Quadratics — Paper 1 · A Level Mathematics 9709

18 1 4 Find the real roots of the equation + = 4. [4] x4 x2

4 marks

Mark scheme: 4 × (x4 ) → 4 x 4 −x 2 − 18 = 0 M1 Recognition of quad in x² or 1÷(x²) ( 4 x 2 − 9)( x 2 + 2) = 0 DM1 Solution of quadratic. x = 1.5 or x = − 1.5 A1 A1√ Positive root. For recognition of (−ve) [4] The A1√ assumes no other real answers. 1

This question in 9709/11 May/June 2007

2 Find the set of values of k for which the line y = kx −4 intersects the curve y = x2 −2x at two distinct points. [4]

4 marks

Mark scheme: 2 kx − 4 = x² − 2x → x² − (2 + k)x + 4 = 0 M1 Complete elimination of y (or x) Use of b² − 4ac M1 Any use ( =0, <0, >0) (2 + k)² = 16 k = 2 or −6 A1 For the values of k, however used k > 2 or k < −6 A1 Correct only. [4] 5

This question in 9709/11 May/June 2009

10 The function f is defined by f : x →2x2 −12x + 13 for 0 ≤x ≤A, where A is a constant. (i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1] (iii) When A has this value, find the range of f. [2] The function g is defined by g : x →2x2 −12x + 13 for x ≥4. (iv) Explain why g has an inverse. [1] (v) Obtain an expression, in terms of x, for g−1(x). [3]

10 marks

Mark scheme: 10 (i) 2x² − 12x + 13 = 2(x − 3)² − 5 3 × B1 Allow even if a, b, c not specifically quoted. [3] (ii) Symmetrical about x = 3. A = 6. B1√ For 2 × his (−b). [1] (iii) One limit is −5 B1√ For his c. Other limit is 13 B1 co. [2] (iv) Inverse since 1:1 (4 > 3). B1 Valid argument. [1] (v) Makes x the subject of the equation M1 Attempts to change the formula. Order of operations correct DM1 “+5”, ÷2, √, +3. Allow for simple algebraic slips such as − 5 for +5 etc. x + 5 → + 3 A1 co – as a function of x, not y. 2 [3] condone ±. dy 2

This question in 9709/11 May/June 2009

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

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

6 (i) Find the first 3 terms in the expansion of (1 + ax)5 in ascending powers of x. [2] (ii) Given that there is no term in x in the expansion of (1 −2x)(1 + ax)5, find the value of the constant a. [2] (iii) For this value of a, find the coefficient of x2 in the expansion of (1 −2x)(1 + ax)5. [3]

7 marks

Mark scheme: → 6 s − 3c = 2 s − 6 c → 4 s = −3c M1 Expanding, collecting, use of t = s ÷ c 3 A1 Answer given. All correct. → tan x = − 4 [2] (ii) x = 180 – 36.9 = 143.1° or B1 co x = 360 – 36.9 = 323.1° B1√ For 180 + first answer. [2] a 2 y = x  a 2  − a2  M1 π. For using correct formula with dx = (π )   Volume = π ∫  x B1 only For correct integration of x–2    x 2  2πa 2 Use of limits 1 to 3 → M1 Must be using y2 or πy2. 3 Equates to 24π → a = 6 A1 Co, allow ±6. [4] 2

This question in 9709/12 May/June 2010

10 The function f : x →2x2 −8x + 14 is defined for x ∈>. (i) Find the values of the constant k for which the line y + kx = 12 is a tangent to the curve y = f(x). [4] (ii) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (iii) Find the range of f. [1] The function g : x →2x2 −8x + 14 is defined for x ≥A. (iv) Find the smallest value of A for which g has an inverse. [1] (v) For this value of A, find an expression for g−1(x) in terms of x. [3]

12 marks

Mark scheme: 10 f : x a 2 x 2 −x8 + 14 (i) y + kx = 12, Sim Eqns. M1 Complete elimination of y (or x) → 2x2 – 8x + kx + 2 = 0 A1 Use of b2 – 4ac M1 Uses b2 – 4ac on eqn = 0, no “x” in a, b, c. → (k – 8)2 =16 → k = 12 or 4. A1 co.co [4] (ii) 2x2 – 8x + 14 = 2(x – 2)2 + 6 B1×3 [3] (iii) Range of f [ 6. B1√ √ for c or from calculus. [1] (iv) Smallest A = 2 B1√ √ to answer to (ii). [1] (v) Makes x the subject M1 Could interchange x, y first. Order of operations correct. M1 Order must be correct. −1 x − 6 g ( x ) = + 2 A1 co 2 [3]

This question in 9709/13 May/June 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

6 A curve has equation y = kx2 + 1 and a line has equation y = kx, where k is a non-zero constant. (i) Find the set of values of k for which the curve and the line have no common points. [3] (ii) State the value of k for which the line is a tangent to the curve and, for this case, find the coordinates of the point where the line touches the curve. [4]

7 marks

Mark scheme: 6 (i) kx2 – kx + 1 = 0 M1 y eliminated k2 – 4k < 0 M1 Applying b2 – 4ac < 0 or = or ≤ or ≥ 0 < k < 4 A1 co [3] (ii) k = 4 only B1√ ft from their k2 – 4k = 0. (Not k = 0) (2x – 1)2 = 0 M1 ft from their k x = ½ , y = 2 or (½, 2) A1, A1 [4] 2

This question in 9709/12 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

3 (i) Sketch the curve y = (x −2)2. [1] (ii) The region enclosed by the curve, the x-axis and the y-axis is rotated through 360◦about the x-axis. Find the volume obtained, giving your answer in terms of π. [4]

5 marks

Mark scheme: 3 (i) Correct shape – touching positive x-axis B1 Ignore intersections with axes [1] (ii) (π ) ∫ ( x − 2 ) 4 d x M1 Use (π ) ∫ y 2 d x & attempt integrate but expansion before integn needs 5 terms  ( x − 2) 5  (π )   A1 5   (π )[0 − ( −32/) 5]) M1 Use of limits 0, 2 on their (π ) y ∫ 2 dx 32π or 6.4π A1 cao Rotation about y-axis max 1/5 5 [4] B1

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

10 (i) Express 2x2 −4x + 1 in the form a(x + b)2 + c and hence state the coordinates of the minimum point, A, on the curve y = 2x2 −4x + 1. [4] The line x −y + 4 = 0 intersects the curve y = 2x2 −4x + 1 at points P and Q. It is given that the coordinates of P are (3, 7). (ii) Find the coordinates of Q. [3] (iii) Find the equation of the line joining Q to the mid-point of AP. [3]

10 marks

Mark scheme: 10 (i) 2(x – 1)2 – 1 OR a = 2, b = –1, c = –1 B1, B1, B1 A = (1, –1) B1√ Allow alt. method for final mark [4] (ii) 2 x 2 − 5 x − 3 = 0 ⇒ ( 2 x + 1)( x − 3) = 0 OE in y M1, M1 Complete elim & simplify, attempt soln. x = − 1 2 , y = 3 1 2 A1 Additional (3, 7) not penalised [3] (iii) Mid-point of AP = (2, 3) B1√ Follow through on their A 1 − 1 Gradient of line = 2 = B1 − 5 5 2 − 1 1 Equation is y − 3 = ( x − 2) OE B1 Or y − 3 1 2 = − 1 5 5( x + 2 ) [3] 2 2

This question in 9709/11 May/June 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

9 A line has equation y = kx + 6 and a curve has equation y = x2 + 3x + 2k, where k is a constant. (i) For the case where k = 2, the line and the curve intersect at points A and B. Find the distance AB and the coordinates of the mid-point of AB. [5] (ii) Find the two values of k for which the line is a tangent to the curve. [4] [Questions 10 and 11 are printed on the next page.]

9 marks

Mark scheme: 9 (i) x 2 + 3 x + 4 = 2 x + 6 ⇒ x 2 + x − 2 (= 0 ) M1 3-term simplification ( x − 1)( x + 2 ) = 0 → (8,1 ), (− 2,2 ) DM1A1 DM1 for attempted solution for x 2 2 cao ( 45 from wrong points scores AB = 3 + 6 = .6 71 or 45 or 3 5 B1 B0) − 1 5,  B1√ [5] Ft their coordinates  2  GCE AS/A LEVEL – October/November 2011 9709 11 (ii) x 2 + (3 − k )x + 2 k − 6(= 0 ) M1 Simplified to 3-term quadratic 2 (3 − k ) 2 − 4(2 k − 6 ) = 0 DM1 Apply b −ac4 = 0 as function of k only (3 − k )(11 − k ) = 0 DM1 Attempt factorisation or use formula Both correct NB Alternative methods for (ii) k = 3 or 11 A1 [4] possible If B0B0 then SCB1 for both y = 1 & ( ) ( )

This question in 9709/11 Oct/Nov 2011

11 Functions f and g are defined by f : x →2x2 −8x + 10 for 0 ≤x ≤2, g : x →x for 0 ≤x ≤10. (i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (ii) State the range of f. [1] (iii) State the domain of f −1. [1] (iv) Sketch on the same diagram the graphs of y = f(x), y = g(x) and y = f −1(x), making clear the relationship between the graphs. [4] (v) Find an expression for f −1(x). [3]

12 marks

Mark scheme: 2 B 1, B1,11 (i) 2( x − 2 ) + 2 [3] For 2 , − 2 , 2 B1 (ii) 2 ≤ f ( x ) ≤ 10 oe B1 [1] Allow < etc. Ignore notation (iii) 2 ≤x ≤ 10 B1√ [1] Ft from part (ii). Ignore notation Or from int with y axis to int with (iv) f ( x ) ≈: half parabola from (,010 ) to (2,2 ) B1 their y = x g ( x ): line through 0 at ≈45 ° B1 f −1 ( x ): reflection of their f ( x ) in g ( x ) B1√ Everything totally correct B1 [4] GCE AS/A LEVEL – October/November 2011 9709 11 2 1 Allow +√or ‒√Dep on final ans as ( ) ( )

This question in 9709/11 Oct/Nov 2011

3 y y = 2x B O x A y = 2 x 5 + 3 x 3 The diagram shows the curve y = 2x5 + 3x3 and the line y = 2x intersecting at points A, O and B. (i) Show that the x-coordinates of A and B satisfy the equation 2x4 + 3x2 −2 = 0. [2] (ii) Solve the equation 2x4 + 3x2 −2 = 0 and hence find the coordinates of A and B, giving your answers in an exact form. [3]

5 marks

Mark scheme: 3 (i) 2x5 + 3x2 = 2x ⇒2x5 + 3x2 – 2x = 0 M1 First line essential [x(2x]4 + 3x2 – 2) = 0 2x4 + 3x2 – 2 = 0 A1 AG Factorising needed for A1 [2] M1 Reasonable attempt at solving a (ii) (x2 + 2)(2x2 – 1) = 0 quadratic in x2 A1 1 x = ± only A1 For a correct pair of solutions, either 2 2 [3] x’s or 1 x and 1 y   SC (±0.707, ±1.41) AWRT B1  1 2 , −1     2, 2   2, −2   2  2

This question in 9709/13 Oct/Nov 2011

5 y y = 6x + k y = 7Ö x B A x O The diagram shows the curve y and the line y 6x k, where k is a constant. The curve and 7√x the line intersect at the points A =and B. = + (i) For the case where k 2, find the x-coordinates of A and B. [4] = (ii) Find the value of k for which y 6x k is a tangent to the curve y [2] 7√x. = + =

