increasing at 4 units per second and the x-coordinate is increasing at 6 units per second. Find the x-coordinate of P. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 d 2 2 d y x x = − B1 d 4 d 6 y x = B1 OE, SOI ( ) 4 2 2 6 their x their − = M1 LHS and RHS must be their d d y x expression and value 4 3 x = oe A1 4

This question in 9709/12 Feb/March 2020

dy 1 10 The gradient of a curve at the point x, y is given by = 2 x + 3 2 −x. The curve has a stationary dx point at a, 14 , where a is a positive constant. (a) Find the value of a. [3] … … … … … … … … … … … … (b) Determine the nature of the stationary point. [3] … … … … … … … … … … (c) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) ( ) 1 2 2 3 0 a a + − = M1 SOI. Set d d y x = 0 when x = a. Can be implied by an answer in terms of a ( ) 2 4 3 a a + = 2 4 12 0 a a → − − = M1 Take a to RHS and square. Form 3-term quadratic ( )( ) 6 2 6 a a a − + → = A1 Must show factors, or formula or completing square. Ignore a = ‒2 SC If a is never used maximum of M1A1 for 6 x = ,with visible solution 3 10(b) ( ) 1 2 2 2 d 3 1 d y x x − = + − B1 Sub their a → 2 2 d 1 2 1 ( 0) 3 3 d y or x = −= − < →MAX M1A1 A mark only if completely correct If the second differential is not 2 3 − correct conclusion must be drawn to award the M1 3 10(c) ( ) ( ) ( ) 3 2 2 3 2 2 3 1 2 x y x c + = − + B1B1 Sub x = their a and y = 14 ( ) 3 2 4 1 4 9 18 3 c → = − + M1 Substitute into an integrated expression. c must be present. Expect c = ‒4 ( ) 3 2 2 4 1 3 4 3 2 y x x = + − − A1 Allow ( ) . f x =… 4

This question in 9709/12 Feb/March 2020

9 The equation of a curve is y = 3 −2x 3 + 24x. dy d2y (a) Find expressions for and . [4] dx dx2 … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of each of the stationary points on the curve. [3] … … … … … … … … … … … … … … … (c) Determine the nature of each stationary point. [2] … … … … … … … … …

9 marks

Mark scheme: 9(a) d d y x = 3(3−2x)² × −2 + 24 = ( ) 2 6 3 2 24 − − + x (B1 without ×−2. B1 for ×−2) B1B1 d² d ² y x = ( ) 12 3 2 2 − − ×− x = 24(3 – 2x) (B1FT from ୢ௬ ୢ௫ without – 2) B1FT B1 4 9(b) d 0 d = y x when ( ) 2 6 3 2 24 − = x → 3 2 2 x − = ± M1 x = ½, y = 20 or x = 2½, y = 52 (A1 for both x values or a correct pair) A1A1 3 9(c) If x = ½, d² d ² y x = 48 Minimum B1FT If x = 2½, d² d ² y x = −48 Maximum B1FT 2

This question in 9709/11 May/June 2020

3 A weather balloon in the shape of a sphere is being inflated by a pump. The volume of the balloon is increasing at a constant rate of 600 cm3 per second. The balloon was empty at the start of pumping. (a) Find the radius of the balloon after 30 seconds. [2] … … … … … … … … … … … (b) Find the rate of increase of the radius after 30 seconds. [3] … … … … … … … … … … …

5 marks

Mark scheme: 3(a) Volume after 30 s = 18000 4 π ³ 18000 3 r = M1 r = 16.3 cm A1 2 3(b) d 4π ² d V r r = B1 d d r t = d d r V × d d V t = 600 4π ²r M1 d d r t = 0.181 cm per second A1 3 Question Answer Marks

This question in 9709/12 May/June 2020

10 The equation of a curve is y = 54x − 2x −7 3. dy d2y (a) Find and . [4] dx dx2 … … … … … … … … (b) Find the coordinates of each of the stationary points on the curve. [3] … … … … … … … … (c) Determine the nature of each of the stationary points. [2] … … … … …

9 marks

Mark scheme: 10(a) d d y x = 54 – 6(2x – 7)² B2,1 d² d ² y x = −24(2x – 7) (FT only for omission of ‘ 2 × ’ from the bracket) B2,1 FT 4 10(b) ( ) 2 d 0 2 7 9 d = → − = y x x M1 x = 5, y = 243 or x = 2, y = 135 A1 A1 3 10(c) x = 5 d² d ² y x = −72 → Maximum (FT only for omission of ‘ 2 × ’ from the bracket) B1FT x = 2 d² d ² y x = 72 → Minimum (FT only for omission of ‘ 2 × ’ from the bracket) B1FT 2 Question Answer Marks

This question in 9709/12 May/June 2020

11 y A y = x3 −2bx2 + b2x x O a b The diagram shows part of the curve with equation y = x3 −2bx2 + b2x and the line OA, where A is the maximum point on the curve. The x-coordinate of A is a and the curve has a minimum point at b, 0 , where a and b are positive constants. (a) Show that b = 3a. [4] … … … … … … … … … … … … … … … … … … (b) Show that the area of the shaded region between the line and the curve is ka4, where k is a fraction to be found. [7] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) 2 2 d 3 4 d y x bx b x = − + B1 ( )( ) ( ) 2 2 3 4 0 3 0 − + = → − − = x bx b x b x b M1 or 3 b x b = A1 3 3 b a b a = → = AG A1 Alternative method for question 11(a) 2 2 d 3 4 d y x bx b x = − + B1 Sub b = 3a & obtain d 0 d y x = when x = a and when x = 3a M1 2 2 d 6 12 d y x a x = − A1 < 0 Max at x = a and > 0 Min at x = 3a. Hence ܾ= 3ܽ AG A1 4 Question Answer Marks 11(b) Area under curve = ( ) 3 2 2 6 9 d  − + x ax a x x M1 4 2 2 3 9 2 4 2 − + x a x ax B2,1,0 4 4 4 4 9 11 2 4 2 4   − + =     a a a a (M1 for applying limits 0 → a) M1 When x = a, 3 3 3 3 6 9 4 = − + = y a a a a B1 Area under line = 3 1 4 2 × a their a M1 Shaded area = 4 4 4 11 3 2 4 4 − = a a a A1 7

This question in 9709/13 May/June 2020

3 Air is being pumped into a balloon in the shape of a sphere so that its volume is increasing at a constant rate of 50 cm3 s−1. Find the rate at which the radius of the balloon is increasing when the radius is 10 cm. [3] … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 3 (Derivative =) 2 4πr (→ 400π) B1 SOI Award this mark for d d r V 50 derivative their M1 Can be in terms of r 1 8π or 0.0398 A1 AWRT 3

This question in 9709/11 Oct/Nov 2020

6 The equation of a curve is y = 2 + 25 −x2. Find the coordinates of the point on the curve at which the gradient is 3.4 [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 6 ( ) [ ] 1/2 2 d 1 25 2 d 2 −   = − × −     y x x x B1 B1 ( ) 2 1/2 2 2 4 16 3 9 25 25 − = → = − − x x x x M1 Set = 4 3 and square both sides ( ) ( ) 2 2 2 16 25 9 25 400 4 − = → = → = ± x x x x A1 When x = ‒ 4, y = 5 → (‒ 4, 5) A1 5

This question in 9709/11 Oct/Nov 2020

12 y B A 4, 0 O x 1 2 −2x y = 4x y = 3 −x C 1 The diagram shows a curve with equation y = 4x 2 −2x for x ≥0, and a straight line with equation y = 3 −x. The curve crosses the x-axis at A 4, 0 and crosses the straight line at B and C. (a) Find, by calculation, the x-coordinates of B and C. [4] … … … … … … … … … (b) Show that B is a stationary point on the curve. [2] … … … … … … (c) Find the area of the shaded region. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 12(a) ( ) 1 1 2 2 4 2 3 4 3 0 − = − → − + = x x x x x *M1 3-term quadratic. Can be expressed as e.g. 2 4 3 − + u u (=0) ( ) ( )( )( ) 1 1 2 2 1 3 0 or 1 3 0 x x u u    − − = − − =          DM1 Or quadratic formula or completing square 1 2 1 , 3 = x A1 SOI 1, 9 x = A1 Alternative method for question 12(a) ( ) 2 1 2 2 4 3   = +       x x *M1 Isolate 1 2 x ( ) 2 2 16 9 6 10 9 0 = + + → − + = x x x x x A1 3-term quadratic ( )( ) ( ) 1 9 0 − − = x x DM1 Or formula or completing square on a quadratic obtained by a correct method 1, 9 = x A1 4 12(b) 1/2 d 2 2 d = − y x x *B1 1/2 d or 2 2 0 d y x x − = when x =1 hence B is a stationary point DB1 2 Question Answer Marks Guidance 12(c) Area of correct triangle = 1 2 (9 ‒ 3) × 6 M1 or ( )( ) 9 2 3 1 3 d 3 18 2 x x x x   − = − →−      ( ) 3 1 2 2 2 4 (4 2 ) d 3 2      − = −       x x x x x B1 B1 ( ) 64 72 81 16 3   − − −     M1 Apply limits 4 → their 9 to an integrated expression 1 3 14 − A1 OE Shaded region = 1 2 3 3 18 14 3 − = A1 OE 6

This question in 9709/11 Oct/Nov 2020

2.7 The point 4, 7 lies on the curve y 2 = f x and it is given that f ′ x = 6x−1 −4x−3 (a) A point moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of increase of the y-coordinate when x 4. [3] = … … … … … … … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … …

7 marks

Mark scheme: 7(a) ( ) f ' 4 5 2   =     *M1 Substituting 4 into ( ) f ' x d d d d d d y y x t x t   = ×     → d d y t       = 5 2 × 0.12 DM1 Multiplies their ( ) f ' 4 by 0.12 d d   =     y t 0.3 A1 OE 3 7(b) ( ) 1 1 2 2 6 4 1 1 2 2 x x c − − + − B1 B1 B1 for each unsimplified integral. Uses (4, 7) leading to c = (-21) M1 Uses (4, 7) to find a c value ( ) 1 1 2 2 8 or f 12 8 21 or 1 2 21 y x x x x x − = + − + − A1 Need to see y or f(x) = somewhere in their solution and 12 and 8 4

This question in 9709/12 Oct/Nov 2020

1 8 The equation of a curve is y 2x 1 for x 2. 2x 1 = + + > −1 + dy d2y (a) Find and . [3] dx dx2 … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the stationary point and determine the nature of the stationary point. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) [ ] ( ) 2 d 2 [ 2 2 1 d − = − + y x x ] B1 B1 ( ) 2 3 2 d 8 2 1 d y x x − = + B1 3 8(b) Set their d d y x = 0 and attempt solution M1 (2x + 1)2 = 1 → 2x + 1 = ( ) ± 1 or 4x2 + 4x = 0 → (4)x(x + 1) = 0 M1 Solving as far as x = … x = 0 A1 WWW. Ignore other solution. (0, 2) A1 One solution only. Accept x = 0, y = 2 only. 2 2 d d y x > 0 from a solution 1 2 > − x hence minimum B1 Ignore other solution. Condone arithmetic slip in value of 2 2 d d y x . Their 2 2 d d y x must be of the form ( ) 3 2 1 − + k x 5

This question in 9709/13 Oct/Nov 2020

1 1 1 2 where x 0 and k is a positive constant.10 A curve has equation y x 2 x−1 k = + + k2 > (a) It is given that when x 14, the gradient of the curve is 3. = Find the value of k. [4] … … … … … … … … … … … … … … … … … … … … … … … k2 @1 1 A 1 13 2 2(b) It is given instead that Ô 1 k x + x−1 + k2 dx = 12. 4k2 Find the value of k. [5] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(a) [ ] ( ) 1/2 3/2 d 0 d 2 2 − −     = − +         y x x x k B2, 1, 0 [ ] ( ) 0 implies that more than 2 terms counts as an error Sub d 3 d = y x when 1 1 Expect 3 4 4 = = − x k M1 k = 1 7 (or 0.143) A1 4 Question Answer Marks Guidance 10(b) 3/2 1/2 1/2 1/2 2 2 1 1 2 2 3 −        + + = + +           x x x x x k k k k B2, 1, 0 OE 2 2 2 1 2 1 3 12 4     + + − + +         k k k k M1 Apply limits 2 2 4 → k k to an integrated expression. Expect 2 7 3 12 4 + + k k 2 7 3 13 12 4 12 k k + + = M1 Equate to 13 12 and simplify to quadratic. OE, ( ) 2 expect 7 12 4 0 k k + − = k = 2 7 only (or 0.286) A1 Dependent on ( )( ) ( ) 7 2 2 0 − + = k k or formula or completing square. 5

This question in 9709/13 Oct/Nov 2020

dy 6 6 A curve is such that = and A 1, −3 lies on the curve. A point is moving along the curve dx 3x −2 3 and at A the y-coordinate of the point is increasing at 3 units per second. (a) Find the rate of increase at A of the x-coordinate of the point. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) At x = 1, d 6 d = y x B1 d d d 1 1 3 d d d 6 2 x x y t y t = × = × =       M1 A1 Chain rule used correctly. Allow alternative and minimal notation. 3 Question Answer Marks Guidance 6(b) [ ] ( ) ( ) [ ] 2 6 3 2 3 2 x y c − − = ÷ + −         B1 B1 3 1 c −= −+ M1 Substitute 1, 3. x y = = − c must be present. ( ) 2 3 2 2 y x − = − − − A1 OE. Allow f(x)= 4

This question in 9709/12 Feb/March 2021

11 y A x O 2 The diagram shows the curve with equation y = 9 x−1 −4x−3 2 . The curve crosses the x-axis at the point A. (a) Find the x-coordinate of A. [2] … … … … … … … (b) Find the equation of the tangent to the curve at A. [4] … … … … … … … … … (c) Find the x-coordinate of the maximum point of the curve. [2] … … … … … … (d) Find the area of the region bounded by the curve, the x-axis and the line x = 9. [4] … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 11(a) 1 3 2 2 9 4 0 x x − −   − =     leading to ( ) 3 2 9 4 0 x x − − = M1 OE. Set y to zero and attempt to solve. 4 x = only A1 From use of a correct method. 2 11(b) 3 5 2 2 d 1 9 6 d 2 − −   = − +     y x x x B2, 1, 0 B2; all 3 terms correct: 9, 3 2 1 2 x − − and 5 2 6x − B1; 2 of the 3 terms correct At x = 4 gradient = 1 6 9 9 16 32 8 − + =       M1 Using their x = 4 in their differentiated expression and attempt to find equation of the tangent. Equation is ( ) 9 4 8 = − y x A1 or 9 9 8 2 = − x y OE 4 11(c) 5 2 1 9 6 0 2 x x − − + =       M1 Set their d d y x to zero and an attempt to solve. 12 = x A1 Condone ( )12 ± from use of a correct method. 2 Question Answer Marks Guidance 11(d) 1 1 1 3 2 2 2 2 d 4 9 4 9 1 1 2 2 x x x x x − − − − = − −                  B2, 1, 0 B2; all 3 terms correct: 9, 1 1 2 2 4 , 1 1 2 2 x x − − − B1; 2 of the 3 terms correct ( ) 8 9 6 4 4 3     + − +         M1 Apply limits their 4 →9 to an integrated expression with no consideration of other areas. 6 A1 Use of π scores A0 4

This question in 9709/12 Feb/March 2021

3 The equation of a curve is y = x −3 x + 1 + 3. The following points lie on the curve. Non-exact values are rounded to 4 decimal places. A 2, k B 2.9, 2.8025 C 2.99, 2.9800 D 2.999, 2.9980 E 3, 3 (a) Find k, giving your answer correct to 4 decimal places. [1] … … … … (b) Find the gradient of AE, giving your answer correct to 4 decimal places. [1] … … … … … … The gradients of BE, CE and DE, rounded to 4 decimal places, are 1.9748, 1.9975 and 1.9997 respectively. (c) State, giving a reason for your answer, what the values of the four gradients suggest about the gradient of the curve at the point E. [2] … … … … … … … …

4 marks

Mark scheme: 3(a) 1.2679 B1 AWRT. ISW if correct answer seen. 3 – 3 scores B0 1 3(b) 1.7321 B1 AWRT. ISW if correct answer seen. 1 3(c) Sight of 2 or 2.0000 or two in reference to the gradient *B1 This is because the gradient at E is the limit of the gradients of the chords as the x-value tends to 3 or ꝺx tends to 0. DB1 Allow it gets nearer/approaches/tends/almost/approximately 2 2

This question in 9709/12 May/June 2021

11 y 1 2 y = x 2 + k2x−1 x O 4k29 4k2 1 The diagram shows part of the curve with equation y = x 2 + k2x−12, where k is a positive constant. (a) Find the coordinates of the minimum point of the curve, giving your answer in terms of k. [4] … … … … … … … … … … … … … … … … The tangent at the point on the curve where x = 4k2 intersects the y-axis at P. (b) Find the y-coordinate of P in terms of k. [4] … … … … … … … … … … … The shaded region is bounded by the curve, the x-axis and the lines x = 94k2 and x = 4k2. (c) Find the area of the shaded region in terms of k. [3] … … … … … … … … … … …

11 marks

Mark scheme: 11(a) 1/2 2 3/2 d 1 1 d 2 2 − − = − y x k x x B1 B1 Allow any correct unsimplified form 1/2 2 3/2 1/2 2 3/2 1 1 1 1 0 leading to 2 2 2 2 x k x x k x − − − − − = = M1 OE. Set to zero and one correct algebraic step towards the solutions. d d y x must only have 2 terms. ( ) 2 , 2 k k A1 4 11(b) When x = 4k2, d 1 1 3 d 4 16 16   = − =     y x k k k B1 OE 2 1 5 2 2 2   = + × =     k y k k k B1 OE. Accept 2 2 + k k Equation of tangent is ( ) 2 5 3 4 2 16 − = − k y x k k or ( ) 2 5 3 4 2 16 = + → = + k y mx c k c k M1 Use of line equation with their gradient and ( 2 4 , ) k their y , When 5 3 7 0, 2 4 4 k k k x y   = = − =     or from 7 , 4 k y mx c c = + = A1 OE 4 Question Answer Marks Guidance 11(c) 3 1 1 1 2 2 2 2 2 2 2 d 2 3 −    + = +       x x k x x k x B1 Any unsimplified form 3 3 3 3 16 9 4 3 3 4     + − +         k k k k M1 Apply limits 2 2 9 4 4 → k k to an integration of y. M0 if volume attempted. 3 49 12 k A1 OE. Accept 4.08 3 k 3

This question in 9709/13 May/June 2021

9 The volume V m3 of a large circular mound of iron ore of radius r m is modelled by the equation V = 3 r −1 3 −1 for r ≥2. Iron ore is added to the mound at a constant rate of 1.5 m3 per second. 2 2 (a) Find the rate at which the radius of the mound is increasing at the instant when the radius is 5.5 m. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the volume of the mound at the instant when the radius is increasing at 0.1 m per second. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 9(a) d d V r   =     2 9 1 2 2 r   −     B1 OE. Accept unsimplified. 2 d d d 1.5 1.5 1.5 d d d d 112.5 9 1 5.5 d 2 2 r r V V t V t their r       = × = = =     −         M1 Correct use of chain rule with 1.5, their differentiated expression for d d V r and using 5.5 r = . 0.0133 or 3 225 or 1 [ 75 metres per second] A1 3 9(b) d d 1.5 or or 1 5 d d 0.1 V V their r r = OR 1.5 0.1 d d V their r = 2 2 1.5 OE 1 9 2 r     ×   =     −         B1 FT Correct statement involving d d V r or their d d V r , 1.5 and 0.1. 2 9 1 15 2 2 r     − =            r = 1 10 2 3 + B1 OE e.g. AWRT 2.3 Can be implied by correct volume. [Volume =] 8.13 AWRT B1 OE e.g. 3 5 30 3 −+ . CAO. 3

This question in 9709/12 Oct/Nov 2021

11 y 1 7 1 y = 2x + 10 − 1 3 x −2 A 3, 65 x O 5 2 1 1 and the normal to the curve The diagram shows the line x = 52, part of the curve y = 12x + 107 − x −2 3 at the point A 3, 6 . 5 (a) Find the x-coordinate of the point where the normal to the curve meets the x-axis. [5] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region, giving your answer correct to 2 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) ( ) 4 3 d 1 1 d 2 3 2 y x x = + − B1 OE. Allow unsimplified. Attempt at evaluating their d d y x at x = 3 ( ) 4 3 1 1 5 2 6 3 3 2     + =   −     *M1 Substituting x = 3 into their differentiated expression – defined by one of 3 original terms with correct power of x. Gradient of normal = 1 dy their dx − 6 5   = −     *DM1 Negative reciprocal of their evaluated d d y x . Equation of normal ( )( ) 6 normal gradient 3 5 y their x − = − 6 4.8 5 6 24 5 y x y x   = − +  = − +     DM1 Using their normal gradient and A in the equation of a straight line. Dependent on *M1 and *DM1. [When y = 0,] x = 4 A1 or (4, 0) 5 Question Answer Marks Guidance 11(b) Area under curve = ( ) [ ] 1 3 1 7 1 d 2 10 2 x x x      + −     −   M1 For intention to integrate the curve (no need for limits). Condone inclusion of π for this mark. ( ) 2 3 2 3 2 1 7 4 10 2 x x x − + − A1 For correct integral. Allow unsimplified. Condone inclusion of π for this mark. 2 3 9 3 6.25 3 0.5 2.1 1.75 4 2 4 2   ×     + − − + −           M1 Clear substitution of 3 and 2.5 into their integrated expression (with at least one correct term) and subtracting. 0.48[24] A1 If M1A1M0 scored then SC B1 can be awarded for correct answer. [Area of triangle =] 0.6 B1 OE [Total area =] 1.08 A1 Dependent on the first M1 and WWW. 6

This question in 9709/12 Oct/Nov 2021

d2y 910 The equation of a curve is such that = 6x2 −4 . The curve has a stationary point at −1, . dx2 x3 2 (a) Determine the nature of the stationary point at −1, 9 . [1] 2 … … … … (b) Find the equation of the curve. [5] … … … … … … … … … … … … … … … … … … (c) Show that the curve has no other stationary points. [3] … … … … … … … … … … … (d) A point A is moving along the curve and the y-coordinate of A is increasing at a rate of 5 units per second. Find the rate of increase of the x-coordinate of A at the point where x = 1. [3] … … … … … … … … … … …

12 marks

Mark scheme: 10(a)     2 2 2 2 3 2 d 4 d 6 1 0 minimum 10 minimum d d 1        y y x x or B1 Sub 1  x into 2 2 d d y x , correct conclusion. WWW 1 10(b)   3 2 d 2 2 d    y x c x x *M1 Integrating 2 2 d d y x (at least one term correct). 0 = −2 + 2 + c leading to c = [0] DM1 Substituting d 1, 0 d   y x x (need to see) to evaluate c. DM0 if simply state 0  c or omit c.   4 1 2 2     y x their c x k x A1 FT Integrated. FT their non-zero value of c if DM1 awarded. 9 1 2 2 2   k leading to k = [2] DM1 Substituting x = –1, y = 9 2 to evaluate k (dep on *M1). 4 1 2 2 2    y x x A1 OE e.g. 1 2  x or 4 2 . A0 (wrong process) if c not evaluated but correct answer obtained. 5 10(c) 3 2 d 2 2 0 d    y x x x M1 Their d 0 d  y x . Leading to 5 1 x  M1 Reaching equation of the form 5 x a  . So only stationary point is when x = −1 A1 1  x and stating e.g. ‘only’ or ‘no other solutions. 3 Question Answer Marks Guidance 10(d) At x = 1,  d 4 d  y x *M1 Substituting 1  x into their d d y x. d d d 1 5 d d d 4     x x y t y t DM1 OE Using chain rule correctly SOI. 5 4 A1 OE e.g. 1.25. 3