6 marks

Mark scheme: M1 Expressing as a clear quadratic soi5 (i) 6 x + 2 = 7 x ⇒ 6 ( x )2 − 7 x + 2 = 0 M1 oe e.g. (3t − 2 )(2t − 1) = 0 (3 x − 2 )(2 x − 1) = 0 2 1 x = or A1 1 solution sufficient. Accept e.g. t = 2/3 3 2 4 1 x = or (or 0.444, 0.25) A1 Both solutions required cao 9 4 OR (6 x + 2 )2 = 49 x → 36 x 2 − 25 x + 4 = 0 M1A1 Attempt to square both sides (9 x − 4 )(4 x − 1) = 0 M1 Attempt to solve (or formula etc.) 4 1 x = or (or 0.444, 0.25) oe A1 9 4 [4] (ii) 72 − 4 × 6 × k (= 0 ) M1 Apply b 2 −ac4 (= 0 ) 49 k = or 2.04 A1 Attempt to equate derivatives 24 d 1 d 7 −1 2 OR 7 x 2 = 6 M1 6 x + k ) → x ) = ( ( dx dx 2 49 49 49 x = , y = → k = or 2.04 A1 144 12 24 [2] GCE AS/A LEVEL – May/June 2012 9709 11

This question in 9709/11 May/June 2012

9 y A y = – x2 + 8x –10 B x O The diagram shows part of the curve y 8x which passes through the points A and B. The = −x2 + −10 curve has a maximum point at A and the gradient of the line BA is 2. (i) Find the coordinates of A and B. [7] (ii) Find y dx and hence evaluate the area of the shaded region. [4] ã

11 marks

Mark scheme: 9 y = − x 2 + 8 x − 10 dy (i) = −2x + 8 B1 co dx = 0 when x = 4, A is (4, 6) M1A1 Sets to 0 and attempt to solve for x. co. Equation of AB is y − 6 = 2 ( x − 4 ) M1 Correct form of equation. Sim eqns with eqn of curve M1 Eliminates x or y completely → x 2 −x6 + 8 = 0 or y 2 −y8 + 12 = 0 A1 Method for quadratic eqn = 0. → B (2, 2) A1 co (Must not be guessed from diagram) [7] 2 x 3 2 (ii) ∫ − x + 8 x − 10 dx = − 3 + 4 x − 10 x B2,1 3 terms, loses 1 for each error Uses his x limits 2 to 4 M1 Uses x limits correctly – allow ± → 9⅓ A1 co – allow ± (2 must have been [4] correctly found, not guessed) GCE AS/A LEVEL – May/June 2012 9709 12 8 10 f : x a 2 x + 5 g : x a x − 3 –1 B1 co (i) f = ½(x − 5) M1 A1 Attempt at x the subject. co but (f(x) 1 8

This question in 9709/12 May/June 2012

1 Solve the inequality x2 −x −2 > 0. [3]

3 marks

Mark scheme: 1 (x + 1) (x – 2) or other valid method M1 Attempt soln of eqn or other method −1, 2 A1 x < –1, x > 2 A1 Penalise ≤ , ≥ [3] 1 1 − −

This question in 9709/13 Oct/Nov 2013

5 Find the set of values of k for which the line y = 2x −k meets the curve y = x2 + kx −2 at two distinct points. [5]

5 marks

Mark scheme: 5 x 2 + x ( k − 2) + ( k − 2)( = 0) M1 Equate and move terms to one side of equ. Apply b 2 − 4 ac (>0). Allow [ at this ( k − 2) 2 − 4( k − 2)( > 0) soi M1 stage. ( k − 2 )( k − 6 )( > 0 ) DM1 k < 2 or k > 6 (condone Y, [) A2 Attempt to factorise or solve or find 2 Allow {–∞, 2}U{6, ∞} etc. [5] solns. SCA1 for 2, 6 seen with wrong inequalities

This question in 9709/11 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

9 y B y = 3 x y = x + 2 A x O The diagram shows parts of the graphs of y = x + 2 and y = 3 x intersecting at points A and B. (i) Write down an equation satisfied by the x-coordinates of A and B. Solve this equation and hence find the coordinates of A and B. [4] (ii) Find by integration the area of the shaded region. [6] [Question 10 is printed on the next page.]

10 marks

Mark scheme: 2 9 (i) x − 3 x + 2 or k 2 − 3k + 2 or (3 x ) = ( x + 2 )2 M1 OR attempt to eliminate x eg sub y 2 x = 9 x = 1 or 2 or k = 1 or 2 or x 2 −x5 + 4 (= 0 ) A1 y 2 − 9 y + 18 = 0 y = 3 or 6 x = 1 or 4 A1 x = 1 or 4 y = 3 or 6 A1 [4] 1 dx –  M1DM1 Attempt to integrate. Subtract at (ii) ∫ 3 x  ∫ ( x + 2 ) dx or attempt at trapezium  2 some stage 3  1 2  1  2x −  x + 2 x  or ( y 2 + y1 )( x 2 − x1 A1A1 Where ( x1 , y 1 ), ( x 2 , y 2 ) is their 2  2  2 ) (1, 3), (4, 6)    1   1  (16 − 2) –  (8 + 8 ) −  + 2  or their × 9 × 3 DM1 Apply their 1→4 limits correctly    2   2  to curve 1 A1 For A mark allow reverse subtn→ 2 −1 →1 but not reversed limits [6] 2 2 OR y 2  M1DM1  ∫ ( y − 2 ) d y or attempt at trap  − ∫ 9 d y  1 2 1  y 3 y − 2 y or ( x1 + x 2 )( y 2 − y1 ) − A1A1  2 2  27   1  1  (18 − 12 ) −  4 − 6  or × 5 × 3 − [8 − ]1 DM1 Apply their 3→6 limits correctly   2  2  to curve 1 A1 2 1 2

This question in 9709/13 Oct/Nov 2014

1 A line has equation y = 2x −7 and a curve has equation y = x2 −4x + c, where c is a constant. Find the set of possible values of c for which the line does not intersect the curve. [3]

3 marks

Mark scheme: 1 x 2 − 4 x + c = 2 x − 7 → x 2 − 6 x + c + 7 ( = 0 ) M1 All terms on one side 36 − 4 ( c + 7 ) < 0 DM1 Apply 4 0. Allow ⩽. c > 2 A1 [3]  x 5   9 2  [ ]      

This question in 9709/13 Oct/Nov 2015

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

1 Find the set of values of k for which the equation 2x2 + 3kx + k = 0 has distinct real roots. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 2 (3 ) 4 2 k k − × × M1 Attempt 2 4 b ac − 2 9 8 k k − > 0 soi Allow 2 9 8 k k − . 0 A1 Must involve correct inequality. Can be implied by correct answers 0, 8/9 soi A1 k < 0, k > 8/9 (or 0.889) A1 Allow (‒∞, 0) , (8/9, ∞) Total: 4

This question in 9709/12 Feb/March 2017

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 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

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

7 Points A and B lie on the curve y = x2 −4x + 7. Point A has coordinates 4, 7 and B is the stationary point of the curve. The equation of a line L is y = mx −2, where m is a constant. (i) In the case where L passes through the mid-point of AB, find the value of m. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the set of values of m for which L does not meet the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) d 2 4 0 d = − = y x x 2, y → = = x 3 B1 B1 Midpoint of AB is (3, 5) B1 FT FT on (their 2, their 3) with (4,7) 7 3 m → = (or 2.33) B1 4 7(ii) Simultaneous equations → ( ) 2 4 9 0 − − + = x x mx *M1 Equates and sets to 0 must contain m Use of b²−4ac → (m + 4)² − 36 DM1 Any use of b²−4ac on equation set to 0 must contain m Solves = 0 → −10 or 2 A1 Correct end-points. −10 < m < 2 A1 Don’t condone ⩽ at either or both end(s). Accept −10 < m, m < 2. 4

This question in 9709/12 Oct/Nov 2017

2 The equation of a curve is y = x2 −6x + k, where k is a constant. (i) Find the set of values of k for which the whole of the curve lies above the x-axis. [2] … … … … … … … … … … … (ii) Find the value of k for which the line y + 2x = 7 is a tangent to the curve. [3] … … … … … … … … … … … …

5 marks

Mark scheme: 2(i) A complete method as far as finding a set of values for k by: Either (x – 3)² + k – 9 >0, k – 9 >0 Either completing the square and using ‘their k – 9’ > or ⩾ 0 OR or 2x – 6 = 0 → (3, k ─ 9), k – 9 >0 M1 Differentiating and setting to 0, using ‘their x=3’ to find y and using ‘their k – 9’ > or ⩾0 OR or b² < 4ac oe → 36 < 4k Use of discriminant < or ⩽ 0. Beware use of > and incorrect algebra. → k > 9 Note: not ⩾ A1 T&I leading to (or no working) correct answer 2/2 otherwise 0/2. 2 Question Answer Marks Guidance 2(ii) EITHER 2 – 6 x x k + = 7 – 2x → 2 – 4 –7 x x k + (= 0) *M1 Equates and collects terms. Use of b² – 4ac = 0 (16 – 4(k – 7) = 0) DM1 Correct use of discriminant = 0, involving k from a 3 term quadratic. OR 2x – 6 = – 2 → x = 2 (y = 3) *M1 Equates their d d y x to ± 2, finds a value for x. (their 3) or 7– 2(their 2) = (their 2)2 – 6(their 2) + k DM1 Substitutes their value(s) into the appropriate equation. → k = 11 A1 3

This question in 9709/12 May/June 2018

1 1 Showing all necessary working, solve the equation 4x −11x 2 + 6 = 0. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 ½ ½ 4 3 2 − − x x oe soi Alt: 2 4 6 11 16 73 36 + = ⇒ − + x x x x M1 Attempt solution for ½ x or sub u = ½ x ½ 3/ 4 or 2 = x ( )( ) 16 9 4 − − x x A1 Reasonable solutions for ½ x implies M1 (ݔ= 2, 3/4, M1A0) 9 /16 oe or 4 = x A1 Little or no working shown scores SCB3, spotting one solution, B0 3

This question in 9709/11 Oct/Nov 2018

2 A line has equation y = x + 1 and a curve has equation y = x2 + bx + 5. Find the set of values of the constant b for which the line meets the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 2 5 1 1 4 0 + + = + → + − + = x bx x x x b M1 ( ) 2 2 ( 4 ) 1 16 − = − − b ac b M1 b associated with ‒3 & +5 or 1 − b associated with 4 ± A1 ( ) ( ) 2 2 2 0 2 0, 2, 1 4 − = + = = ± −= ± x or x x b (M1A1) Association can be an equality or an inequality b ⩾ 5, b ⩽ –3 A1 4

This question in 9709/11 Oct/Nov 2018

7 y y = x A 2, 2 y = k x3 −7x2 + 12x O x The diagram shows part of the curve with equation y = k x3 −7x2 + 12x for some constant k. The curve intersects the line y = x at the origin O and at the point A 2, 2 . (i) Find the value of k. [1] … … … … … … (ii) Verify that the curve meets the line y = x again when x = 5. [2] … … … … … … … … … … (iii) Find, showing all necessary working, the area of the shaded region. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) 2 = k(8 ‒ 28 + 24) → k = 1/2 B1 1 7(ii) When x = 5, y = [½](125 ‒ 175 + 60) = 5 M1 Or solve [ ]( ) [ ] 3 2 ½ 7 12 5 0, 2 − + = ⇒ = = x x x x x x Which lies on y = x, oe A1 2 7(iii) 3 2 1 [ ( 7 12 ) ] 2 x x x x dx − + − ∫ . M1 Expect 3 2 1 7 5 2 2 ∫ − + x x x 4 3 2 1 7 5 8 6 2 − + x x x B2,1,0FT Ft on their k 2 ‒ 28/3 +10 DM1 Apply limits 0 → 2 8/3 A1 OR 4 3 2 1 7 3 8 6 − + x x x B2,1,0FT Integrate to find area under curve, Ft on their k 2 ‒ 28/3 +12 M1 Apply limits 0 → 2. Dep on integration attempted Area ∆ = ½ × 2 × 2 or 2 2 0 d ½ 2   = =   ∫x x x M1 8/3 A1 5

This question in 9709/11 Oct/Nov 2018

12 10 The equation of a curve is y 2x and the equation of a line is y x k, where k is a constant. x = + + = (i) Find the set of values of k for which the line does not meet the curve. [3] … … … … … … … … … … … … … In the case where k 15, the curve intersects the line at points A and B. = (ii) Find the coordinates of A and B. [3] … … … … … … … … … … … … … … … (iii) Find the equation of the perpendicular bisector of the line joining A and B. [3] … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(i) 12 2x k x x + = − or ( ) 12 2 y k y k y = − + − → 3 term quadratic. Expect 3x² − kx + 12 or 3y2 – 5ky + (2k2 + 12) (= 0) Use of b² − 4ac → k² − 144 < 0 DM1 Using the discriminant, allow ⩽ , = 0; expect 12 and −12 − 12 < k < 12 A1 Do NOT accept ⩽ . Separate statements OK. 3 10(ii) Using k = 15 in their 3 term quadratic M1 From (i) or restart. Expect 3x² − 15x + 12 or 3y2 – 75y + 462 (= 0) x = 1,4 or y = 11, 14 A1 Either pair of x or y values correct.. (1, 14) and (4, 11) A1 Both pairs of coordinates 3 10(iii) Gradient of AB = −1 → Perpendicular gradient = +1 B1FT Use of m1m2=−1 to give +1 or ft from their A and B. Finding their midpoint using their (1, 14) and (4, 11) M1 Expect (2½, 12½) Equation: y – 12½ = (x – 2½) [y = x + 10] A1 Accept correct unsimplified and isw 3

This question in 9709/12 Oct/Nov 2018

9 A curve has equation y = 2x2 −3x + 1 and a line has equation y = kx + k2, where k is a constant. (i) Show that, for all values of k, the curve and the line meet. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) State the value of k for which the line is a tangent to the curve and find the coordinates of the point where the line touches the curve. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(i) For their 3-term quad a recognisable application of 2 4 b ac − M1 Expect 2 2 2 3 1 0 x x k k − + + − = oe for the 3-term quad. ( ) ( ) ( )( ) 2 2 2 4 3 4 2 1 b ac k k − = + − − oe A1 Must be correct. Ignore any RHS 2 9 6 1 k k + + A1 Ignore any RHS ( ) 2 3 1 0 k + . Do not allow > 0. Hence curve and line meet. AG A1 Allow 2 1 (9) 0 3 k   +     . . Conclusion required. ALT Attempt solution of 3-term quadratic M1 Solutions ( ) 1, ½ 1 x k k = + − A1A1 Which exist for all values of k. Hence curve and line meet. AG A1 4 Question Answer Marks Guidance 9(ii) 1/ 3 k = − B1 ALT d / d 4 3 4 3 y x x x k = − ⇒ − = Sub (one of) their 13 k = − into either line 1 → ( ) 2 8 8 2 0 3 9 x x − + = Or into the derivative of line 1 → 4x ( )( ) 3 0 k − + = M1 Sub 4 3 k x = − into line 1 → ( ) ( ) ( ) 2 2 2 4 1 4 3 0 x x x x − + − − = x = 2/3 Do not allow unsubstantiated 2 1 , 3 9   −     following 1 3 k = − A1 x = 2/3, y = ‒1/9 (both required) [from 2 18 24 8 x x − + − (=0) oe] y = ‒1/9 Do not allow unsubstantiated 2 1 , 3 9   −     following 1 3 k = − A1 1/ 3 k = − 4

This question in 9709/13 Oct/Nov 2018

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

6 A line has equation y = 3kx −2k and a curve has equation y = x2 −kx + 2, where k is a constant. (i) Find the set of values of k for which the line and curve meet at two distinct points. [4] … … … … … … … … … … … … … … … … … … … … … … … … (ii) For each of two particular values of k, the line is a tangent to the curve. Show that these two tangents meet on the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(i) 2 2 3 2 2 4 2 2 0 kx k x kx x kx k − = − + → − + + = B1 kx terms combined correctly-implied by correct 2 4 b ac − Attempt to find 2 4 b ac − M1 Form a quadratic equation in k 1 and 1 2 − A1 SOI 1 2 1, k k > < − A1 Allow 1, 1/ 2 x x > < − 4 6(ii) 3 3 2, 1 2 y x y x = − = − + M1 Use of their k values (twice) in 3 2 y kx k = − 3 3 2 1 2 x x −= − + OR 2 2 2 y y + = − M1 Equate their tangent equations OR substitute y = 0 into both lines x = 2 3 , → y = 0 in one or both lines A1 Substitute x = 2 3 in one or both lines 3

This question in 9709/13 Oct/Nov 2019

5 The equation of a line is y = mx + c, where m and c are constants, and the equation of a curve is xy = 16. (a) Given that the line is a tangent to the curve, express m in terms of c. [3] … … … … … … … … … … … … (b) Given instead that m = −4, find the set of values of c for which the line intersects the curve at two distinct points. [3] … … … … … … … … … …

6 marks

Mark scheme: 5(a) ( ) + x mx c = 16 → 2 16 0 + − = mx cx B1 Use of b² − 4ac = c² + 64m M1 Sets to 0 → m = ² 64 −c A1 3 5(b) ( ) 4 − + x x c = 16 Use of b² − 4ac → c² − 256 M1 c > 16 and c < −16 A1 A1 3

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

1 Find the set of values of m for which the line with equation y = mx −3 and the curve with equation y = 2x2 + 5 do not meet. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 ( ) 2 2 2 5 3 2 8 0 + = − → − + = x mx x mx B1 Form 3-term quadratic 2 64 − m M1 Find 2 4 − b ac . 8 8 −< < m A1 Accept (-8, 8) and equality included 3

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

1 (a) Express x2 6x 5 in the form x a 2 b, where a and b are constants. [2] + + + + … … … … … … … … … … … … (b) The curve with equation y x2 is transformed to the curve with equation y x2 6x 5. = = + + Describe fully the transformation(s) involved. [2] … … … … … … … … … … …

4 marks

Mark scheme: 1(a) ( ) [ ] 2 3 4   + −   x B1 B1 2 1(b) [Translation or shift] 3 4 −     −   B1 B1 FT Accept [translation/shift] −       their a theirb OR translation ‒3 units in x-direction and (translation) ‒4 units in y-direction. 2

This question in 9709/13 Oct/Nov 2020

4 A curve has equation y 3x2 4 and a straight line has equation y mx m where m is a constant. = −4x + = + −1, Find the set of values of m for which the curve and the line have two distinct points of intersection. [5] … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 ( ) ( ) ( ) 2 2 3 4 4 1 3 4 5 0 x x mx m x m x m − + = + − → − + + − = M1 3-term quadratic ( ) ( ) 2 2 4 4 4 3 5 − = + −× × − b ac m m M1 Find 2 4 − b ac for their quadratic 2 20 44 + − m m A1 (m + 22)(m ‒ 2) A1 Or use of formula or completing square. This step must be seen 2 , 22 > < − m m A1 Allow 2 , 22 > < − x x 5

This question in 9709/13 Oct/Nov 2020

4 A line has equation y = 3x + k and a curve has equation y = x2 + kx + 6, where k is a constant. Find the set of values of k for which the line and curve have two distinct points of intersection. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 2 6 3 x kx x k + + = + leading to ( ) ( ) [ ] 2 3 6 0 x x k k + − + − = ( ) ( )[ ] 2 3 4 6 0 k k − − − > M1 OE. Apply 2 4 − b ac . [ ] 2 2 15 0 k k − − > A1 Form 3-term quadratic. ( )( )[ ] 3 5 0 k k + − > A1 Or 3, 5 = − k from use of formula or completing square. 3, 5 k k < − > A1 FT Or any correct alternative notation, do not allow , . FT for their outside regions. 5

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

1 (a) Express 16x2 −24x + 10 in the form 4x + a 2 + b. [2] … … … … … … … … … … … … (b) It is given that the equation 16x2 −24x + 10 = k, where k is a constant, has exactly one root. Find the value of this root. [2] … … … … … … … … … … …

4 marks

Mark scheme: 1(a) recovery. + 1 or b = 1 B1 2 1(b) [For one root] k = 1 or ‘their b’ B1 FT Either by inspection or solving or from 242 – 4 × 16 × (10 – k) = 0 WWW [ ] 3 Root or or 0.75 4 = x B1 SC B2 for correct final answer WWW. 2

This question in 9709/12 May/June 2021

8 The first, second and third terms of an arithmetic progression are a, 32a and b respectively, where a and b are positive constants. The first, second and third terms of a geometric progression are a, 18 and b + 3 respectively. (a) Find the values of a and b. [5] … … … … … … … … … … … … … … … (b) Find the sum of the first 20 terms of the arithmetic progression. [3] … … … … … … …

8 marks

Mark scheme: 8(a) 3 2 2 2   + = ×  =     a b a b a 182 = a(b + 3) OE or 2 correct statements about r from the GP, e.g. 18 = r a and b + 3 = 18r or 2 3 + = b r a B1 SOI 324 = a(2a + 3) ⇒ 2a2 + 3a – 324[= 0] or b2 + 3b – 648[= 0] or 6r2 – r – 12[= 0] or 4d2 + 3d – 162[= 0] M1 Using the correct connection between AP and GP to form a 3-term quadratic with all terms on one side. (a – 12)(2a + 27)[= 0] or ( )( )[ ] 24 27 0 − + = b b or ( )( )[ ] 2 3 3 4 0 − + = r r or ( )( )[ ] 6 4 27 0 − + = d d M1 Solving their 3-term quadratic by factorisation, formula or completing the square to obtain answers for a, b, r or d. a = 12, b = 24 A1 WWW. Condone extra ‘solution’ 13.5, 27 =− = − a b only. 5 Question Answer Marks Guidance 8(b) Common difference d = 6 B1 FT SOI. FT their 2 a ( ) 20 20 S 2 12 19 6 2 = × + × M1 Using correct sum formula with their a, their calculated d and 20. 1380 A1 3