This question in 9709/11 May/June 2022

8 y 1 2 2 + 4x−1 y = x A 1, 5 B 16, 5 x O 1 5 intersects the curve at the The diagram shows the curve with equation y x 2 2. The line y = + 4x−1 = points A 1, 5 and B 16, 5 . (a) Find the equation of the tangent to the curve at the point A. [4] … … … … … … … … … … … … … … … … … (b) Calculate the area of the shaded region. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) 1/2 3/2 d ½ 2 d           y x x x At x = 1, d 1 3 2 d 2 2    y x M1 Substitute x = 1 into a differentiated y. Equation of tangent is   3 5 1 2    y x A1 WWW Or 3 13 2 2   y x . 4 Question Answer Marks Guidance 8(b) 3/2 1/2 8 3 / 2  x x B1 OE Integrate to find area under curve, allow unsimplified versions. 128 2 32 8 3 3                      M1 Apply limits 1 → 16 to an integrated expression. Area under line = 15  5 = 75 B1 Or by 16 1 5d  x . Required area = 75 ‒ 66 = 9 A1 4

This question in 9709/13 May/June 2022

11 The point P lies on the line with equation y mx c, where m and c are positive constants. A curve = + has equation y . There is a single point P on the curve such that the straight line is a tangent to = −mx the curve at P. (a) Find the coordinates of P, giving the y-coordinate in terms of m. [6] … … … … … … … … … … … … … … … … … … … … … … … The normal to the curve at P intersects the curve again at the point Q. (b) Find the coordinates of Q in terms of m. [4] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a) 2 0 m mx c mx cx m x       M1 All x terms in the numerator. OE e.g.  mx cx m . 2 2 2 4 0 4 0 b ac c m      M1 OE b2 – 4ac = 0 is implied by c2 – 4m2 = 0.  2  c m A1 SOI. Allow  at this stage. 2 mx [  2 ] 2 0 2 1 0 mx m x x       M1 Sub c = +2m Ignore substitution of -2m.   2 1 0 1 x x     only A1  y m only or (‒1, m) only A1 Alternative method to question 11(a) 2 d d  y m x x M1 As this is a method mark a sign error is allowed. 2  m m x 2 1 x   M1 A1 Equating their d d y x and m and attempt to solve. x = ±1 or 1  x A1 If 1 x  and y m  are the only answers offered here award the final M1 A1. Selecting x = –1 as the only answer and attempt to find y M1  y m or (‒1, m) A1 6 Question Answer Marks Guidance 11(b) Equation of normal is   1 1     y m x m *M1 Through their P with gradient 1  m , OE e.g. 2 1 1    m y x m m . Allow use of the gradient of the curve as   2 1 their m x        with their P. Coordinates of P must be in terms of m only.     2 2 2 1 1 0 x m m x x m m m m x           DM1 OE Equating their normal equation to the equation of the curve and removing x from the denominator.     2 2 1 0 x x m x m      A1 or   2 2 2 2 4 2 2 1 1 1 1 2 4 2 2          m m m m m m x m 2 1     m y m m A1 or 2 1 ,        m m , ignore the coordinates of P. 4

This question in 9709/13 May/June 2022

dy 12 The equation of a curve is such that = 12 −1 −4. It is given that the curve passes through the 2x dx point P 6, 4 . (a) Find the equation of the tangent to the curve at P. [2] … … … … … … … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … … …

6 marks

Mark scheme: −42(a) M1 d y 3  1   −4 3  SOI by gradient used. Substitute x = 6 into −6 1 = 12 ( 2 ) = 12     4 d x  2   4  3 A1 3 1 3 y − 4 = ( x − 6 ) OE e.g. y = x − or evaluates c in y = x + c 4 4 2 4 1 3 OR evaluates c = − using (6, 4) and gradient . ISW 2 4 2 2(b)  −3  B2, 1, 0  1    12  x − 1   −3   1   2    1 x − 1  y =    = −8     −3  2   2         −3 M1 Must have +c .  1  12    6 − 1  Substitute y = 4, x = 6 and solve for c in an integrated  2  −3 4 = + c   4 = −8 2 + c   c = 5 expression. May be unsimplified.   1 − 3 2 −3 A1 OE Must see ‘ y = ’ or ‘ f ( x ) = ’ in the working.  1  x − 1 + 5  y =  − 8    2  4

This question in 9709/11 Oct/Nov 2022

1 3 A curve has equation y = ax 2 −2x, where x > 0 and a is a constant. The curve has a stationary point at the point P, which has x-coordinate 9. Find the y-coordinate of P. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 dy 1 − 12 B2, 1, 0 = ax −2 dx 2 1 − 1 a M1 dy 0 = a ( 9 ) 2 − 2  − 2 = 0  a = 12  Substitute x = 9 and = 0 into their derivative and 2 6 dx solve a linear equation for a.  a = 1 2 A1  1  A1 FT FT on their a. 2 − 18 = 9 )  y = their a  (  18   5 .4 2  2  5  2  2 10  2 2  40  B1 Accept with x 2 present. Must evaluate 5C2 Coefficient of x in  1 + x  is 10   = 2  = 2   p   p  p  p  Coefficient of x 2 in (1 + px ) 6 is 15 ( p ) 2  = 15 p 2  B1 Accept with x 2 present. Must evaluate 6C2   40 2 *M1 Forming an equation in p with their coefficients, the + 15 p = 70 p 2 given 70, no x terms and no extra terms. 15 p 4 − 70 p 2 + 40  = 0 or 3 p 4 − 14 p 2 + 8  = 0 DM1 Forming a 3-term equation in p (or another variable) with all terms on one side and their coefficients. 2 DM1 Attempt to solve 3-term quartic (or quadratic in another 70  70 − 4 (15 )( 40 ) 2 2 5 p − 4 3 p − 2 = 0  or or variable) by factorisation, formula or completing the ( )( )  30 square. 14  14 2 − 4 ( 3 )( 8 ) 6 2 A1 6 p = 2 ,  OE e.g.  or AWRT 0.816 3 3 If *M1 DM1 DM0, allow SC B1 for 4 correct values. 6

This question in 9709/11 Oct/Nov 2022

dy 18 The equation of a curve is such that = 3x 2 −3x−12. The curve passes through the point 3, 5 . dx (a) Find the equation of the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the x-coordinate of the stationary point. [2] … … … … … … … … … … … … … … … … … … … (c) State the set of values of x for which y increases as x increases. [1] … … … … …

7 marks

Mark scheme: 8(a)  3   1  B1 B1 Marks can be awarded for correct unsimplified expressions, 1  3 x 2   3 x 2   32 12  mark each for contents of { } ISW.  y =    +  −   + c   = 2 x − 6 x  3 1        2   2  3 1 M1 Correct use of (3,5) in an integrated expression (defined by at least 5 = 2  3 2 −6 3 2 + c one correct power) including + c. 3 1 A1 Condone c = 5 as their final line if either y = or f(x) = seen y = 2 x 2 − 6 x 2 + 5 elsewhere in the solution, but coefficients must not contain unresolved double fractions. 4 8(b) 1 − 1 M1 Setting given differential to 0. 3 x 2 − 3 x 2 = 0 [x=] 1 A1 CAO WWW Condone extra solution of —1 only if it is rejected. 2 8(c) x>1 or x> “their 8(b)” B1FT Allow ⩾ 1

This question in 9709/12 Oct/Nov 2022

5 The line with equation y = kx −k, where k is a positive constant, is a tangent to the curve with equation y = −1 2x. Find, in either order, the value of k and the coordinates of the point where the tangent meets the curve. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 5 1 2   kx k x ⇒   2 2 2 1 0    kx kx OR quadratic in 2 1 : 2 2 0 2                y k y x y y ky k y k k k *M1 OE e.g.   2 1 0 2    kx kx ,   2 1 0 2    x x k Equate line and curve to form 3-term quadratic (all terms on one side).   2 4 0   b ac ⇒       2 2 4 2 1 0    k k or 2 4 8 [ 0   k k ⇒   4 2 0   k k ] OR using equation in : y    2 2 4 2 0   k k DM1 Use discriminant correctly with their , , a b c not in quadratic formula. DM0 if x still present. May see  2 1 4 0 2         k k or 1 1 4 0 2         k . k = 2 only A1 If DM0 then k = 2, award A0 XP then B0 B0 Allow A1 even if divides by k to solve. If 0  k also present but uses 2  k , award A1.   2 2 4 4 1 0 2 1 0          x x x ⇒ 1 2  x B1 1 2 2 1 2     y B1 Question Answer Marks Guidance 5 Alternative method for Q5 2 d 1 d 2  y x x or 2 1 2  x *M1 Differentiate 1 2 x M0 for 2 2  x . No errors.   2 2 1 1 1 2 2 2 y x x x x    or 2 2 1 1 2 0 2         x x x x DM1 Sub their d d y x into equation of line or set gradient = k to form equation in x. 1 2  x only A1 If DM0 then 1 2  x , award A0XP then B0 B0. 1 2 2 1 2           y B1 2  k B1 5

This question in 9709/11 May/June 2023

10 y A 1, 4 4 y = 2 2x −1 1 B 32, 1 x O 1 4 The diagram shows part of the curve with equation y = and parts of the lines x = 1 and y = 1. 2x −1 2 The curve passes through the points A 1, 4 and B, 32, 1 . (a) Find the exact volume generated when the shaded region is rotated through 360Å about the x-axis. [5] … … … … … … … … … … … … … … … (b) A triangle is formed from the tangent to the curve at B, the normal to the curve at B and the x-axis. Find the area of this triangle. [6] … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(a)           4 4 3 16 16 π d π 16 2 1 d π 2 1 3 2 2 1                   x x x x x *M1 Integrate 2 y (power incr. by 1 or div by their new power). M0 if more than 1 error or   3 16 2 1 6    x x .    3 16 π 3 2 2 1            x A1 OE e.g.   3 8 2 1 3          x .  16 16 π 6 8 6 1             112 7 π π 48 3         DM1 Sub correct limits into their integral: F 3 2       F(1). Must see at least 1 8 . 3 3         Allow 1 sign error. Decimal: 2.33 π or 7.33 . Volume of cylinder 2 1 1 π 1 π 2 2           OR    1.5 1 1 π 1 d π 2   x B1 1 π 2 or 3 π 1 2         seen. Volume of revolution 7 1 π π 3 2         11π 6  A1 A0 for 5.76 (not exact). If DM0 for insufficient substitution, or B0, SC B1 for 11π 6 . 5 Question Answer Marks Guidance 10(b)      3 d 8 2 1 2 d            y x x B2, 1, 0 OE B1 for each correct element in {}. At B gradient = 2  B1 Eqn of tangent 3 1 " 2" 2          y their x OR Eqn of normal 1 3 1 " " 2 2         y their x M1 SOI Following differentiation OE e.g. 2 4   y x or 1 1 2 4   y x . (Must have 1  N T m m for M1). Tangent crosses x-axis at 2 or normal crosses x-axis at 1 2  A1 SOI For at least one intercept correct or correct integration. Area = 5 4 A1 From intercepts: 1 5 5 1 2 2 4   or 1 5 1 4 4   , from lengths: 1 5 5 5 2 2 4    or by integration. 6

This question in 9709/11 May/June 2023

dy 11 The equation of a curve is such that = 6x2 −30x + 6a, where a is a positive constant. The curve dx has a stationary point at a, −15 . (a) Find the value of a. [2] … … … … … … … … … … (b) Determine the nature of this stationary point. [2] … … … … … … … … … … … … (c) Find the equation of the curve. [3] … … … … … … … … … … … … … (d) Find the coordinates of any other stationary points on the curve. [2] … … … … … … … … … … …

9 marks

Mark scheme: 11(a) 2 6 30 6 0    a a a [ ⇒ 6 4 0]   a a B1 Sub  x a into d 0 d  y x . May see 2 5 0    a a a . a = 4 only B1 2 Question Answer Marks Guidance 11(b) 2 2 d 12 30 d   y x x or correct values of d d y x either side of 4  x M1 Differentiate d d y x (mult. by power or dec. power by 1) M0 if no values of d d y x , only signs. At 2 2 2 2 d d 4, 0 minimum or 18 minimum d d y y x x x     or concludes minimum from d d y x values A1 WWW A0 XP if 4  a obtained incorrectly in (a) Must see ‘minimum’. If M0, SC B1 for ‘minimum’ from d d y x sign diagram. 2 11(c)    y    3 2 6 30 6 3 2    x x their a x c B1 FT Expect   3 2 2 15 24    x x x c . B1 poss. even if uses ‘ a ’ – no value in (a) – max 1/3.       3 2 2 15 2 "4" 15 "4" 6 "4"      their their their c M1 Sub x = their"4", y = –15 into integral (must incl +c ) Look for –15 = 128 – 240 + 96 + c [⇒ c = 1]. 3 2 2 15 24 1     y x x x A1 Coefficients must be correct and simplified. Need to see ‘  y ’ or ‘  f  x ’ in the working. 3 11(d)    2 d 6 30 6 "4" 0 d     y x x their x If correct,     6 1 4 0    x x or     2 30 30 4 6 24 12    M1 OE Forming a 3-term quadratic using the given d d y x and solving by factorisation, formula or completing the square. Check for working in (b). Coordinates   1,1 2 A1 Allow 1, 12   x y (ignore 4  x if present). If M0, award SC B1 for   1,1 2 . 2

This question in 9709/11 May/June 2023

10 y 3 y = 9x − 2x + 1 2 A 112, 512 B 712, 312 x O The diagram shows the points A 112, 512 and B 712, 312 lying on the curve with equation 3 y = 9x − 2x + 1 2. (a) Find the coordinates of the maximum point of the curve. [4] … … … … … … … … … … … … … … (b) Verify that the line AB is the normal to the curve at A. [3] … … … … … … … … … (c) Find the area of the shaded region. [5] … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(a)    1/2 d 3 9 2 1 2 d 2                 y x x B1, B1 Including ‘+c’ makes the second term B0.   1/2 9 3 2 1 0 x    leading to 2 1 9 x  M1 Set differential to zero and solve by squaring SOI. Beware   2 2 9 3 2 1 0    x M0A0. 2 1 3 2 1 9   or x x get M0. Max point = (4, 9) A1 WWW y = 9 must come from original equation. 4 10(b) When x = 1½, shows substitution or d 3 d  y x M1 Substituting x = 1½ into their d d y x . Gradient of AB is 5½ 3½ 1 1½ 7½ 3           M1 Substituting into a correct expression for mAB. 1 x3 1 3  . [Hence AB is the normal] A1 Alternative method for Question 10(b) When x = 1½ d 3 d  y x ,[ perpendicular gradient is -1/3] M1 Perpendicular through A has equation 3   x y + 6 which contains B(7.5,3.5) leading to AB is a normal to the curve at A M1 A1 3 Question Answer Marks Guidance 10(c)   5 2 2 2 1 9 5 2 2 2                     x x B1 B1 Integrating y with respect to x.     2.5 2.5 2 2 9 1 9 1 7.5 2 7.5 1 1.5 2 1.5 1 2 5 2 5                    or 9 225 1024 81 32 2 4 5 8 5                 or 1933 149 40 40  or 48.325 – 3.725 M1 OE Apply limits 1½ to 7½ to an integral. Working must be seen. Expect 44.6 . 1 1 1 5 3 6 2 2 2         or 15 2 3 2 1 ( 6)d 3    x x = 2 2 1 15 15 1 3 3 6 6 6 2 2 6 2 2                                      or 285 69 [ 8 8  = 27] B1 SOI Area of trapezium. May be seen combined with the area under the curve integral. [Shaded area = 44.6 – 27 =] 17.6 A1 SC B1 if no substitution of the limits seen. 5 Question Answer Marks Guidance 10(c) Alternative method for Question 10(c) A =   15 2 3 2 3 2 1 ((9 2 1 ) 6 )d 3              x x x x   15 2 3 2 3 2 28 (( 2 1 6)d 3     x x x M1 Finding the equation of AB and subtracting from the equation of the curve.   5 2 2 2 1 28 6 5 3 2 2 2                        x x x A1 A1 127 49 10 10   M1 Apply limits 1½ to 7½ to an integral. Working must be seen. 17.6 A1 SC B1 if no substitution of limits seen. 5

This question in 9709/13 May/June 2023

3 x x x The diagram shows a cubical closed container made of a thin elastic material which is filled with water and frozen. During the freezing process the length, xcm, of each edge of the container increases at the constant rate of 0.01cm per minute. The volume of the container at time t minutes is V cm3. Find the rate of increase of V when x = 20. [3] … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 3 dV 2 B1 SOI = 3 x dx dV  dV dx  2 M1 Correct use of chain rule with x = 20 substituted into =  = 3  20  0.01   dt  dx dt  dV . dx 12 A1 3

This question in 9709/11 Oct/Nov 2023

dy 1 72 3 The equation of a curve is such that = 2x + . The curve passes through the point P 2, 8 . dx x4 (a) Find the equation of the normal to the curve at P. [2] … … … … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(a)   M1 Tangent gradient must come from x = 2 substituted into the − 1  1 2  given expression. [Gradient of normal =]  −=−  11 11 11 Their   2  2  y − 8 2 2 x 92 A1 OE = − or 11 y + 2 x = 92 or y =− + x − 2 11 11 11 2 3(b)  1 2  72   x 2 24  B1, B1 One mark for each correct unsimplified { }. + c   y =   x  2  + 3 −3   + c   − 3  2  x   4 x  1 24 M1 Substitution of x = 2, y = 8 into their integrated expression, 8 = −4 + c defined by at least one correct power. Two terms and + c 4 8 needed.  1  2 24 A1 Both coefficients must be simplified but allow x− 3. Condone y = or 0.25 x − + 10   3  4  x c = 10 as line as long as either y or f(x) = is seen elsewhere. 4

This question in 9709/12 Oct/Nov 2023

5 10 The equation of a curve is y = f x , where f x = 4x −3 3 −20 x. 3 (a) Find the x-coordinates of the stationary points of the curve and determine their nature. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) State the set of values for which the function f is increasing. [1] … … … … … … … … … … … …

7 marks

Mark scheme: 10(a) dy  5 2   20  B2,1,0 B2 Three correct unsimplified { } and no others. =  ( 4 x − 3 ) 3   4  −  B1 Two correct { }or three correct { } and an additional term dx  3   3  e.g. + c. B0 More than one error.  20 2 20  2 M1 d y 3 − = 0 4 x − 3 ) 4 x − 3) = k , k  0 leading to Equating their to 0 and using a valid method to arrive at  (  leading to (  3 3  d x 2 answers. 4 x −=3  m A1  4 x −=3 1  x = 1 ,1 2 d 2 y 40 − 1 B1 OE 3  4 = ( 4 x − 3 ) 2 dx 9  1  d 2 y  160  − 1 160 B1 d 2 y x = 3  0 or − or − 17.8 so max If evaluated the answers for both must be correct OR =   ( 4 x − 3 ) 2   2  2  d x  9  9 d x d 2 y  160  − 1 160 d y 3  0 or or 17.8 so min Clear use of change in sign of correctly for both B1’s. =   ( 4 x − 3 )  x = 1 2 dx  9  9 d x If B1M1A0B0B0 scored then SCB1 can be awarded for: dy  5 2   20  2 =  ( 4 x − 3 ) 3  −   leading to ( 4 x − 3) = 64 leading dx  3   3  5 11 to x = − , . 4 4 d 2 y 10 − 1 5 d 2 y = ( 4 x − 3 ) 2 3 , x = − , 2  0 so max, dx 9 4 dx 11 d 2 y x = ,  0 so min. 2 4 dx 6 10(b) x  12 , x  1 B1 Allow ⩽ and/or ⩾. FT only from special case x − 54 , x  114 Condone: 1  x  12 . 1

This question in 9709/12 Oct/Nov 2023

11 y P 2 y = x + 2 2x −1 Q R x O 1 2 2 The diagram shows part of the curve with equation y = x + . The lines x = 1 and x = 2 2x −1 2 intersect the curve at P and Q respectively and R is the stationary point on the curve. (a) Verify that the x-coordinate of R is 3 and find the y-coordinate of R. [4] 2 … … … … … … … … … … … … … … … (b) Find the exact value of the area of the shaded region. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a) dy −3 B1B1 −8 = + 1 Expect + 1 . −2 2 ( 2x − 1)  2 3   dx ( 2 x − 1) 3 dy −8 DB1 AG. Substitute x = leading to = + 1 = 0 . 2 dx 8 dy Or correct solution of = 0 . 3 dx Hence x-coordinate of R is 2 3 2 3 B1 Answer only is acceptable. When x = , y = + = 2 2 4 2 4 11(b) 20 B1 Both required. y-coordinate of P = 3, y-coordinate of Q = 9   2 ( 2 x − 1) −1   1 2  B1 B1 Area below curve.   +  x  −1 2    2   1   1   5  1  M1 13 − + 2 −−+1 = −− Apply limits 1→2 to an integral. Expect .        6  3   2   3  2  1  20  47 M1 Area of trapezium, only allow errors in y-coordinate  3 +  = of Q. 2  9  18 47 13 4 A1 Shaded region. − = 18 6 9 6 Alternative method 1: Changes the award of the first M1 −7 M1 Must be some evidence of use of limits. Their equation of line PQ:[ y = x + 34] . Integrating between 1 and 2. 9 9 Alternative method 2: Changes the award of the first M1, a B1 and the second M1   M1 34 −16 34 2 For area under the line if their is seen integrated Combining line and curve:   x + −  dx  9 9 ( 2 x − 1) 2  correctly and limits used. Correct9 first and 3rd terms. −8 2 34 1 B1 B1 = x + x + 9 9 ( 2 x − 1) Use of limits on the whole integral M1

This question in 9709/13 Oct/Nov 2023

1 dy 23 A curve is such that = 3 ( 4x + 5) . It is given that the points (1, 9) and (5, a) lie on the curve. d x Find the value of a. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 32 *M1 Integrate to obtain form k (4 x + 5) 1 32 A1 Or (unsimplified) equivalent. Obtain correct 2 (4 x + 5) Condone missing ... + c so far. Substitute x = 1, y = 9 to form an equation in c DM1 1 3 9 A1 1 3 9 2 − . Obtain or imply  y =  ( 4 x + 5 ) 2 − May be implied by  a =  ( 4 (1) + 5 ) 2 2 2 2 Substitute x = 5 to obtain a = 58 A1 5

This question in 9709/12 Feb/March 2024

35 A curve has the equation y = 2 . 2x - 5 Find the equation of the normal to the curve at the point (2, 1), giving your answer in the form ax + by + c = 0 , where a, b and c are integers. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5 Differentiate to obtain form kx (2 x 2 − 5) −2 M1 Obtain correct − 12 x (2 x 2 − 5) −2 A1 OE Substitute (2, 1) to obtain gradient − 249 A1 8 OE e.g. − . Allow −2.67. 3 Apply negative reciprocal to their numerical gradient to obtain gradient of *M1 3 Must have been some attempt at differentiation. Expect normal 8 Attempt equation of normal using their gradient of the normal and (2, 1) DM1 3 Expect y −=1 ( x − 2 ) . 8 Obtain 3 x − 8 y + 2 = 0 (allow multiples) A1 Or equivalent of requested form e.g. 8 y − 3 x − 2 = 0 . 6

This question in 9709/12 Feb/March 2024

11 y A B O x M 1 3 The diagram shows the curve with equation y = 2x - 2 3 - 3x - + 1 for x 2 0 . The curve crosses the x-axis at points A and B and has a minimum point M. (a) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … … … (b) Find the area of the region bounded by the curve and the line segment AB. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) 4 − 53 − 34 B1 1 2 1 −  − Differentiate to obtain − 3 x + x  3 3 − 3 x + 1 OE Expect quadratic 2  x  1 1   − 3 3   or rewrite as a quadratic equation in x or x Allow 2 x 2 − 3 x + 1 . 1 1 M1 Substitution SOI if dealt with correctly later − Equate first derivative to zero and reach a solution for x 3 or x 3 with no error in use of indices −  3  2 1 1 or complete square to find minimum point 2  a −  − where a = x 3  4  8 Obtain x = 6427 A1 Or exact equivalent. SC B1 if no working shown. Ignore extra solution x = 0 . y = − 18 seen B1 Or exact equivalent. Allow −0.125 . 4 1 − 1311(b) M1 − or equivalent and attempt solution Recognise equation as quadratic in x 2 2a − 3a + 1 = 0 where a = x 3 . 1 1 A1 OE 3 1 3 Obtain x −= 1 and x −= 2 SC B1 if no M mark awarded. Obtain 1 and 8 A1 SC B1 if no M mark awarded. 1 2 1 2 *M1 9 3 + x or 2 out of 3 correct terms Integrate to obtain form k1 x 3 + k 2 x Expect 6 x 3 − x 3 + x . 2 1 2 A1 No other terms from a second integral. Obtain correct 6x 3 − 9 x 3 + x 2 Apply their limits correctly DM1 Their limits must be from their working. [Obtain –0.5 and conclude area is] 0.5 A1 7