This question in 9709/12 May/June 2021

3 A line with equation y = mx −6 is a tangent to the curve with equation y = x2 −4x + 3. Find the possible values of the constant m, and the corresponding coordinates of the points at which the line touches the curve. [6] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 ( ) 2 2 4 3 6 leading to 4 9 x x mx x x m − + = − − + + May be implied on the next line. ( ) 2 2 4 leading to 4 4 9 b ac m − + − × DM1 SOI. Use of the discriminant with their a, b and c ( )( ) 4 6 or 2 10 0 leading to 2 or 10 m m m m + = ± − + = = − A1 Must come from 2 4 0 − = b ac SOI Substitute both their m values into their equation in line 1 DM1 m = 2 leading to x =3 ; m = ‒10 leading to x = ‒3 A1 (3, 0), (‒3, 24) A1 Accept 'when x = 3, y = 0; when x = ‒3, y = 24' If final A0A0 scored, SC B1 for one point correct WWW Alternative method for Question 3 2 4 2 4 = − → − = dy x x m dx *M1 ( ) 2 4 3 2 4 6 − + = − − x x x x DM1 2 2 2 4 3 2 4 6 9 3 − + = − −→= → = ± x x x x x x A1 0, = y 24 or (3, 0), (‒3, 24) A1 Substitute both their x values into their equation in line 1 DM1 Or substitute both their ( ) , x y into 6 = − y mx When x = 3, m = 2; when x = ‒3, m = ‒10 A1 If A0, DM1, A0 scored, SC B1 for one point correct WWW 6

This question in 9709/13 May/June 2021

2 A curve has equation y kx2 2x and a line has equation y kx where k is a constant. = + −k = −2, Find the set of values of k for which the curve and line do not intersect. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 ( ) [ ] 2 2 2 2 leading to 2 2 0 kx x k kx kx k x k + − = − + −+ − + = ( ) ( ) 2 2 4 2 −+ − −+ k k k DM1 Apply 2 4 − b ac ; allow 1 error but a, b and c must be correct for their quadratic. ( )( ) 2 5 12 4 or 2 2 4 − + −+ −+ − k k k k k A1 May be shown in quadratic formula. ( )( ) 2 5 2 −+ − + k k DM1 Solving a 3-term quadratic in k (all terms on one side) by factorising, use of formula or completing the square. Factors must expand to give their coefficient of k2. 2 2 5 < < k A1 WWW, accept two separate correct inequalities. If M0 for solving quadratic, SC B1 can be awarded for correct final answer. 5

This question in 9709/11 Oct/Nov 2021

4 The first term of an arithmetic progression is a and the common difference is The first term of a geometric progression is 5a and the common ratio is The sum to infinity−4.of the geometric 4. progression is equal to the sum of the first eight terms of the−1arithmetic progression. (a) Find the value of a. [4] … … … … … … … … … … … … … The kth term of the arithmetic progression is zero. (b) Find the value of k. [2] … … … … … … … …

6 marks

Mark scheme: 4(a) ( ) 1 4 5 1 a −± B1 Use of correct formula for sum to infinity. ( ) 8 2 7 4 2 a  + −    *M1 Use of correct formula for sum of 8 terms and form equation; allow 1 error. 4 8 112 = − a a leading to [ ] 28 = a DM1 Solve equation to reach a value of a. 28 = a A1 Correct value. 4 4(b) ( )( ) 28 1 4 0 + − − = their k M1 Use of correct method with their a. [ ] 8 = k A1 2

This question in 9709/11 Oct/Nov 2021

2 A curve has equation y = x2 + 2cx + 4 and a straight line has equation y = 4x + c, where c is a constant. Find the set of values of c for which the curve and line intersect at two distinct points. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 [ ] 2 2 2 4 4 leading to 2 4 4 0 + + = + + − + − = x cx x c x cx x c *M1 Equate ys and move terms to one side of equation. ( ) ( ) 2 2 4 2 4 4 4 − = − − − b ac c c DM1 Use of discriminant with their correct coefficients. 2 2 4 16 16 16 4 4 12   − + − + = −   c c c c c A1 ( ) ( ) 2 4 0 leading to 4 3 0 − > − > b ac c c M1 Correctly apply ‘> 0’ considering both regions. 0, 3 < > c c A1 Must be in terms of c. SC B1 instead of M1A1 for c ⩽ 0, c ⩾ 3 5

This question in 9709/12 Feb/March 2022

1 (a) Express x2 −8x + 11 in the form x + p 2 + q where p and q are constants. [2] … … … … … … … … … … … … (b) Hence find the exact solutions of the equation x2 −8x + 11 = 1. [2] … … … … … … … … … … … …

4 marks

Mark scheme: 1(a) B1 If p and q-values given after their completed square expression, mark the expression and ISW. … 5  or q = − 5 B1 2 1(b) (x – 4)2 − 5 = 1 so (x – 4)2 = 6 so  4 6  x M1 Using their p and q values or by quadratic formula 4 6   x or 8 24 2  A1 Or exact equivalent. No FT; must have  for this mark. ISW decimals 1.55, 6.45 if exact answers seen. If M0, SC B1 possible for correct answers. 2

This question in 9709/11 May/June 2022

k2 5 3 The coefficient of x4 in the expansion of 2x2 + is a. The coefficient of x2 in the expansion of x 2kx −1 4 is b. (a) Find a and b in terms of the constant k. [3] … … … … … … … … … … … … … … … … … … … … … … … (b) Given that a + b = 216, find the possible values of k. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(a) 4 x term is    2 2 3 2 10 2        k x x M1 For selecting the term in 4 x . 4 4 4 80 80 k x a k   A1 For correct value of a. Allow 4 4 8 0 k x . [ 2 x term is [6 ](2kx)2  1 = 24k2x2 ] 2 2 4  b k B1 For correct value of b. Allow 2 2 2 4 k x . 3 3(b)   4 2 4 2 80 24 216 0 10 3 27 0            k k k k M1 Forming a 3-term equation in k (all terms on one side) with their a and b and no x’s.     2 2 2 3 5 9 0    k k [⇒ 2 3 9 or ] 2 5   k M1 Attempt to solve 3-term quartic (or quadratic in another variable) by factorisation, formula or completing the square – see guidance.  3 2  k A1 OE e.g. 6 2  , 1 .5  , AWRT 1 .22  Omission of  A0. Additional answers A0. If M1 M0, SC B1 can be awarded for correct final answer, max 2/3. 3

This question in 9709/11 May/June 2022

5 The equation of a curve is y = 4x2 −kx + 12k2 and the equation of a line is y = x −a, where k and a are constants. (a) Given that the curve and the line intersect at the points with x-coordinates 0 and 34, find the values of k and a. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that a = −72, find the values of k for which the line is a tangent to the curve. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) 2 2 1 4 0 0 0 2 k a      M1 Equating the equations of curve and line and substituting 0 Condone slight errors e.g. ± sign errors. 2 2 3 3 1 3 4 4 4 2 4 k k a            M1 Equating the equations of curve and line and substituting 3 4 x  . Condone slight errors e.g. ± sign errors. k = 2, 2 a  A1 A1 WWW Alternative method for question 5(a)   3 0 0 4 x x          or   4 3 0 x x   [⟹ 2 4 3 0] x x   *M1 Use 0, 3 4 to form a quadratic equation. Do not allow   3 0 0 4 x x          .     2 2 2 2 1 1 4 leading to 4 1 0 2 2 x kx k x a x k x k a          DM1 Equating the equations of curve and line and rearranging so that terms are all on same side. Condone slight errors e.g. ± sign errors. k = 2, 2 a  A1 A1 WWW Alternative method for question 5(a) 3 0 4 b a    and 3 0 4 c a   *M1 Using sum and product of roots. Condone ± sign errors. 1 3 4 4 k  and 2 1 2 0 4 k a   DM1 Equating the equations of curve and line and equating to 3 4 and 0. k = 2, 2 a  A1 A1 WWW 4 Question Answer Marks Guidance 5(b)   2 2 2 2 1 7 1 7 4 4 0 2 2 2 2 x kx k x x kx x k           *M1 OE Substitute 7 2 a  and rearrange so that terms are all on same side, condone ± sign errors. Watch for multiples.   2 2 1 7 1 4 4 2 2 k k          *DM1 Use of b2 – 4ac with the coefficients from their 3-term quadratic. Both coefficients ‘b’ and ‘c’ must consist of two components. 2 7 2 57 k k    A1 OE    3 7 19 k k   or other valid method DM1 Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. k = 3, k = 19 7  A1 OE e.g. AWRT 2.71. No ISW if inequalities used. SC: If second DM1 not scored, SC B1 available for correct final answers. Alternative method for question 5(b) 8 1 x k   and 2 2 1 7 4 2 2 x kx k x     *M1 Equating gradients and equating line and curve.   2 2 1 7 4 8 1 (8 1) 2 2 x x x x x       or 2 2 1 1 1 1 7 4( ) 8 8 2 8 2 k k k k k              *DM1 Forming an equation in x or k only. 2 28 8 3 x x   or 2 7 2 57 k k   A1 OE A correct 3 term quadratic in x or k only.    14 3 2 1 x x   or    3 7 19 k k   or other valid method DM1 OE Factorising or use of the formula or completing the square. Must be evidence of an attempt to solve for this mark. Dependent upon both previous method marks. Question Answer Marks Guidance 5(b) k = 3, k = 19 7  A1 OE e.g. AWRT 2.71. No ISW if inequalities used. SC: If second DM1 not scored, SC B1 available for correct final answers. 5

This question in 9709/12 May/June 2022

@ A 4 (a) The curve with equation y x2 2x is translated by −1 . = + −5 3 Find the equation of the translated curve, giving your answer in the form y ax2 bx c. [3] = + + … … … … … … … … … … … … … (b) The curve with equation y x2 2x is transformed to a curve with equation y 4x2 4x = + −5 = + −5. Describe fully the single transformation that has been applied. [2] … … … … … … … …

5 marks

Mark scheme: 4(a)     2 1 2 1 5 3      x x , or     2 1 1 6 3   x M1 for dealing with 1 0        and M1 for dealing with 0 3    .   2 4 1    y x x A1 Answer only given full marks. 3 Question Answer Marks Guidance 4(b) {Stretch}{x direction or horizontally or y-axis invariant}{ factor ½} B2, 1, 0 Additional transformation B0. 2

This question in 9709/13 May/June 2022

2 5 (a) Solve the equation 6 y 0. [4] + y −7 = … … … … … … … … … … … … … … 2 (b) Hence solve the equation 6 tan x 0 for [3] + tan x −7 = 0Å ≤x ≤360Å. … … … … … … … … …

7 marks

Mark scheme: 5(a) 1/2 6 2 7   y y [= 0] 1 1 2 2 2 1 3 2               y y [= 0] or e.g.    2 1 3 2   u u [= 0] DM1 Or use of formula or completing the square. 1/2 1 2 [ ] , 2 3 y  A1 Answers only SC B1 if DM1 not scored.  1 4 , 4 9 y  A1 Answers only SC B1 if DM1 not scored. 4 5(b) Use of tan x = their y values M1 Must have at least 2 values of y from part (a).  x 14[.0], 24[.0],  x 194[.0], 204[.0] A1 A1 FT FT for 180 + angle (twice). AWRT 3

This question in 9709/13 May/June 2022

2 1 Solve the equation 3x + 2 = x −1. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 ( 3x + 2 )( x − 1) = 2  3x 2 −−x 4  = 0 M1 OE Multiply by denominator and obtain a quadratic. ( 3 x − 4 )( x + 1) = 0  M1 Solve by factorising, formula or completing the square. 4 A1 Allow 1.33 [x =] − 1, If M1 M0, SC B1 possible for two correct answers. 3 3