This question in 9709/12 Feb/March 2024

4 The equation of a curve is y = f ( x) , where f ( x) = ( 2x - 1) 3x - 2 - 2 . The following points lie on the curve. Non-exact values have been given correct to 5 decimal places. A(2, 4), B(2.0001, k), C(2.001, 4.00625), D(2.01, 4.06261), E(2.1, 4.63566), F(3, 11.22876) (a) Find the value of k. Give your answer correct to 5 decimal places. [1] … … … … The table shows the gradients of the chords AB, AC, AD and AF. Chord AB AC AD AE AF Gradient of 6.2501 6.2511 6.2608 7.2288 chord (b) Find the gradient of the chord AE. Give your answer correct to 4 decimal places. [1] … … … … … … … … (c) Deduce the value of f l ( 2) using the values in the table. [1] … … … … … … …

3 marks

Mark scheme: 4(a) [k] = 4.00063 B1 CAO 1 4(b) [Gradient AE] = 6.3566 B1 CAO 1 4(c) Suggests that  f' 2 6.25     B1 CAO 1

This question in 9709/11 May/June 2024

11 y O x 4 3 A function is defined by f ( x) = 3 - + 2 for x ! 0 . The graph of y = f ( x) is shown in the diagram. x x (a) Find the set of values of x for which f ( x) is decreasing. [5] … … … … … … … … … … … … … … (b) A triangle is bounded by the y-axis, the normal to the curve at the point where x = 1 and the tangent to the curve at the point where x =- 1. Find the area of the triangle. Give your answer correct to 3 significant figures. [8] … … … … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 11(a) 4 2 d 12 3 d   y x x x 4 2 d 12 3 0 d y x x x    leading to 4 2 3 12 0 x x   or -12 + 3x2 = 0 M1 Set = 0 or uses , ⩽ and simplifies. Must be from 4 2 d d   y A B x x x .   2 2 3 4 0 x x   leading to 2 x only A1 SC B1 for 2  x if M0 scored. 2 0 and 0 2     x x or (-2, 0) and (0, 2) or 2 2   x and x 0  B1FT Allow and/or. B1FT Allow 2 0 and / or 0 2    x x   but only B1B0 if 0 included in either or both. Allow [–2, 0) and (0, 2]. Allow B1B0 for 2 2 x   or (–2, 2). Must be from 4 2 d d   y A B x x x . 5 B marks only available if d d y x 4 2   A B x x . Question Answer Marks Guidance 11(b) [At 1] 3 and tan 9 x y m    *M1 Using their d d y x . 1 1 norm 9 9    m DM1 Equation of normal is   1 1 26 3 1 leading to 9 9 9 y x y x            A1 At 1, 1, 9    x y m M1 Equation of tangent is    1 9 1 leading to 9 8 y x y x     A1 Meet when 1 26 49 9 8 leading to 1.19512 , 9 9 41 x x x             M1 Equates their tangent and their normal. Area = 1 26 1 .19512 8 2 9 their their          M1 If 2 1   y y is used integration must be correct and substitution shown. 6.51 A1 AWRT Accept fraction wrt 6.51 8

This question in 9709/11 May/June 2024

2 has a minimum point at A and intersects the positive x-axis at B.6 The curve with equation y = 2x - 8x 1 (a) Find the coordinates of A and B. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) y O B x 2x - 32 y = 3 1 2 y = 22x - 88x A 2 and the line AB. It is given that the The diagram shows the curve with equation y = 2x - 8x 1 2x - 32 equation of AB is y = . 3 Find the area of the shaded region between the curve and the line. [5] … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 1 2 d 1 2 8 d 2     y x x 1 2 2 4 0    x M1 Equating their two term d d y x , with at least one term correct, to 0. [A is]   4, 8  or 4, 8 x y   A1 [B is]   16,0 or 16, 0 x y   B1 4 Note: Correct answers without use of d d y x can be awarded 4/4. Question Answer Marks Guidance 6(b)    3 2 2 2 8 3 2 2 x x C    B1 Seen correct in unsimplified form or better.     2 2 2 32 32 or 3 12     x x x C B1 Seen correct in unsimplified form or better. Attempt to integrate, defined by at least one correct power in each expression, and then subtract. M1 Multiplying by 3 before integration scores M0. 3 3 2 2 2 2 8 8 16 .16 4 .4 3 3 2 2                                        2 2 16 32 16 4 32 4 3 3                            M1 Use of their x values, > 0, from (a) as limits in their integrated expressions. Allow, for correct limits, sight of  256 80 256 112 3 3 3 3                                                 . If incorrect limits are used, then clear substitution must be seen. Question Answer Marks Guidance 6(b) Alternative Method 1 for first 4 marks of Question 6(b)    3 1 2 2 2 8 2 8 3 2 x x dx x x C              (B1) Seen correct in unsimplified form or better. [Area of triangle =] 48 (B1) Attempt to integrate, defined by at least one correct power, and then subtract their triangle area. (M1) 3 3 2 2 2 2 8 8 16 .16 4 .4 3 3 2 2                                      (M1) Use of their x values, > 0, from (a) as limits in their integrated expression. Allow sight of 256 80 3 3                            . If incorrect limits are used, then clear substitution must be seen. Question Answer Marks Guidance 6(b) Alternative Method 2 for first 4 marks of Question 6(b) Subtract and then integrate, defined by at least two correct powers. Condone functions being the wrong way round. (M1) If terms in x have not been combined use the first scheme.  3 2 2 4 8 32 3 3 2 3 2               x x x (B2,1,0) B2 for 3 correct terms, B1 for any 2 correct terms.  3 3 2 2 2 2 4 8 32 16 4 8 32 4 16 16 4 4 3 3 3 2 3 3 2 3 2 2                                               (M1) Use of their x values, >0, from (a) as limits in their integrated expression. Allow sight of 32 0 3         . If incorrect limits are used, then clear substitution must be seen. 32 3 , 10 2 3 or 10.7 (B1) AWRT Allow 32 3  or 32 3  changed to + 32 3 for this mark. (5) Condone the inclusion of π for the first 4 marks but use of 2 y  scores a maximum of B1 for the triangle.

This question in 9709/12 May/June 2024

9 A function f is such that f l ( x) = 6 ( 2x - 3) 2 - 6x for x ! R . (a) Determine the set of values of x for which f ( x) is decreasing. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Given that f ( 1) = - 1, find f ( x) . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) 2 6 2 3 6 0 x x    or = 0 2 6 2 3  x 2 6 2 3 6   x is used, do not treat as a MR.    2 2 24 78 54 or 4 13 9 or 1 4 9 x x x x x x       OR   2 6 2 3 6   x x   leading to 2 3 leading to 2 3 x x x x     M1 Expanding brackets and collecting terms to arrive at a three term quadratic, only condone sign errors.   9 1 , 4  x B1 9 1 4 x   or 9 1 and 4 x x   or 9 1, 4       DB1FT OE Condone consistent use of ⩽ and ⩾ or [ ]. Do not allow 9 1 or 4   x x nor 9 1, 4   x x . FT on their values coming from a correct initial statement. 4 Question Answer Marks Guidance 9(b)      3 2 6 6 f 2 3 3 2 2 x x x C                  B1 B1 B1 for each   Correct integral .   3 2 1 1 3 1 C    M1  f 1  x equated to their integrated expression, defined by two terms with at least one correct power + C, with x = 1.    3 2 2 3 3 3 f x x x        A1 CAO Only condone C = 3 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form   y mx c . Alternative method for Question 9(b)     2 3 2 24 78 54 leading to f 8 39 54 f x x x x x x x C                (B2,1,0) B2 completely correct, B1 any two correct terms. 1 8 39 54 C    (M1)  f 1  x equated to their integrated expression, defined by three terms with at least one correct power + C, with x = 1.  3 2 8 39 54 24 f x x x x        (A1) Only condone C = 24 as final answer if coefficients have been simplified earlier. Do not ISW if the result is of the form . y mx c   4

This question in 9709/12 May/June 2024

3 210 The equation of a curve is y = ( 5 - 2 x) + 5 for x 1 52 . (a) A point P is moving along the curve in such a way that the y-coordinate of point P is decreasing at 5 units per second. Find the rate at which the x-coordinate of point P is increasing when y = 32 . [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Point A on the curve has y-coordinate 32. Point B on the curve is such that the gradient of the curve at B is - 3 . Find the equation of the perpendicular bisector of AB. Give your answer in the form ax + by + c = 0 , where a, b and c are integers. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) 2  x     1 1 2 2 d 3 5 2 2 5 2 d 2 y k x x x           M1* OE Differentiating to get   1 2 5 2  k x only. d d d leading to d d d y y t x t x         d 9 5 d t x   DM1 Correct statement linking their numerical expression for d d y x with d d t x and 5.  5 9 or 0.556 = A1 AWRT 4 Question Answer Marks Guidance 10(b)   1 2 5 2 3   k x M1 Equating their d d y x of the form   1 2 5 2  k x to 3 . [B is]   2, 6 A1 1 32 6 Gradient 2 2 AB m    1 1 4 , gradient of perpendicular 26   m M1* For A, y must be 32. Clear use of difference in y co-ordinates difference in x co-ordinates for points A and B, condone inconsistent order, and using m1m2 = 1 . If incorrect values or another complete method used, then working must be clear.   2 2 6 32 Mid point is , 0,19 2 2         M1* Finding the midpoint of AB using A and B. If incorrect values used then all working must be clear. For A, y must be 32.   2 19 0 13    y x DM1 Finding the equation of the perpendicular bisector using their midpoint and their perpendicular gradient. 2 13 247 0 x y    or integer multiples of this. A1 6

This question in 9709/12 May/June 2024

2 1 5 The equation of a curve is y = 2x - + 3 . 2x (a) Find the coordinates of the stationary point. [3] … … … … … … … … … … (b) Determine the nature of the stationary point. [2] … … … … … … (c) For positive values of x, determine whether the curve shows a function that is increasing, decreasing or neither. Give a reason for your answer. [2] … … … … … … …

7 marks

Mark scheme: 5(a) Differentiate to obtain 2 1 2 4   x x B1 OE Condone ‘+c’. Equate first derivative to zero and solve 2 4 0   K x x as far as 3 , and x k K  k non- zero M1 Not given if ‘+c’ used. 1 2  x and 9 2  y A1 OE B1 SC if no visible solution of the cubic. 3 Question Answer Marks Guidance 5(b) Differentiate their first derivative, substitute their x value. Substitution may be implied by a correct inequality or correct value, M1 Must differentiate one term correctly. Expect 3 1 4 12 at 2 x x      Alternative: substitute values of x into d d y x . One value 1 2  x and one value 1 0. 2 x    conclude minimum A1 Following correct work only 2 5(c) State increasing … B1 … with clear reference to first derivative always being positive [for 0] x  B1 Dependent on first derivative being correct. It is not sufficient to substitute values of x. 2

This question in 9709/13 May/June 2024

9 y x O 1 3 The diagram shows the curve with equation y = 2x 3 + 10 . (a) Find the equation of the tangent to the curve at the point where x = 3 . Give your answer in the form ax + by + c = 0 where a, b and c are integers. [5] … … … … … … … … … … … … … … … … (b) The region shaded in the diagram is enclosed by the curve and the straight lines x = 1, x = 3 and y = 0 . Find the volume of the solid obtained when the shaded region is rotated through 360° about the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) Differentiate to obtain form 1 2 2 3 (2 10)   kx x M1 OE 1 2 2 3 3 (2 10)   x x A1 Or unsimplified equivalent. Substitute 3  x in first derivative and evaluate to find gradient *M1 Expect 27 8 . Allow if first derivative of forms 1 3 2 (2x 10)   k , 1 3 2 (2x 10)   kx or 1 2 3 2 (2x 10)   kx . Attempt equation of tangent at   3, 8 with numerical gradient DM1 Use of gradient of the normal is DM0. [±]( 27 8 17) 0    x y or integer multiples A1 5 9(b) State or imply volume is 3 π (2 10) d   x x B1 Implied if π appears only at the end. Do not allow an unsimplified:     2 1/2 3 π 2 10 x   . Integrate to obtain 4 1 2  k x k x and evaluate using limits 1 and 3 M1 Where 1 2 0  k k . 60π A1 OE Allow from a correct integral and sight of limits. Allow numerical answers in the range 188-189. 3