This question in 9709/11 Oct/Nov 2022

3 (a) Find the set of values of k for which the equation 8x2 + kx + 2 = 0 has no real roots. [2] … … … … … … … … … … … … (b) Solve the equation 8 cos21 −10 cos 1 + 2 = 0 for 0Å ≤1 ≤180Å. [3] … … … … … … … … … … … …

5 marks

Mark scheme: 3(a) k 2 − 4 8 2 [  0] M1 Use of b2 – 4 a c but not just in the quadratic formula. −8 < k < 8 or −8 < k , k < 8 or k  8 or (–8, 8) A1 Condone ‘ − 8 < k or k < 8’, ‘ − 8 < k and k < 8’ but not 64 . 2 3(b) 2 ( 4cos− 1)( cos− 1) or ( 4cos− 1)( cos− 1) M1 OE Or use of formula or completing the square. Allow use of replacement variable. 2 A1 OE For both answers. cos= , cos= 1 2 8 SC: If M0, SC B1 available for sight of cos= and 1 8 [θ =] 0°, 75.5° A1 AWRT ISW rejection of 0°. For both answers and no others in the range 0  180, must be in degrees. SC: If M0 B1 scored, SC B1 available for correct answers. 2 SC: If M1 A0 scored, SC B1 available for cos= and θ = 8 75.5° only, WWW. 3

This question in 9709/12 Oct/Nov 2022

6 The equation of a curve is y = 4x2 + 20x + 6. (a) Express the equation in the form y = a x + b 2 + c, where a, b and c are constants. [3] … … … … … … … … … … … (b) Hence solve the equation 4x2 + 20x + 6 = 45. [3] … … … … … … … … … … … … (c) Sketch the graph of y = 4x2 + 20x + 6 showing the coordinates of the stationary point. You are not required to indicate where the curve crosses the x- and y-axes. [3]

9 marks

Mark scheme: 6(a) 2 There is no requirement for the candidate to list a , b and c.  5  y = 4  x +  − 19 Look at values in their final expression, condone omission of 2 ,  2  and award marks as follows: B1 a = 4 B1 5 b = OE 2 B1 c = −19 3 6(b)   5  2    5  2  *M1 Equate their quadratic completed square form from 6(a) to 45 or  Their 4  x +  − 19  = 45    x +  = 16  re-start and use completing the square.    2      2   Solve as far as x = DM1 Any valid method leading to two answers. 13 A1 SC: If M0 or M1 DM0 awarded, B1 available for correct final  x = 3 , − answers. 2 2 3 6(c) Quadratic curve that is the right way up (must be seen either side B1 No axes required, ignore any axes even if incorrect. of stationary point) Stationary point stated using any valid method or correctly B1 FT FT their values from 6(a) as long as their expression is of the labelled on their diagram. 2  5  Expect  − , −19  . B1 FT form p ( qx + r ) + s .  2  Condone if stated correctly but plotted incorrectly. 3

This question in 9709/12 Oct/Nov 2022

1 A line has equation y = 3x −2k and a curve has equation y = x2 −kx + 2, where k is a constant. Show that the line and the curve meet for all values of k. [4] … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 x 2 − kx + 2 = 3 x − 2k leading to x 2 − x ( k + 3) + ( 2 + 2k )  = 0 M1 3-term quadratic, may be implied in the discriminant. 2 2 DM1 Cannot just be seen in the quadratic formula. b − 4ac = ( k + 3) − 8 (1 + k ) (ignore ‘= 0’ at this stage) = ( k − 1) 2 accept ( k − 1)( k − 1) A1 Or use of calculus to show minimum2 of zero at k = 1 or sketch of f ( k ) = k − 2k + 1. 0 Hence will meet for all values of k A1 Clear conclusion. 4

This question in 9709/12 Feb/March 2023

p2 6 The first three terms of an arithmetic progression are , 2p −6 and p. 6 (a) Given that the common difference of the progression is not zero, find the value of p. [3] … … … … … … … … … … … … … … … (b) Using this value, find the sum to infinity of the geometric progression with first two terms p2 and 2p −6. [2] 6 … … … … … …

5 marks

Mark scheme: 6(a)     2 2 2 2 6 3 12 0 6 6        p p p p p OR       2 2 2 6 2 6 3 12 0 6 6          p p p p p p OR   2 1 0 6   d d quadratic in p (all terms on one side) or 2-term quadratic in . d OE e.g.   2 18 72 0    p p ,   2 1 9 36 0 2    p p .   2 18 72 0    p p ⇒     6 12 0    p p or     2 18 18 4 1 72 2    OR   1 1 0 6 6           d d d DM1 Solve a 3-term quadratic in p by factorisation, formula or completing the square or solve a 2-term quadratic in d by factorisation. p = 12 only A1 Since p = 6 gives d = 0. If *M1 DM0 then p = 12 only, award SC B1, max 2/3 marks. A0 XP if error in either factor and 12  p only. 12  p only by trial and improvement 3/3. 3 Question Answer Marks Guidance 6(b) For GP r = 2 2 6 6              p p = 18 3 24 4        B1 OE SOI. Sum to infinity = 24 3 1 4  = 96 B1 FT FT their value of p if used correctly to find r (B0 if ‘ p ’ used) provided 1  r . e.g. 18  p ⇒  54 121.5 5 1 9    S . 2

This question in 9709/11 May/June 2023

3 (a) Express 4x2 −24x + p in the form a x + b 2 + c, where a and b are integers and c is to be given in terms of the constant p. [2] … … … … … … … … … … … … (b) Hence or otherwise find the set of values of p for which the equation 4x2 −24x + p = 0 has no real roots. [1] … … … … … … … … … … …

3 marks

Mark scheme: 3(a) 2 4 3 seen or 4 and 3 x a b    B1 OE Award marks for the correct expression or their values a, b and c. Condone 4 ( 3) 36    x p = 0 and 4 9 4        p . –36 + p or p – 36 seen or c = p – 36 B1 2 3(b) 36 0   p leading to 36  p or 2 24 4 4 0 36    p p or 36 p B1 Allow (36, )  or 36 p  . Consider final answer only. 1

This question in 9709/12 May/June 2023

4 Solve the equation 8x6 + 215x3 −27 = 0. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 4 6 3 [8 215 27 0] x x    3 3 8 1 27 0 x x    OR 2 215 215 4.8. 27 215 47089 or 2.8 2.8       If a substitution is used then the correct coefficients must be retained. Condone substitution of 3  x x . 1 , 27 8  A1 Both correct values seen. SC: if M0 scored SC B1 is available for sight of 1 8 and 27  OE 1 or 0.5 , 3 2  A1 SC: if M0SCB1 scored then SCB1 is available for the correct answers and no others. Do not ISW if answers given as a range. 3

This question in 9709/12 May/June 2023

2 A line has equation y = 2cx + 3 and a curve has equation y = cx2 + 3x −c, where c is a constant. Showing all necessary working, determine which of the following statements is correct. A The line and curve intersect only for a particular set of values of c. B The line and curve intersect for all values of c. C The line and curve do not intersect for any values of c. [4] … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 2 M1 Forming a 3-term quadratic, all terms on one side. cx + 3 x − c = 2cx + 3 leading to cx + ( 3 − 2c ) x − ( c + 3)  = 0 2 2 M1 2nd M1 for b 2 − 4ac correct for their a, b, c b − 4ac = ( 3 − 2c ) + 4c ( c + 3) i.e. no sign errors. = 8c 2 + 9 A1 > 0 [for all values of c] leading to B [Intersects for all values of c] A1 WWW 4

This question in 9709/11 Oct/Nov 2023

6 The equation of a curve is y = x2 −8x + 5. (a) Find the coordinates of the minimum point of the curve. [2] … … … … … … … … … … … @ A 4 The curve is stretched by a factor of 2 parallel to the y-axis and then translated by . 1 (b) Find the coordinates of the minimum point of the transformed curve. [2] … … … … … … … … … … (c) Find the equation of the transformed curve. Give the answer in the form y = ax2 + bx + c, where a, b and c are integers to be found. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: x − 8 x + 5 = 0  2 x − 8 = 0  Correct differentiation of x2 and equating their to 0.6(a) d ( 2 M1 d y ) dx d x Alternative method 1 for first mark of Question 6(a) 2 M1 2 y = ( x − 4 ) − 11 Attempt to complete the square as far as y = ( x − 4 )  k . Alternative method 2 for first mark of Question 6(a) −b 8 M1 x = = 2 a 2 x = 4, y = -11 A1 8  64 − 20 Answers from x = leading to x = 4  11 2 scores M0A0 2 6(b) x = ( their x value from a ) + 4 [=8] B1 FT Can be from finding the equation of the transformed curve, d y differentiating and putting = 0. d x y = ( their y value from a )  2 +1 [ −21] B1 FT Can be from putting x = 8 in the equation of the transformed curve. 2 If B0B0 scored, SC B1 for sight of ( 4, − 22 ) . 6(c) 2 2 B1 Can be implied if both transformations done together: 2 x − 8 x + 5 or 2{( x − 4 ) − 11} ( ) 2 ( x − 4 ) 2 − 8 ( x − 4 ) + 5 + 1 OE. ( ) M1 For the x translation, each x becomes ( x − 4 ) . + 1 or + 1 ( x − 4 ) 2 − 8 ( x − 4 ) + 5 ( x − 4 − 4 ) 2 − their11 ( )   M1 For the y translation of +1. y = 2 x 2 − 32 x + 107 or a = 2, b = −32, c = 107 A1 Evidence to support their answer may be in (b) but answer must be seen in (c). 4

This question in 9709/12 Oct/Nov 2023

7 (a) Verify the identity 2x −1 4x2 + 2x −1  8x3 −4x + 1. [1] … … … … … tan21 + 1 1 (b) Prove the identity  [3] tan21 −1 1 −2 cos21. … … … … … … … … … … … … … … … … … … (c) Using the results of (a) and (b), solve the equation tan21 + 1 = 4 cos 1, tan21 −1 for 0Å ≤1 ≤180Å. [5] … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 4 x 2 + 2 x − 1 = 8 x 3 + 4 x 2 − 2 x − 4 x 2 − 2 x + 1 = 8 x 3 − 4 x + 1 B1 AG7(a) ( 2 x − 1)( ) Six correct terms leading to the correct answer. 1 7(b) sin 2  *M1 sin in the numerator and denominator. For use of tan= + 1  sin 2 + cos 2 cos cos 2  Starting with the LHS 2  = 2 2  sin   sin − cos  − 1 cos 2  1 DM1 For use of sin 2 + cos 2 = 1 twice, in a correct expression, = need to see clear evidence of this step 1 − cos 2 − cos 2  resulting in an expression in cos 2 . 1 A1 AG = 1 − 2cos 2  Alternative method 1 for Question 7(b) sin 2 + cos 2  sin 2 + cos 2 *M1 For use of sin 2 + cos 2 = 1 twice. Starting with the RHS 2 2 2  = 2 2  sin + cos − 2cos   sin − cos  sin 2  DM1 Dividing throughout by cos 2 . + 1 cos 2  = need to see clear evidence of this step sin 2  − 1 cos 2  tan 2 + 1 A1 AG = tan 2 − 1 7(b) Alternative method 2 for Question 7(b) sec 2  *M1 For use of 1+ tan 2 = sec 2  twice. Starting with the LHS sec 2 − 2 1 DM1 AG For multiplying throughout by cos 2  to give the RHS. Clear statement  1 − 2cos 2  A1 3 7(c) 1 2 B1 Replace LHS with RHS from (b) and clear fractions. 1 − 2cos  = 4cos leading to 1 = 4cos 2 ( ) 1 − 2cos  8cos 3 − 4cos+ 1 = 0    4cos + 2cos− 1 = 0  ( 2cos− 1)( 2 *B1 Use of the expression from (a) with x = cos. ) −2 4 + 16 DB1 OE  x or cos= 1 and OR 0.31, −0.81 AWRT For all three values. 2 8 = 60,72,144 B2,1,0 B2 for three correct answers only, B1 for two correct answers and no others (but allow 36 instead of 144) in the given range or 3 correct answers plus other values in the given range. Ignore answers outside of the given range. Accept AWRT 72.0, 144.0 . SC B1 for all 3 correct answers in radians and no others: π 2π 4π , and . 3 5 5 5