This question in 9709/13 May/June 2024

2 a 2 The curve y = x - has a stationary point at (-3, b). x Find the values of the constants a and b. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Differentiate to obtain 2 x + ax −2 or equivalent B1 Equate first derivative to zero, substitute x = −3 and attempt value of a M1 Must be an attempt at differentiation. Obtain a = 54 A1 Obtain b = 27 A1 4

This question in 9709/11 Oct/Nov 2024

7 y A 7 x O 2 12 The diagram shows part of the curve with equation y = . The point A on the curve has 3 2x + 1 coordinates 7b , 6l. 2 (a) Find the equation of the tangent to the curve at A. Give your answer in the form y = mx + c . [4] … … … … … … … … … … … … … … … … … … … (b) Find the area of the region bounded by the curve and the lines x = 0 , x = 7 and y = 0 . [4] 2 … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) − 4 M1 Differentiate to obtain form k1(2 x + 1) 3 4 A1 − Obtain correct − 8(2 x + 1) 3 or unsimplified equivalent  7  M1 Gradient must come from a differentiated Attempt equation of tangent at  , 6  with numerical gradient expression.  2  1 31 A1 Obtain y = − x + or equivalent of requested form 2 4 4 7(b) 2 M1 Integrate to obtain form k 2(2 x + 1) 3 2 A1 Obtain correct 9(2 x + 1) 3 or unsimplified equivalent Use correct limits correctly to find area M1 Substitute correct limits into an integrated expression. 36 – 9 minimum working required. Obtain 27 A1 SC B1 if M1 A1 M0 scored. 4

This question in 9709/11 Oct/Nov 2024

9 The equation of a curve is y = 4 + 5x + 6 x 2 - 3x 3. (a) Find the set of values of x for which y decreases as x increases. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) It is given that y = 9x + k is a tangent to the curve. Find the value of the constant k. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) Differentiate to obtain 5 + 12 x − 9 x 2 B1 Attempt to find two critical values by solving quadratic equation or inequality M1 1 5 A1 SC B1 if no method for solving the quadratic. Obtain values − and 3 3 1 5 A1FT SC B1 if no method for solving the quadratic. Conclude x − , x  3 3 4 9(b) Equate first derivative to 9 and simplify to 3 term quadratic *M1 2 A1 SC B1 for solving 5 + 12 x − 9 x 2 = 9 without Obtain x = 3 simplifying to a 3-term quadratic. Use x-value and corresponding y-value to determine value of k DM1 28 A1 28 2 Obtain k = SC B1 for k = from solving 5 + 12 x − 9 x = 9 9 9 without simplifying to a 3-term quadratic. 4

This question in 9709/11 Oct/Nov 2024

3 The equation of a curve is y = 2x 2 - 3 . Two points A and B with x-coordinates 2 and ( 2+ h) respectively lie on the curve. (a) Find and simplify an expression for the gradient of the chord AB in terms of h. [3] … … … … … … … … … … (b) Explain how the gradient of the curve at the point A can be deduced from the answer to part (a), and state the value of this gradient. [2] … … … … … … … … … … … … … …

5 marks

Mark scheme: 3(a)  2 ( 2 + h ) 2 − 3 B1 SOI  f ( 2 + h ) = 2 M1 2 their − their 5 − 5 2 ( 2 + h ) − 3 2 ( 2 + h ) − 3 ( ( )  2 h 2 + 8h   )  =  can be implied by the ( 2 + h ) − 2  h  ( 2 + h ) − 2 simplified expression or the correct answer. 2 Their 5 must come from 2 ( 2 ) − 3. 2h + 8 or 2 ( h + 4 ) A1 3 3(b) h → 0 , or chord [AB] → tangent [at A] B1 Either of these statements or any sight of h = 0. 8 B1FT Could come from anywhere except wrong working. Either correct or FT their linear expression from (a). 2

This question in 9709/12 Oct/Nov 2024

7 (a) By expressing - 2x 2 + 8x + 11 in the form - a ( x - b) 2 + c , where a, b and c are positive integers, find the coordinates of the vertex of the graph with equation y =- 2x 2 + 8 x + 11. [3] … … … … … … … … (b) y O x The diagram shows part of the curve with equation y =- 2x 2 + 8 x + 11 and the line with equation y = 8x + 9 . Find the area of the shaded region. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) 2 2 M1* p  0. −2 ( x  p )  q or −2 ( x  p )  q ( ) 2 2 DM1 −2 ( x − 2 )  q or −2 ( x − 2 )  q ( ) 2 A1 Accept x = 2, y = 19 or 2, 19. −2 ( x − 2 ) + 19 and (2, 19) 3 7(b) Method 1  x = 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* 2 2 3 B1* Both terms correct.  −2 x + 2 dx  − x + 2 x  ( )  3  2   2  M1 Apply their limits, one positive and one negative, obtained  − + 2  −  − 2  from equating the line and the curve to their integrated  3   3  expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 2  x = 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium.  −2 x 3 8 2   8 2  B1* OE  + x + 11x  −  x + 9 x  All terms correct.  3 2   2  1 The second integral can be replaced by (1 + 17 )  2 OE. 2 7(b)  −2   2   M1 Apply their limits, one positive and one negative, obtained −  4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated  + 4 + 11 −  + 4 − 11  (  3   3   expressions. If the trapezium has been used, the second integral can be replaced by their 18. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. Method 3  x = 1 B1* Both x co-ordinates for the points of intersection. Subtract and attempt to integrate M1* 2 3 8 2 B1* All terms correct. − ( x − 2 ) − x + 10 x 3 2  2  M1 Apply their limits, one positive and one negative, obtained − 4 + 10   − (18 − 4 − 10 ) from equating the line and the curve, to their integrated  3  expression. 8 2 DB1 AWRT 2.67 WWW. = , 2 3 3 7(b) Method 4  x = 1 B1* Both x co-ordinates for the points of intersection. Attempt to integrate and subtract M1* The second integral can be replaced with what is clearly their area of a trapezium.  2 3   8 2  B1* All terms correct.  − ( x − 2 ) + 19 x  −  x + 9 x  1  3   2  The second integral can be replaced with (1 + 17 )  2 OE. 2  2   M1 Apply their limits, one positive and one negative, obtained 18 − 19 )  (−  4 + 9 ) − ( 4 − 9 ) from equating the line and the curve, to their integrated  + 19  − (  3   expression. If the trapezium has been used the second integral can be replaced with their 18 OE. 8 2 DB1 AWRT 2.67 WWW. = , 2 8 8 3 3 Condone −→ . 3 3 11 SC B1 for mistaking triangle for trapezium leading to , i.e. 3 a total of 2/5. 5

This question in 9709/12 Oct/Nov 2024

10 A function f with domain x 2 0 is such that f l (x) = 8 ( 2x - 3 ) 3 - 10x 3 . It is given that the curve with equation y = f ( x) passes through the point (1, 0). (a) Find the equation of the normal to the curve at the point (1, 0). [3] … … … … … (b) Find f ( x) . [4] … … … … … … … … … … … … … … … … … … … It is given that the equation f l ( x) = 0 can be expressed in the form 125x 2 - 128 x + 192 = 0 . (c) Determine, making your reasoning clear, whether f is an increasing function, a decreasing function or neither. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) −18 B1 SOI 1 M1 Use of m1m2 = −from1 f ( x ) with x = 1. 18 y  − 0  1 A1 OE = ISW x − 1 18 3 10(b)    B1B1 B1 for each unsimplified {}. 5 4    Can be implied by equivalent simplified or partly simplified 1 1 1  3 . . 3 . x ) = 8 ( 2 x − 3 )  −10 x   + c   f ( versions. 4 5 2     3  3   4 5  3 − 6 x 3 + c   3 ( 2 x − 3 )   5 M1 Use of x = 1 and y = 0 in their integrated f ( x ) , defined as an 3 − 6 (1) 3 + c  0 = 3 − 6 + c  0 = 3 ( 2 (1) − 3 ) 4 expression with at least one correct power, which must contain + c. 4 5 A1 Only condone c = 3 as their final answer if all coefficients have  3 ( 2 x − 3 ) 3 − 6 x 3 + 3 previously been simplified in a correct statement.  f ( x ) or y = 4 10(c) b 2 − 4ac = 1282 −4 125  192 and stating “< 0” M1* b 2 − 4ac = −79616 can be accepted in place of working. OR use of the quadratic formula and stating “No solutions” OR completing the square for the given quadratic and stating positive or > 0. OR sketch of the given quadratic and stating positive. No turning points [in the original function.] DM1 Decreasing because f  ( any positive x value )  0 A1 WWW e.g. f ' (1) = −18. 3

This question in 9709/12 Oct/Nov 2024

1 2 211 The equation of a curve is y = kx - 4 x + 2 , where k is a constant. dy d 2 y (a) Find and in terms of k. [2] dx dx 2 … … … … … … (b) It is given that k = 2 . Find the coordinates of the stationary point and determine its nature. [4] … … … … … … … … … … … … … … … … … … (c) Points A and B on the curve have x-coordinates 0.25 and 1 respectively. For a different value of k, the tangents to the curve at the points A and B meet at a point with x-coordinate 0.6. Find this value of k. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 11(a) dy 1 − 12 B1 = kx − 8 x dx 2 d 2 y 1 − 32 B1 = − kx − 8 d x 2 4 2 311(b) 1  2 2 −1 2 3 1 32  M1 OE x −− 8 x = 0 ⇒ 1 − 8 x = 0 or x = 64 x   x = or 8 x = 1  1  64  Award if working leads to x = WWW. 4 d y 1 Setting their to zero and solving, providing their only error(s) are 2 2 1 2 d x Squaring x −− 8 x = 0 to x −− 64 x = 0 gets M0. incorrect coefficients 1 A1 If x = 0 included, A0 and max of 3/4. x = only 1 4 1 2 2 SC B1 only for x = only from squaring x −− 8 x = 0 4 directly to x −−1 64 x 2 = 0 (SC B1 replacing the M1A1). 1 11 A1 11 y = from squaring x −− 2 8 x 2 = 0 to SC B1 for y = 4 4 x −−1 64 x 2 = 0. 3 B1 FT WWW − d 2 y 1 = − x 2 − 8 which is negative, so maximum 2 d 2 y dx 2 FT their x-value and their . dx 2 No FT if x = 0 is the only solution. 4 11(c) 1 M1* OE When x = 1, attempting to find y = k − 2 and gradient = k − 8 SC B1 if both correct gradients only, or both correct 2 y-coordinates only.  1  A1  k  k k k Equation of tangent is y − k + 2 =  k − 8  ( x − 1) OE, e.g. y =  − 8  x + + 6 or y = x − 8 x + + 6.  2   2  2 2 2 1 1 M1* OE When x = ,attempting to find y = k + 1.75 and gradient = k − 2 4 2 1 A1 k 9 k 9 Equation of tangent is y − k − 1.75 = ( k − 2 )( x − 0.25 ) OE, e.g. y = ( k − 2 ) x + + or y = kx − 2 x + + . 2 4 4 4 4  1  1 DM1  k  k k 9 Meet at  k − 8  ( 0.6 − 1) + k − 2 = ( k − 2 )( 0.6 − 0.25 ) + k + 1.75 OE, e.g.  − 8  0.6 + + 6 = ( k − 2 ) 0.6 + + .  2  2  2  2 4 4 Equate two tangent equations and substitute x = 0.6 M0 if constants in both equations are the same. ⇒  −0.2 k + k + 3.2 − 2 = 0.35 k − 0.7 + 0.5 k + 1.75  A1 ⇒ 0.05k = 0.15 k = 3 6