This question in 9709/12 Oct/Nov 2023

5 The first, second and third terms of a geometric progression are 2p + 6, 5p and 8p + 2 respectively. (a) Find the possible values of the constant p. [3] … … … … … … … … … … … (b) One of the values of p found in (a) is a negative fraction. Use this value of p to find the sum to infinity of this progression. [4] … … … … … … … … … … …

7 marks

Mark scheme: 5(a) 5 p 8 p + 2 M1 OE. Setting up a valid relationship in terms of p. = 2 p + 6 5 p 9 p 2 − 52 p − 12 [ = 0] DM1 OE. Simplifying to a 3 term quadratic equation, only condone sign errors.  leading to p = −92 and 6 A1 ( 9 p + 2 )( p − 6 ) = 0  3 5(b)  2   50  *M1 2 a = 2  −  + 6  =  FT their − , allow any negative non-integer.  9   9  9 10 50  1  *M1 2 r = −  =− Ft their − , allow any negative non-integer.   9 9  5  9 50  1  125 DM1 A1 Can only get DM1 if |r| < 1. S =   1 −−  = Accept AWRT 4.63 . 9  5  27 4

This question in 9709/13 Oct/Nov 2023

6 It is given that the coefficient of x3 in the expansion of ( 2 + ax) 4 ( 5 - ax) is 432. Find the value of the constant a. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 6 4 2 2 4 1 3 B1 B1 OE 2( ax )  2 ( ax ) ,  Expect 24 a 2 x 2 ,8a 3 x 3 (may be seen in an expansion). 2 3 Multiply terms involving x 2 and 3x by 5 − ax to obtain 3x term *M1 Must find two products only (may be seen in an expansion). DM1 Equate coefficient of 3x to 432 and solve for a Ignore inclusion of 3x at this stage. Obtain a = 3 only A1 5

This question in 9709/12 Feb/March 2024

1 (a) Express 3y 2 - 12y - 15 in the form 3 ( y + a) 2 + b , where a and b are constants. [2] … … … … … … … … … … … (b) Hence find the exact solutions of the equation 3x 4 - 12x 2 - 15 = 0 . [3] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a) 2 3 2 27   y or 2, 27 a b   B1 B1 2 1(b)   2 2 2 9 x   leading to 2 2 3 x   M1 Must be 2 x unless substitution is clear. 2 2 1 or 5 x x   M1 Allow omission of -1 if ±3 seen. 5  x A1 B1 SC if M1M1 not awarded. Ignore ± i, i, –i, √–1. Use of calculator with no working scores 0/3. Alternative method for Question 1(b) 3 x 4 – 12 x 2 – 15 = 0 leading to    2 2 3 5 1 [   x x = 0] (M1) 2 2 1 or 5 x x   (M1) Allow omission of –1 if factors seen. Factorising or other valid method. 5  x (A1) B1 SC if M1M1 not scored. Ignore ± i, i, –i, √–1. Use of calculator with no working scores 0/3. 3

This question in 9709/11 May/June 2024

11 The function f is defined by f ( x) = 10 + 6x - x 2 for x d R . (a) By completing the square, find the range of f. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … The function g is defined by g( )x = 4 x + k for x d R where k is a constant. (b) It is given that the graph of y = g -1 f ( x) meets the graph of y = g ( x) at a single point P. Determine the coordinates of P. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11(a) Express f ( )x as: 2 ( 3)   a x or 2 3   a x where 19 or 1 a   If the form   2 6 10     f x x x is used the form must be returned to  f x Completed square form must give 2 x . Answers must come from completion of the square (not calculus or graphs). 2 19 (3 )  x or 2 19 ( 3)   x A1 OE  19 f x  or 19 y  with ⩽, not < or –∞ < f(x) 19  or –∞ ⩽ f(x) 19  or (–∞, 19] or [–∞, 19] A1 FT Using their constant following the award of M1. SC B1 answer only or answer from a method not involving completion of the square. 3 Question Answer Marks Guidance 11(b) 1 1 4 g ( ) ( )    x x k B1    1 2 1 10 6 4 4        g f x x x k x k M1 OE May use their completed square form for f(x). Simplify the quadratic equation obtained from 1 g f ( ) g( )   x x provided k is present and apply 2 4 0   b ac to this quadratic equation *M1 Expect 2 10 10 5 0. x x k     Obtain   100 4 5 10 0    k and hence 7  k A1 Use their k to form and solve a quadratic in x DM1 Allow if their quadratic has two solutions.   5, 13   only A1 SC B1 if no method seen. Alternative Method for first 4 marks State f ( ) gg( )  x x (B1) gg( ) 16 5   x x k (M1) Apply 2 4 0   b ac to quadratic equation obtained from f ( ) gg( )  x x (*M1) Provided k is present. 100 4(5 10) 0    k and hence 7  k (A1) 6

This question in 9709/13 May/June 2024

11 The function f is defined by f ( )x = 3 + 6 x - 2x 2 for x ! R . (a) Express f ( )x in the form a - b ( x - c) 2 , where a, b and c are constants, and state the range of f. [3] … … … … … … … … … … … … … … (b) The graph of y = f ( x) is transformed to the graph of y = h ( x) by a reflection in one of the axes followed by a translation. It is given that the graph of y = h ( x) has a minimum point at the origin. Give details of the reflection and translation involved. [2] … … … … … … … … … The function g is defined by g ( )x = 3 + 6x - 2x 2 for x G 0 . (c) Sketch the graph of y = g ( x) and explain why g is a one-one function. You are not required to find the coordinates of any intersections with the axes. [2] … … … (d) Sketch the graph of y = g -1 ( x) on your diagram in (c), and find an expression for g -1 ( )x . You should label the two graphs in your diagram appropriately and show any relevant mirror line. [4] … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) 3 B1 Obtain b = 2 and c = 2 2 B1 15  3  Obtain − 2 x −   2  2  15 15 B1 FT Following their value of a. State range is y or f ( x ) with ⩽ given or clearly implied (not <) 2 2 3 11(b) State that reflection is in x-axis B1 Accept transformations in any order.  3  B1 FT Following their values of a and c in part (a). −   Accept transformations in any order. 2 State or imply that translation is by   or equivalent  15     2  2 11(c) Sketch the correct graph appearing in second and third quadrants only B1 State that each y-value is associated with a single x-value or equivalent B1 Accept passes the horizontal line test. Ignore passes the vertical line test. 2 11(d) Sketch the correct graph with suitable labelling to distinguish the two curves B1 Appearing in third and fourth quadrants only. Draw the line y = x B1 See above; no need to label the line. Attempt correct process for finding the inverse function M1 Allowing use of  and y so far. 3 15 1 A1 Must involve x at the conclusion. Obtain − − x or equivalent 2 4 2 4

This question in 9709/11 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

9 The equation of a curve is y = 1 k 2 x 2 - 2 kx + 2 and the equation of a line is y = kx + p , where k and p 2 are constants with 0 1 k 1 1. (a) It is given that one of the points of intersection of the curve and the line has coordinates b 5 , 1 l. 2 2 Find the values of k and p, and find the coordinates of the other point of intersection. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) It is given instead that the line and the curve do not intersect. Find the set of possible values of p. [3] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a)  1 2 25 5 1  M1*  5 1   k  − 2 k  + 2 =  Using  ,  in the curve equation or equating the line and 2 4 2 2  2 2     OR  5 1 5 the curve and then using x = and p = − k .    1 2 25 5 5  1 5   2 2 2 k  − 2 k  + 2 = k  + − k   Simplify to get a three-term quadratic in k. Condone errors in   2 4 2 2  2 2   simplification. 25k 2 − 40k + 12  = 0  2 A1 OE k = 5 Condone inclusion of k = 6. 5 1  2  5  DM1*  5 1  =  their   + p  p = Using  ,  and their k in an equation in p. 2  5  2   2 2  Either the line (as shown) or 4 p 2 + 12 p + 5 = 0 are the most likely and solving for p. 1 A1 OE p = − 2 Condone inclusion of p = − 5. 2 2 2 6 5 2 DM1 Equating the line and curve using their k and p and simplify to x − x +  = 0   4 x − 60 x + 125  = 0   get a three-term quadratic [= 0]. 25 5 2  25 9  A1 A1 OE  ,  25 9  2 2  Accept x = , y = . 2 2 9(a) Alternative Method for Question 9(a)  1 2 25 5 5  M1* OE k  − 2 k  + 2 = k  + p   5  1   2 4 2 2 ,  Using   in the curve equation or equating the line and 2  2 2  4 p + 12 p + 5  = 0 5 1 2 the curve and then using x = and k = − p. 2 5 5 Simplify to get a three-term quadratic in p [= 0]. 1 A1 p = − OE Condone inclusion of p = − 5. 2 2 1  5   1  DM1*  5 1  =  k  +  their −   k = Using  ,  and their p in the line equation and solving for 2  2   2   2 2  k. 2 A1 OE k = 5 Condone inclusion of k = 6. 5 2 2 6 5 2 DM1 Equating the line and curve using their k and p and simplify to x − x +  = 0   4 x − 60 x + 125  = 0    get a three-term quadratic [= 0]. 25 5 2  25 9  A1 A1 OE  ,  25 9  2 2  Accept x = , y = . 2 2 7 9(b)  1 2 2  1 2 2 M1* Equate the original equations of the curve and the line and k x − 2 kx + 2 = kx + p  k x − 3kx + 2 − p   collect like terms; k and p must still be present.  2  2 2 1 2 DM1 Use of b 2 − 4 ac for their quadratic in x to give an expression in 9 k −4 k ( 2 − p ) 2 k and p. This expression can come from their equation in (a). 5 A1 p − 2 3

This question in 9709/12 Oct/Nov 2024

3 (a) Find the coefficients of x3 and x4 in the expansion of ( 3 - ax) 5 , where a is a constant. Give your answers in terms of a. [3] … … … … … … … … … … … (b) Given that the coefficient of x4 in the expansion of ( ax + 7)( 3 - ax) 5 is 240, find the positive value of a. [3] … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3(a) 3 5 2 3 2 3 4 5 4 4 M1 Allow for either term, allow sign error and combination 10  3  a  5 3 a  x :  3 ( − ax )  −  or x :  3 ( − ax )   notation. 3 4 x 3 : −90 a 3 A1 Allow in the full expansion. x 4 : 15a 4 A1 Allow in the full expansion. 3 3(b) Coefficient of x 4 is a  their − 90a 3 + 7  their 15a 4  = 15a 4  M1 Must select two appropriate terms only.   15 a 4 = 240 DM1 Reducing to a simple quartic equation in a.  a 4 = 16  a = 2 A1 A0 if a = −2 is given as a solution. 3