This question in 9709/13 Oct/Nov 2024

2 y M P O x 2 5 The diagram shows the curve with equation y = 2 x - + 3 . The curve crosses the x-axis at the point x P (1, 0) and M is a minimum point. (a) Find the gradient of the curve at P. [2] … … … … … … (b) Find the coordinates of M. Give each coordinate correct to 3 significant figures. [3] … … … … … … … … … …

5 marks

Mark scheme: 2(a) 5 B1 OE 4 x + x 2  d y  B1 FT Correct use of x = 1 in their two-term differentiated = 9    d x  expression, defined as an expression with one correct power. 2 2(b)  5  M1 to zero, where Their  4 x + 2  = 0 and valid method as far as ' x = ...' Equate their derivative of the form Ax B2  x  x A, B ≠ 0, and solve. If no working is seen, this can be implied by a correct answer for x. x =−1.08 A1 AWRT y = 9.96 A1 AWRT 3

This question in 9709/12 Feb/March 2025

1 dy 3 22 The equation of a curve is such that = 4 ( 2x - 5) - 9x . The curve passes through the point dx A b,4 - 11 l. 2 (a) Find the gradient of the normal to the curve at the point A. [2] … … … … … … … … … … (b) Find the equation of the curve. [4] … … … … … … … … … … … … … …

6 marks

Mark scheme: 2(a) 1 M1 dy 3 2 [Gradient of tangent] = 4 ( 2 −4 5 ) −9 4  = 90  Substitute x = 4 into . dx −11 3 1 2 is M0 unless they = 4 ( 2 −4 5 ) −9 4 2 1 reach − . 90 1 A1 AWRT −0.0111. [Gradient of normal] = − 90 2 2(b)  1 4   32  B1 B1 Accept unsimplified. y =  ( 2 x − 5 )  −6 x   + c   2   3 M1 11 11 1 Sub x = 4, y = − into an integrated expression − = 2 + c  ( 2  4 − 5 ) 4 −6 4 2 2 2 and attempt to find c. 3 A1 Condone c = 2 as final answer if ‘y = …’ seen 1 4 2 y = ( 2 x − 5 ) − 6 x + 2 previously. 2 Fractions must be simplified. Accept f(x) in place of y. 4

This question in 9709/11 May/June 2025

2 9 7 The equation of a curve is y = 4 x + - 8 . x 2 (a) A point P is moving along the curve in such a way that its y-coordinate is decreasing at 5 units per second. Find the rate at which the x-coordinate of point P is changing when x = 2 . [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the stationary points of the curve and determine their nature. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a)  dy  18 B1 OE = 8 x −   3 Accept unsimplified.  dx  x d y 55 M1 OE At x = 2, = d x 4 d y dx For evaluating their or . d x dy dy dy dt 55 d M1 For correct use of chain rule with ±5 and their =  ⇒ = −t5 dx dt dx 4 d x d y (may be algebraic). d x Condone missing brackets.  d x  4 A1 4 = − Or decreasing at a rate of .    d t  11 11 AWRT – 0.364. 4 7(b) 18  4 9  M1 d y 8 x − = 0  x = Equating their 2-term to zero. 3   x  4  d x 3 6 A1 AWRT 1.22. x =  or ± 2 2 y = 4 (for both) A1 A0 A1 if one point correct. AWRT 4.00.  d 2 y  2 = 8 + 544 M1 ForAt leastdifferentiation.one correct term needed.  d x  x So both are minima A1 No need for reason. WWW on x-values. 5

This question in 9709/11 May/June 2025

4 A point P is moving along the curve with equation y = ax 2 - 12 x in such a way that the x-coordinate of P is increasing at a constant rate of 5 units per second. (a) Find the rate at which the y-coordinate of P is changing when x = 9 . Give your answer in terms of the constant a. [3] … … … … … … … … … … … … … … … … (b) Given that the curve has a minimum point when x = 1 , find the value of a. [2] 4 … … … … … … … …

5 marks

Mark scheme: 4(a) 1 *M1 For attempt at differentiation; at least one correct term 3  dy = ax 2 − 12 needed.    dx  2 Condone poor notation throughout.  dy dy dx   3 12  DM1 For correct use of chain rule with 5, x = 9 and their dy . =  =  a  9 − 12   5  dx    dt dx dt  2  Condone missing brackets and allow errors in their working.  dy   9  45 45 a − 120 A1 OE simplified form. = 5  a − 12  or a − 60 or 22.5a − 60 or    dt   2  2 2 15 or ( 3a − 8 ) 2 3 4(b) 1 M1 d y 3  1  2 For setting their 2 term with at least one term correct a    − 12 = 0 d x 2  4  = 0 and substituting x = 0.25. Condone missing brackets. d y Allow a restart for if 2 terms seen and at least one term d x correct.  a = 16 A1 2

This question in 9709/12 May/June 2025

12 1 A curve has equation y = 2 x + . x 2 Find the equation of the tangent to the curve at the point (-2, -1). Give your answer in the form y = mx + c . [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 2 + (12 )( −2 ) x− 3 B1 Correct differential but can be unsimplified. −3 *M1 Substitute x = −2 into their differential, which must contain 2 + (12 )( −2 )( −2 )  = 5  x−3. y −−( 1) DM1 Attempt to find equation of tangent through ( −2, − 1) with Either ( their 5 ) = x −−( 2 ) their numerical gradient obtained as described above. or −=1 ( their 5 ) −( 2 ) + c  c = y = 5x + 9 A1 4

This question in 9709/13 May/June 2025

9 10 A curve C has equation y = + 2 x - 5 . 2x - 5 (a) Find the coordinates of the two stationary points. [4] … … … … … … … … … … … … … … … d 2 y (b) Find and hence determine the nature of each stationary point. [3] dx 2 … … … … … … … … … - 3 (c) The curve C is transformed to the curve C1 using a translation of e o followed by reflection in 7 the x-axis. (i) State the coordinates of the maximum point of C1. [1] … … … … … … a (ii) Find the equation of C1 in the form y = + dx + e , where a, b, c, d and e are integers. bx + c [3] … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) −9 2(2 x − 5) −2 + 2 B1 Correct differential. M1 d y ( their − 18(2 x − 5) −2 + 2 ) = 0 and rearrange to form a quadratic. Equating a two term to 0 and dealing correctly with the 2 2 d x  ( 2 x − 5 ) = 9 or 8 x − 40 x + 32 = 0    negative power. d y −2 Their two term must contain (2 x − 5) . d x (1, − 6 ) and ( 4, 6 ) A1, A1 A1 for two correct x-values or one correct point, second A1 for all correct. 4 10(b)   B1 FT Following through on their first derivative which must −3 −3 144 x − 360 −2 −18 −2 2(2 x − 5)  = 72(2 x − 5) or 4  contain (2 x − 5) .   ( 2 x − 5 )  M1 Substitute x-coordinate of each stationary point and determine 2 x − 5) −3 Use ( their x = 1 and x = 4 ) in (their 72 ( ) their nature. Nature of the turning points must correctly To determine the nature of both turning points. 2 d y follow from their values of 2 . d x  d 2 y  72 A1 CWO For x = 1 ,  2 =  − or  0 ⟹ maximum  d x  27  d 2 y  72 For x = 4 ,  2 =  or  0 ⟹ minimum  d x  27 3 10(c)(i) (1, − 13 ) B1 1 10(c)(ii) 9 M1  − 3   y = + 2 ( x  3 ) − 5  7 Application of to the original expression for C but   (2 x  3 ) − 5  7  condone +/−sign errors.  9  M1  9  y =  −  + 2 x − 5  .  y =  −  + 2 ( x  3 ) −5 7  SC B1 for   2 ( x  3 ) − 5   2 x − 5  9 A1 Answer must be in this format; the ' y = ' can be implied by y = − − 2 x − 8 2 x + 1 earlier inclusion. 3

This question in 9709/13 May/June 2025

8 The equation of a curve is y = x 3 + ax 2 + bx + 5 . The curve has a stationary point at (1, 9). (a) Find the values of the constants a and b. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the other stationary point. [3] … … … … … … … … … … … … … (c) A point P is moving along part of the curve in such a way that the y-coordinate of P is increasing at a constant rate of 6 units per second. Find the rate at which the x-coordinate of P is increasing when x = 5 . [3] … … … … … … … … … … … …

11 marks

Mark scheme: 8(a) 9 = 1 + a + b + 5 B1 dy 2 B1 = 3 x + 2 ax + b dx Gradient = 0 at (1, 9) so 0 = 3 + 2a + b M1 d y Setting their to zero and substituting x = 1. d x Attempt to solve their linear equations simultaneously DM1 Can be implied by their answers. a = −6, b = 9 A1 WWW 5 8(b) dy 2 M1 d y = 3 x − 12 x + 9 = 0 Setting their to zero. dx d x Solution [leading to x = 1 or x = 3] DM1 Solving their 3-term quadratic (3, 5) or x = 3, y = 5 A1 WWW Ignore (1, 9 ) if given as a second answer. Only dependent on the first M1. 3 8(c) dy 2 M1 dy At x = 5, = 3  5 − 12 +5 9 Substituting x = 5 into their . May be implied. dx dx dx dt M1 OE 6 = their 24  or their 24 = 6  dt dx  dy dy dx  Use of chain rule SOI  =   .  dt dx dt  d x d t Linking correctly (or ), their 24 and 6. d t d x d x 1 A1 OE = d t 4 3

This question in 9709/15 May/June 2025

11 A curve passes through the point P (4, 3) and is such that dy 8 10 = - . dx x 2 ( 2 x - 3 ) 2 (a) Find the equation of the normal to the curve at P. Give your answer in the form y = mx + c . [3] … … … … … … … … … … (b) Find the rate of change of the gradient of the curve when x = 4 . [3] … … … … … … … … … … … … … (c) Given that the curve also passes through the point (-1, q), find the value of q. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Substitute 4 to obtain gradient of curve is 1 B1 10 Attempt equation of normal using (4, 3) and −m1 for their gradient M1 y = −10 x + 43 A1 3 11(b) −16 x −3 +40(2 x − 3) −3 B1 B1 OE    Substitute 4 to obtain 1007 B1 3 −8 x +5(2 x − 3) + c 11(c)  y =  −1 −1 B1 B1   Substitute x = 4, y = 3 in an integrated expression to find value of c M1 Obtain 3 = −+2 1 +c and hence y = −8 x −1 + 5(2 x − 3) −1 + 4 A1 OE For finding c = 4. Substitute x = −1 to obtain q = 11 A1 Not y = 11. 5

This question in 9709/11 Oct/Nov 2025

5 y O a b x 1 The equation of a curve is y = 4x 2 - x . The curve has a maximum point when x = a and crosses the x-axis at the point with coordinates (b, 0), where b 2 0 . The shaded region is bounded by the curve, the line x = a and the x-axis (see diagram). (a) Find the value of a. [3] … … … … … … … … … … … … … … … … (b) Find the exact area of the shaded region. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 1 B15(a)  dy  − 2 = 2 x − 1    dx  1 2 2 x −−=1 1 2 1 0  x = 2 M1 d y Setting their of the form k x −−2 1 equal to 0 and d x 2 solving as far as x 1 = d , condoning sign errors only. Can be implied by correct final answer but not if clearly following wrong working.  a = 4 A1 Alternative Method for Question 5(a): y = 4 x − ( x ) 2 B1 Recognising the quadratic in .x −b −4 M1 Condone sign errors only. Max at x = = = 2 2 2 a −2 Allow z = x , − ( z − 2) + 4  max at 2. 2 = 2 or − ( x − 2) 2 + 4  x 1  a = 4 A1 3 5(b)  x = or b =  16 B1 SOI It may be found in 5(a), but must be seen in 5(b). Condone extra ‘solution’ x = 0. 3  4   1 2  B2, 1, 0 B2 for both correct components and no other x terms. 2  12   4 x − x dx = −  x  +  x  B1 for one correct term. ( )   2    32  Allow any correct unsimplified form. 3 3  8 2 1 2   8 2 1 2  M1 Substituting their a (from part (a)) and their b (from [Area =]   16 −  16  −   4 −  4  1  3 2   3 2  2 an attempt to solve 4 x − x = 0 ) into an integrated expression (defined by having at least one correct power) and subtracting. If correct limits and integration, then minimum 128 40 acceptable working is − . 3 3 If incorrect limits or integration, then full substitution of every term must be seen. Note: needs 0  a  b, otherwise M0, but allow limits applied either way round. Allow missing brackets if recovered.  128 40  88 1 DB1 88  = −  = or 29 Must be exact. Allow − if it becomes 88.  3 3  3 3 3 3 Do not ISW if a further area is added or subtracted. Dependent upon B1B2 scored earlier. 5