This question in 9709/13 Oct/Nov 2024

8 (a) Express 3x 2 - 12 x + 14 in the form 3 ( x + a) 2 + b , where a and b are constants to be found. [2] … … … … … … … The function f ( )x = 3x 2 - 12x + 14 is defined for x H k , where k is a constant. (b) Find the least value of k for which the function f -1 exists. [1] … … … For the rest of this question, you should assume that k has the value found in part (b). (c) Find an expression for f -1 ( )x . [3] … … … … … … … … … … … … … … … … … … (d) Hence or otherwise solve the equation ff ( )x = 29 . [3] … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) 2 B1 B1 3 ( x − 2 ) + 2 or a = −2, b = 2 2 8(b) 2 or k = 2 or k  2 B1FT FT on their a. Do not accept x = 2 or x ⩾ 2. 1 8(c) 2 2 y − 2 M1 Using their completed square form. 3 ( x − 2 ) + 14 − 12 = y  ( x − 2 ) = 3 y − 2 DM1 x =  + 2 3 −1 x − 2 A1 3 x − 6 f ( x ) = + 2 OE, e.g. y = + 2. 3 3 3 8(d) Finding f −1 ( 29 ) [= 5] M1 Or solving f(x) = 29 [using their completed square form, OE]. Finding f − 1 ( their 5) M1 Or solving f(x) = their 5. x = 3 A1 If using f(x) method, x = 1 must be discarded. Alternative solution for Question 8(d) 2 2 M1 2 2 2 3 3 x − 12 x + 14 − 12 3 x − 12 x + 14 + 14 = 29. − 2) + 2 = 29 using their completed square form Or 3 3 ( x − 2 ) + 2 ( ) ( ) ( ) Allow if the ' = 29' appears later in the working. 4 2 DM1 OE Solving as far as 9 ( x − 2 ) = 9 or x − 4 x + 3 = 0 x 4 − 8 x 3 + 24 x 2 − 32 x + 15 = 0. Or  27 ( ) x = 3 only A1 WWW Only dependent on the first M1. 3

This question in 9709/13 Oct/Nov 2024

9 y y = x 3 - 3x + 3 y = 2x 3 - 4x 2 + 3 x O The diagram shows the curves with equations y = x 3 - 3x + 3 and y = 2x 3 - 4x 2 + 3 . (a) Find the x-coordinates of the points of intersection of the curves. [3] … … … … … … … … … … … … … … … (b) Find the area of the shaded region. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 9(a) y = x 3 − 3 x + 3 and y = 2 x3 − 4 x 2 + 3  x3 − 4 x 2 + 3x  = 0 M1 Reducing to 3-term cubic or quadratic if x cancelled. (x x − 1)( x − 3 ) = 0  DM1 Factorising the cubic or quadratic. x = 0, 1 and 3 {x = 0 may be seen in the working} A1 SC B1 for x = 1, 3 only, with no M marks awarded. 3 9(b) Attempt at integration of both functions. Can be before or after subtraction M1 3 3 2 Expect integration of x − 3 x + 3 − 2 x − 4 x + 3 dx or ( ( ) ( ) ) of the functions or integrals   ( − x 3 + 4 x 2 − 3 x ) dx. At this stage, subtraction can be done either way.  x 4 4 x 3 3 x 2   x 4 3 2   2 4 4 3   A1 OE = − + −  or   − x + 3 x  −  x − x + 3 x   ± covers A1 being awarded to those who subtract the ‘other’  4 3 2   4 2   4 3   way.  81 108 27   1 4 3   DM1 OE = − + − −− + − ,       63 7   27 13   4 3 2   4 3 2   Minimum required is  −  −  −  , i.e. four fractions.  4 4   2 6  or Correctly apply limits their 1 and 3.  81 27   1 3   81 108   1 4   − + 9 − − +3     −  − + 9  −  − + 3   Do not allow if x = 0 used.  4 2   4 2   2 3   2 3   Need at least one correct substitution in every bracket. If two integrals, need to see substitution into both. Allow one sign error only in each expression, if brackets are not shown. 8 A1 Accept if this comes from use of limits f (1) − f ( 3 ) or = 3 3 2 − 8  ( x − 4 x + 3 x ) dx, if 3 used. Only dependent on the first method mark. Accept AWRT 2.67. 4

This question in 9709/13 Oct/Nov 2024

1 A curve has equation y = 5 + 3x - 2x 2 and a straight line has equation y = kx + 13 , where k is a constant. Find the set of values of k for which the curve and the line do not meet. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 2 2 B1 OE [kx + 13 = 5 + 3x − 2 x ] 2 x + ( k − 3) x + 8  = 0 Eliminate y to obtain a three-term quadratic. Use of b 2 − 4ac  0 or b 2 − 4ac = 0 with their coefficients of their new M1 OE quadratic equation. Condone  errors only. Use of ‘ ’0 scores M0, unless recovered. –5 and 11 A1 Identification of correct critical values, may only be seen in their final answer. −5 k  11 A1 CWO Do not allow ‘or’. A0 if ⩽ sign or signs used. 4

This question in 9709/12 Feb/March 2025

5 (a) Find the first three terms, in ascending powers of x, in the expansion of each of the following expressions. (i) ( 2 - px) 5 [2] … … … … … … 4 (ii) b1 - 1 xl [2] 2 … … … … … … 4 (b) Given that the coefficient of x2 in the expansion of ( 2 - px) 5 b1 - 1 xl is 93, find the possible 2 values of the constant p. [3] … … … … … … … … … … …

7 marks

Mark scheme: 5(a)(i) 32 − 80 px + 80 p 2 x 2 B2,1,0 B2 for all correct. B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(a)(ii) 3 2 B2,1,0 OE 1 − 2 x + x B2 for all correct. 2 B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(b) 3 2 *M1 3 terms FT their values. Coefficient of 2x = 32  + ( −−2 80 p ) + 80 p 2 48 + 160 p + 80 p 2 = 93 ⇒ 5 ( 4 p + 9 )( 4 p − 1) = 0  DM1 Set their 3-term coefficient to 93 and attempt to solve by factorising or other accepted method for solving their 3-term quadratic. 9 1 A1 SC B1 following M0 if method for solving p = − and p = quadratic is not shown. 4 4 3

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

11 (a) Express x 2 + 4 x + 2 in the form ( x + a ) 2 + b , where a and b are integers. [2] … … … … … … … … The functions f and g are defined as follows. f ( )x = x 2 + 4x + 2 for x G-2 g ( )x =- x - 4 for x H-2 (b) (i) Find an expression for f -1 ( )x . [3] … … … … … … … … … … … … … … … (ii) Find an expression for ( gf ) -1 ( )x . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 11(a) 2 B1B1 B1 for each correct { }. {( x + 2 ) }−2 Allow a = 2, b = −2. If contradictory, give preference to the expression. 2 11(b)(i) y = ( x + 2 ) 2 − 2  y + 2 = ( x + 2 ) 2 *M1 Equating y or f −1 ( x ) or f −1 or f ( x ) to their completed square form and first step. x and y may be interchanged at this stage. Condone  errors. x =  y + 2 − 2 DM1 Condone  errors during simplification. [f −1 ( x ) = ] − x + 2 − 2 A1 Do not condone x = or f ( x ) = . 3 x 2 + 4 x + 211(b)(ii)  − 4 *M1 Using fg ( x ) = {( −−x 4 ) + 2}2 − 2 scores 0/4. their completed square form − 4 or − ( )  gf ( x ) =−   x + 2 ) 2 − 2 A1  gf ( x ) =− ( x =  −−y 2 − 2 DM1 Finding x from their completed square form, which must contain a − ( x + k ) 2 term. Condone  errors only during simplification. x = − y + 2 − 2 is DM0 (Square rooting then  or −)1 x and y may be interchanged at this stage. −1 A1 Do not condone x = . [( gf ) ( x ) =] − −−x 2 − 2 Alternative Method for Question 11(b)(ii) ( gf ) −1 = f −1g −1 *M1 SOI Allow with their f − 1 and their g − 1 . Using g −1f −1 ( x ) scores 0/4. g −1 ( x ) = −−x 4 A1 ( gf ) −1 ( x ) = −−−+x 4 2 − 2 DM1 Allow with their f − 1 and their g − 1 . −1 A1 Do not condone x = [( gf ) ( x ) =] − −−x 2 − 2 4

This question in 9709/12 May/June 2025

11 The function f is defined by f ( )x = x 2 + 4ax + a for x ! R , where a is a constant. The function g is such that g -1 ( )x = 3 2 x - 4 for x ! R . (a) Given that the range of f is f ( )x H - 33 , find the possible values of a. [4] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Given instead that fgg ( 0 ) = 96 , find the value of a. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 11(a)  ( x + 2 a ) 2 − 4 a 2 + a B1  f ( x ) =  2 M1 Condone or  . their −4a + a = −33 ( ( ) ) 4a 2 −−a 33 [= 0] B1 Condone or . A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. Alternative Method for Question 11(a) x = −2 a , y = −4 a 2 + a B1 The co-ordinates of the minimum point (likely to be either from differentiation or completing the square). 2 M1 Their y-coordinate equated to −33. Condone or  . their − 4 a + a = −33 ( ) −4a 2 + a + 33  = 0 B1 OE Condone or  . A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. 11(a) Alternative Method 2 for Question 11(a) x 2 + 4ax + a + 33 [=0] B1 Condone or  . ( 4 a ) 2 − 4 (1 a + 33 ) [ = 0] M1 Condone or  . 16a 2 − 4a − 132 [=0] B1 OE Condone or  . Accept 4a 2 − a − 33 or multiples thereof. A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. 4 11(b) 1 3 B1 Expression for g ( x ) . SOI  g ( x ) =  ( x + 4 ) 2 Either 1 3 M1 Replacing x with 0 in their g ( x ) . g ( 0 ) = 2   g ( 0 ) =  ( ( 0 )  4 )  2 gg ( 0 ) = 6 A1 f ( their 6 )  = 36 + 24a + a  M1 Or 3 2 3 M1 Either composite function using their g ( x ) .  x + 4   x + 4  fg ( x ) =   + 4 a   + a  2   2   x 3 + 4  3 + 4   3   ( x 3 + 4 )  2  or gg ( x ) =  = + 2  2  16    A1 Complete algebraic expression for fgg ( x ) .  3  2  3     ( x 3 + 4 )  ( x 3 + 4 ) fgg ( x ) =  + 2  + 4 a  + 2  + a  16   16      x ) .  ( 0 + 4 ) 3  2  ( 0 + 4 ) 3  M1 Replacing x with 0 in their fgg ( fgg ( 0 ) =  + 2  + 4 a  + 2  + a  16   16  11(b) Then 36 + 24 a + a = 96 A1 OE A1 OE  a = 12 5 6

This question in 9709/13 May/June 2025

3 (a) Use completing the square to find the exact solutions of the equation 4x 2 - 4 x - 1 = 0 . [2] … … … … … … … … … … … … … 1 (b) Hence solve the equation 4 tan i = 4 + for 0° 1 i 1 180° . [3] tan i … … … … … … … … … … … … …

5 marks

Mark scheme: 3(a) 2 2 M1 OE  1  2  1  1  0 4  x −   −=2 0  or ( 2 x − 1)  −=2 0 or  x −  − =  Must deal with the coefficient of x2 and x correctly to  2   2  2  2 produce an ( ax + b ) term. 1 1 A1 1 1  2 . x =  OE, e.g. x = ( ) 2 2 2 SC B1 only for correct solutions from another method. 2 3(b) 1 M1 Setting tan= their x for at least one value of their x. tan= 1  2 ( ) May restart and solve the quadratic in tan. 2 [ =] 50.4, 168.3 AWRT and no other answers in the range 0   180 A1 A1 SC A1 only for AWRT 0.879 and 2.94 radians. SC M0 B1 B1 for answers only. SC M0 B1(only) for both answers only in radians. 3

This question in 9709/15 May/June 2025

1 Find the set of values of the constant k for which the quadratic equation 3kx 2 + ( k + 8) x + 3 = 0 has two distinct real roots. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Use of b2 – 4ac *M1 Obtain k 2 − 20 k + 64 A1 Attempt solution of quadratic equation or inequality DM1 Solve quadratic using suitable method. E.g., factorisation, quadratic formula, completing the square. Obtain k  4, k  16 or clear equivalent B1 B0 for use of ⩽ and/or ⩾. 4

This question in 9709/11 Oct/Nov 2025

4 (a) Express 1 - 6x - x 2 in the form a - ( x + b ) 2 , where a and b are constants. [3] … … … … … … … … … … … … … (b) The graph of y = x2 is transformed to the graph of y = 1 - 6x - x 2 by a reflection followed by a m translation of e o. Give details of the reflection and determine the values of m and n. [3] n … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) {10} – (x {+}{3})2 B1 B1 B1 3 4(b) State reflection in x-axis B1 Obtain m = −3 B1 FT Following their value of b. Obtain n = 10 B1 FT Following their value of a. 3

This question in 9709/11 Oct/Nov 2025

1 (a) Express 9x 2 - 36 x + 8 in the form p ( x + q) 2 + r , where p, q and r are constants. [2] … … … … … … … (b) Hence find the set of values of the constant k for which the equation 9x 2 - 36 x + 8 = k has no real roots. [1] … … … … … (c) Find the exact roots of the equation 9x 2 - 36 x + 8 =-15 . [2] … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1(a) 9 {(x − 2) 2 }−28 B2,1,0 B2 for all 3 correct values. B1 for any two correct. Allow p = 9, q = −2, r = −28 . If contradictory, give preference to the expression. Condone correct values in an expression of the correct form, even if missing the square on the bracket. Allow B1 for 9{(x − 2) 2 + k }. 2 1(b) k −28 B1FT k  theirr. 1 1(c) 2 M1 2 2 2 13 36  36 − 4 9 23 Solving as far as ' ( x + q ) = '. 9 ( x − their 2 ) − their 28 = −15  ( x − 2 ) = or x = 9 2  9 Allow sign errors only in the rearrangement, or rearranging and using the quadratic formula (see quadratics guidance). 13 6  13 13 36  468 A1 ISW if decimals given after the exact answers. x = 2  or or 2  or oe 3 3 9 18 2

This question in 9709/12 Oct/Nov 2025

6 y 6 4 y = x3 2 – 1 0 1 2 3 4 5 x y = f(x) – 2 – 4 – 6 – 8 The diagram shows the graphs of y = x3 and y = f ( x) . The graph of y = x3 is transformed to the graph of y = f ( x) by a sequence of transformations. (a) Describe fully a suitable sequence of transformations. Make clear the order in which the transformations are applied. [5] … … … … … … (b) You are given that f ( x) = a ( x + b ) 3 + c . State the values of the constants a, b and c. [3] … … … … … … …

8 marks

Mark scheme: 6(a) {Stretch} {[scale] factor 2} {[parallel to] in/on y[-axis] or x-axis B2,1,0 B2 for all three {} elements correct. B1 for two {} elements invariant} correct.  3  B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation}   or {[+]3 in/on [the] x[direction]} and correct.   − 4  Do not condone use of ‘Right’ and ‘Down’ in place of x and y {– 4 in/on [the] y [direction]} directions. 3  0  Note: The vectors can be seen as  and   . 0  −4  Correct transformations in the correct order without any extra B1 Combined translations (condone transformations accompanied by transformations. Down and right can be condoned for this mark if  0  a vector) must come after the stretch or, if separate, the the candidate’s intention is clear.    −4  must come after the stretch. Alternative Method for Question 6(a)  3  B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation}   or {[+]3 in/on [the] x [direction]} and correct.   − 2  Do not condone use of ‘Right’ and ‘Down’ in place of x and y {-2 in/on [the] y [direction]} directions. 3  0  Note: The vectors can be seen as  and   . 0  −2  {Stretch} {[scale]factor 2} {[parallel to] in/on y-[axis] or x axis B2,1,0 invariant OE} Numerically correct transformations in the correct order without B1 Combined translations (condone transformations accompanied by any extra transformations. Down and right can be condoned for  0  this mark if the candidate’s intention is clear. a vector) must come before the stretch or, if separate, the    −2  must come before the stretch. 5 6(b) a = 2 B1 Individual values are considered the final answer and NOT those in a ( x + b )3 + c, unless individual values are not stated. b = −3 B1 c = − 4 B1 3

This question in 9709/13 Oct/Nov 2025

10 A function f is defined by f ( )x = px 2 + 4 x + q for x ! R , where p and q are constants. (a) It is given that p = 2 and q = 10 . (i) Express f ( )x in the form a ( x + b ) 2 + c , where a, b and c are constants. [3] … … … … … … … … … … … … … … … … … … (ii) State the range of f. [1] … … … … … … (b) It is given instead that q =-5 and the roots of f ( )x = 0 are 5m and -9m , where m is a constant. Find the values of p and m. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(a)(i) (2 x + 1) 2 +8 B1 B1 for each {} element.   B1 B1 3 10(a)(ii)  Range or f  or y  ( x )    8 B1 FT OE. FT their value of c. x 8 scores B0. 1 10(b) 2 2 *M1 Allow one error for this mark if the intention is clear. p ( 5m ) + 4 ( 5m ) − 5 = 0 and p ( −9m ) + 4 ( −9m ) − 5  = 0  5 − 20m  2 *DM1 Eliminate p by substitution or equating two expressions for p. 81  m + 4 ( −9 m ) − 5  = 0   2   (5m)    36 m + 5 2 or 25   m + 4 ( 5 m ) − 5  = 0  2    ( − 9 m )   5 − 20 m   36 m + 5  or  2  =  2   (5 m )   ( −9 m )  2520m = 280 DM1 OE Simplify as far as a linear equation or a quadratic where m = 0 is ignored or dismissed. m = 19 , p = 9 A1, A1 Alternative Method for Question 10(b) p ( 5m )2 + 4 ( 5m ) − 5 = 0 and p ( −9m )2 + 4 ( −9m ) − 5  = 0 *M1 Allow one error for this mark if the intention is clear. 2 2 *DM1 Equating their two expressions. p ( 5m ) + 4 ( 5m ) −=5 p ( −9m ) + 4 ( −9m ) − 5 DM1 Simplify as far as a quadratic where m = 0 is ignored or dismissed  2 1  45 56 pm − 56m = 0  m =   5 = or 45m = 5 and substitute for m or p.  p  p p = 9, m = 19 A1, A1 10(b) Alternative Method 2 for Question 10(b) 2 2 *M1 Equating 5m and − 9m to the two solutions from the quadratic −+4 4 − 4 ( p )( −5 ) −−4 4 − 4 ( p )( −5 ) = 5m and = −9 m formula. 2 p 2 p Allow one error for this mark if the intention is clear. 2 2 *DM1 Attempt to solve simultaneously by equating their expressions for −+4 4 − 4 ( p )( −5 ) −−4 4 − 4 ( p )( −5 ) 5 = m, or dividing one by the other and equating to −.9 ( 5  2 ) p ( −9 2 ) p 16 + 20 p = 196 DM1 OE Simplify as far as a linear expression – no square root. p = 9, m = 19 A1, A1 Alternative Method 3 for Question 10(b)  x 2 + 4mx − 45m2  = 0 *M1 Recognising the relationship between 5m and − 9m and the ( x − 5m )( x + 9m )  quadratic equation. Or +  = −4 m and  = −45 m 2 4 2 5 *DM1 2 4 5 4 m = , −45m = − Using x + x −  = 0  . Comparing coefficients or using p p p p b c +  = − and = . a a 1 2 DM1 Forming an equation in one unknown. m =  − 45 m = − 5 m p 1 2 or p =  5 p = 45 p m m = 19 , p = 9 A1, A1 5

This question in 9709/13 Oct/Nov 2025

3 (a) Express 4x 2 + 10x + 6 in the form a ( x + b) 2 + c , where a, b and c are rational constants to be determined. [2] … … … … … … … … … … … … … … … … (b) The curve with equation y = 4x 2 + 10x + 6 and the line y = k have exactly one point of intersection. Using your answer to part (a) or otherwise, state the value of the constant k. [1] … … … … … … … …

3 marks

Mark scheme: 3(a)   5  2   1  B2,1,0 B2 All 3 components correct, 4  x +   −  B1 Any 2 components correct.  4    4  5 1 Allow a = 4, b = , c = − unless contradicted 4 4 by the algebraic expression. Allow decimals. 2 3(b) 1  1 1  B1 FT Follow through their c from 3(a). k = − or −0.25  allow − or y = −  4  4 4  Don’t allow x = − 1. 4 If solved through restarting answer must be WWW. 1

This question in 9709/15 Oct/Nov 2025

5 Solve the equation 3 27 x - 28 + = 0 . [3] x 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 5 [ x 6 − 28 x 3 + 27 = 0 ] → u 2 − 28u + 27  = 0 M1* Multiply by 3x and recognise as a quadratic in x 3. quadratic using suitable method. E.g.  x 3 − 1 x 3 − 27 = 0  → ( u − 27 )( u − 1) = 0  DM1 Solve )( )  (  factorisation, quadratic formula, completing the square. x = 1, 3 B1 WWW 3

This question in 9709/15 Oct/Nov 2025

8 y O a 2a x The diagram shows the curve with equation y = 6x + 5 . The shaded region is bounded by the curve, the x-axis and the lines x = a and x = 2a , where a is a positive constant. The shaded region is rotated through 360° about the x-axis to form a solid. The solid has volume, V, such that V H 46 r . (a) Show that 9a 2 + 5a - 46 H 0 . [4] … … … … … … … … … … (b) Find the range of possible values of a. [3] … … … … … … … …

7 marks

Mark scheme: 8(a) 2 a B1 CAO SOI V = π  a ( 6 x + 5 ) d x  Must include π and the correct limits. 2 a 2 M1* Correct integral of 6 x + 5 (can be awarded if π is = π 3 x + 5 x  a missing and the limits are missing or incorrect).  2 a  6 x + 5 ) 2  or = (π    12  a = π 12a 2 + 10a − 3a 2 − 5a 46π  DM1 Correct substitution of correct limits. ( )  9 a 2 + 5 a − 46  0 A1 AG Can only be awarded if the argument leading to the statement is complete and clear. 4 8(b) Solve inequality (or equation) M1 Solve quadratic by suitable method. B1 Obtain − 23, 2 Condone absence of − 23. 9 9 Condone use of x. Final answer a  2 B1 WWW Don’t allow x 2. 3

This question in 9709/15 Oct/Nov 2025