This question in 9709/12 Oct/Nov 2025

4 1 9 The function f is defined by f ( x) = + for x 2 2 . ( 3x - 6) 2 ( 3 x - 6) 3 (a) Find an expression for fl( )x and hence determine whether f is an increasing function, a decreasing function or neither. [4] … … … … … … … … … … … … … … (b) State whether f - 1 exists. Give a reason for your answer. [1] … … … … … … … … … … The function g is defined by g ( )x = 4x - 3 for x 2 a . (c) Find the range of g in terms of the constant a. [1] … … … … … … … … … (d) Find the set of values of a for which the composite function fg exists. [2] … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) 4 −3 ( 2 ) 1 −3 ( 3 ) B1B1B1 B1 for the correct powers, B1 for ×3 in at least one + 3 4 term, B1 for all correct which can be unsimplified. ( 3 x − 6 ) ( 3 x − 6 ) −3 −4 or 4 −3 ( 2 )( 3x − 6 ) + 1 −3 ( 3)( 3 x − 6 ) Decreasing. B1* This mark is only available if f ' ( x ) is of the form ( − p )( 3x − 6 ) −3 + ( −q )( 3x − 6 ) −4 , where p and q are positive coefficients. 4 9(b) f −1 exists because f is a decreasing function DB1 Or f −1 exists because it is one-to-one, or passes the horizontal line test. 1 9(c) g ( x )  4 a − 3 B1 Allow ‘ y  ’ or ‘g ’ only. Accept 4a −3 y  or ( 4 a − 3 ,  ) . Condone ( 4 a − 3 ,   . Accept g ( x )  g ( a ) , but not if they make an error ‘simplifying’ it. 1 9(d) Either 4a −3 2 allow with x or a or  M1 Do not allow = or , unless they reach a correct inequality later. 5 5 A1 5 5  5   5  a  or a  oe Accept  a ,  a  ,  ,   or  ,   . 4 4 4 4  4   4  Or 3  ( 4 a − 3 ) − 6  0 allow with x or a or  M1 Do not allow = or , unless they reach a correct inequality later. 5 5 A1 5 5  5   5  a  or a  oe Accept  a ,  a  ,  ,   or  ,   . 4 4 4 4  4   4  2

This question in 9709/12 Oct/Nov 2025

1 23 23 The equation of a curve is y = f ( x) , where f ( x) = x ( x - 2) . The following points lie on the curve. 2 Non-exact values of the y-coordinates are given correct to 6 decimal places. A(8, 72), B(8.001, k), C(8.01, 72.300388), D(8.1, 75.038882) (a) Find the value of k. Give your answer correct to 6 decimal places. [1] … … … … … The table below shows the gradients of the chords AB and AC, given correct to 4 decimal places. Chord AB AC AD Gradient of chord 30.0039 30.0388 (b) Find the gradient of the chord AD. Give your answer correct to 4 decimal places. [1] … … … … … … … (c) State what the values in the table suggest about the value of fl ( 8 ) . [1] … … … … … … …

3 marks

Mark scheme: 3(a) 72.030004 B1 CAO. Not AWRT. 1 3(b) 30.3888 B1 CAO. Not AWRT. 1 3(c) 30[.0] B1 CAO. 30 may be accompanied by ‘around’, ‘approximately’ etc. 1

This question in 9709/13 Oct/Nov 2025

8 6 11 The equation of a curve is y = - . 3x - 8 x - 1 (a) Find the coordinates of the point at which the tangent to the curve at the point (3, 5) intersects the line y =-8x . [6] … … … … … … … … … … … … … … … … … … … … … … … … … (b) (i) Find the x-coordinates of each of the stationary points of the curve. [3] … … … … … … … … … … … d 2 y (ii) Find and hence determine the nature of each of the stationary points. [4] dx 2 … … … … … … … … … … … … … … …

13 marks

Mark scheme: 11(a)    B1 B1 for each {} element. d y  −8 3  6  =  2  + 2  d x  ( 3 x − 8 )  ( x − 1)  B1    *M1 Using their differentiated expression, which must contain  −8 3  6  −45  −2 −2  2  + 2  =  ( 3x − 8 ) and ( x − 1) , and x = 3 . This may be seen in their line  ( 3.3 − 8 )  ( 3 − 1)   2  equation.  −45  *DM1 Correct form of a line equation with their gradient, but not the their ( x − 3 ) y − 5 =   negative reciprocal, and ( 3, 5 ) . 2    −45   145  or 5 =  their  ( 3 ) + c  c =    2   2   −45  DM1 Replacing y with −8x and collecting terms. 29 x = 145  −8 x − 5 = ( x − 3 )    2  x = 5, y = −40 A1 Accept ( 5, − 40 ) . 6 11(b)(i) −24 6 2 2 *M1 d y −2 −2 + = 0 ⇒ 24 ( x − 1) = 6( 3 x − 8 ) Equate their , which must contain ( 3x − 8 ) and ( x − 1) to 0, 2 2 ( 3 x − 8 ) ( x − 1) d x and clear of fractions.   4 ( x − 1) 2 = ( 3 x − 8 ) 2  Only condone  errors.   5 x 2 − 40 x + 60 [= 0] or  2 ( x − 1) = 3 x − 8 DM1 OE Forming a three-term quadratic. Condone only  errors. Or taking square roots. For this method, the  must be present. 2, 6 A1 Both values. 3 11(b)(ii) 2     B1FT d y d y  −24 −2 3   6 −2  Correct differentials of the elements of their , which must 2 =  3  +  3  d x B1FT −2 − d x  ( 3 x − 8 )   ( x − 1)  contain ( 3 x − 8 ) and ( x − 1) 2. d 2 y their 144 their 12 M1 d 2 y At x = 2, 2 = 3 − 3 Replacing x with their 2 and their 6 in their 2 , which must dx ( 6 − 8 ) ( 2 − 1) dx −3 −3 contain ( 3 x − 8 ) and ( x − 1) . d 2 y their 144 their 12 and at x = 6, 2 = 3 − 3 Correct final answers. dx (18 − 8 ) ( 6 − 1) d 2 y A1 WWW At x = 2, 2 = −30  0, therefore max[imum] Correct values, or working, and  0 and  0 are required. dx d 2 y 48  6  At x = 6, 2 = 3    0, therefore min[imum] dx 10  125  4

This question in 9709/13 Oct/Nov 2025

7 r cm h cm A manufacturer wishes to design an open cylindrical tank, as shown in the diagram. The tank will have a base but no top. The outside of the tank will have a fixed surface area of 600r cm2. The radius r cm and height h cm of the tank can vary. (a) Show that the volume, V cm3, of the tank is given by 2 rr ( 600 - r ) V = . 2 [3] … … … … … … … … … … … … … … … … … … … (b) Find the exact value of r which corresponds to the maximum value of V. [3] … … … … … … … … … … … … … (c) Hence, find the maximum value of V. [2] … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) 2  600π − πr 2 600 − r 2  M1* Uses given info and correct formulae to form 2πrh + πr = 600π oe  h = =  equation in h and r.  2πr 2 r  600 − r 2 DM1 Sub their expression for h into V = πr 2 h. ( ) 2 V =  πr oe 2r 2 A1 AG πr 600 − r ( ) = CAO 2 WWW 3 7(b) 1 3 dV 1 2  3πr 2  M1* Differentiate given expression for V. V = π 600r − r = π 600 − 3r or  300π −  1 2 ( ) ( ) 2 dr 2  2  Condone missing π (must be of form a − br ). 2 Alternative: h = r at max V, so 2πr ( r ) + πr 2 = 600π oe 600 − 3r 2 = 0 DM1 Equate their derivative to zero and attempt to solve as far as an equation of the form ‘ r = ’. Alternative: Solving 2πr ( r ) + πr 2 = 600π r = 10 2 A1 CAO (accept 200 ). 3 7(c) π  3  M1 Sub their value of r from 7(b) into the given 10 2 − 10 2 V =  600 ( ) ( )  expression for V, providing their value of r > 0 2   and gives V  0. 8890 ( 3sf ) Accept exact answer 2000 π 2 A1 AWRT 8890 (3sf). 2

This question in 9709/15 Oct/Nov 2025

11 y P O x The diagram shows the curve with equation y = 4x 2 - x 3 and the tangent to the curve at the point P. The point P has x-coordinate 3. (a) Find the equation of the tangent to the curve at the point P. Give your answer in the form y = mx + c . [5] … … … … … … … … … … … … … … … … … … … … (b) The shaded region is bounded by the curve, the x-axis and the tangent to the curve at P. Find the exact area of the shaded region. [6] … … … … … … … … … … … … … … The graph of y = 4x 2 - x 3 is transformed by a stretch of scale factor 1 in the x-direction. The point Q is 3 the image of P under this transformation. The transformed shaded region is bounded by the transformed curve, the x-axis and the tangent to the transformed curve at Q. (c) (i) Find the equation of the transformed curve in the form y = mx 2 + nx 3 , where m and n are integers to be found. [1] … … … (c) (ii) State the coordinates of Q and the area of the transformed shaded region. [2] … … … …

14 marks

Mark scheme: 11(a) dy 2 B1 CAO = 8 x − 3 x dx dy B1 = 24 − 27 = −3  when x = 3 dx y = 9 [when x = 3] B1 SOI y − 9 = −3 ( x − 3 ) or y = −3 x + c → 9 = −+9 c →=c 18 oe M1 d y Uses their y and their numerical to find d x equation of the tangent; condone one sign error. y = −3 x + 18 A1 5 11(b) 4 M1* Must obtain ax 3 + bx 4 and indicate the limits 3 2 3 Area between curve and x-axis = 4 x − x dx  and attempt to integrate )   ( and 4. 3 4 A1 SC B1 for use of wrong or no limits (only for 3 4  4 x x  =  −  correct integral).  3 4  3  256   81  DM1 Correct sub of correct limits (allow one slip).  − 64  −  36 −  64  3   4  Minimum acceptable: − 63. 3 4 67 A1 SOI = May be implied by a correct final answer if the 12 two areas are combined. SC B1 if substitution of the limits is not seen. 9 67 DM1 27 Shaded region = ( 6 − 3)  −their Expect −‘their integral’, but must be ‘area 2 12 2 6 3 under their line’ minus ‘their area under the  −3 x  67 or  + 18 x  – their curve’, where ‘their integral’ is an attempt at the  2  3 12 area under the curve between x = 3 and x = 4. May use the lengths from their tangent equation. 95 A1 Calculating area of triangle – (correct) area under = any equivalent exact answer the curve. 12 11(b) Alternative Method for Question 11(b): Finds area between curve and tangent between x = 3 and x = 4 M1* Integrate at least two of the four terms correctly. 4 Area under the line could be found from the 4 x 2 − x 3 3 x + 18 ) − ( ) dx ( 4 − 3  ( − 3 trapezium area )( 9 + 6 ) . 2 4 2 3 4 A1 Integrating all four terms correctly.  3 x 4 x x  =  − + 18 x − +   4 x 3 x 4   2 3 4  3 SC B1 for the correct integral − +  .  3 4   256   27 81  DM1 Correct sub of limits (allow one slip). =  −24 + 72 − + 64  −− + 54 − 36 +   3   2 4  23 A1 SOI = SC B1 if substitution of the limits is not seen. 12 1 23 DM1 Calculating area of triangle between x = 4 and Shaded region = ( 6 − 4 ) 6 + 2 12 x = 6 + their area, providing limits of 3 and 4 are used to find the area between the curve and the tangent. 95 A1 Must be exact. = 12 6 11(c)(i) y = 36 x 2 − 27 x 3 or state m = 36, n = −27 B1 CAO (must be expanded) 1 11(c)(ii) Q(1, 9) B1 CAO coordinates of Q. 95 B1 FT 1 Area = of their area from 11(b). 36 3 Allow 2.64. 2

This question in 9709/15 Oct/Nov 2025