Use differentiation to find gradients, tangents — Paper 2 · IGCSE Mathematics - Additional 0606

4 The point P lies on the curve y = 3 x 2 - 7x + 11. The normal to the curve at P has equation 5y + x = k . Find the coordinates of P and the value of k. [6]

6 marks

Mark scheme: 4 dy B1 = 6 x − 7 soi dx 1 B1 finds or uses correct gradient of normal mnormal = − soi 5  1  M1 uses m1 m2 = − 1 with numerical gradients oe mtangent = 5 soi or ( 6 x − 7 )  −  = − 1  5  6 x − 7 = 5 oe ⇒ x = 2 A1 y = 9 A1 k = 47 A1 Alternative method 1 B1 mnormal = − 5 mtangent = 5 M1 3 x 2 − 12 x + 11 − c = 0 oe A1 solving 3 x 2 − 12 x + 12 = 0 oe to find x = 2 A1 y = 9 A1 k = 47 A1

This question in 0606/22 May/June 2017

12 The function g is defined, for x 2- , by g ( x) = . 2 2x + 1 (i) Show that g l ( x) is always negative. [2] (ii) Write down the range of g. [1] The function h is defined, for all real x, by h ( x) = kx + 3 , where k is a constant. (iii) Find an expression for hg ( x) . [1] (iv) Given that hg ( 0) = 5 , find the value of k. [2] (v) State the domain of hg. [1]

7 marks

Mark scheme: 12(i) B1 −2 −×3 2 − 2 − 6 Allow − 3(2 x + 1) × 2 or 2 oe −6(2 x + 1) or 2 oe isw ( 2 x + 1) ( 2 x + 1) Denominator or (2 x + 1) 2 is positive [and B1 − k FT their g ′( x ) of the form 2 oe numerator negative therefore g ′( x ) is always ( 2 x + 1) negative] oe where k > 0; Allow (2 x + 1) −2 is always positive 12(ii) g > 0 B1 12(iii) 3k B1 + 3 oe isw 2 x + 1 12(iv) 3k B1 + 3 = 5 2(0) + 1 2 B1 implies the first B1 k = isw 3 12(v) 1 B1 x > − 2

This question in 0606/22 May/June 2017

7 The gradient of the normal to a curve at the point with coordinates ,x y is given by . 1 - 3x (i) Find the equation of the curve, given that the curve passes through the point (1, −10). [5] (ii) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 4 . [4]

9 marks

Mark scheme: 7(i) d y 3 x − 1 B1 Gradient = Negative reciprocal. (Gradient or ) = Can be implied. d x x 1 1 B1 ± One correct term − = 3x 2 − x 2 3 1 M1 at least 1 fractional power increased y = 2 x 2 − 2 x 2 ( + C ) by1. − 10 = 2 − 2 + C → C = − 10 A1 one term correct with simplified coefficients 3 1 A1 For C from correct working. y = 2 x 2 − 2 x 2 − 10 7(ii) x = 4 → y = 16 − 4 − 10 = 2 B1 dy 1 B1 → = 6 − = 5.5 dx 2 Eqn with their grad and point (4, ...) M1 y − 2 A1 Must be in the form y = mx + c but Eqn of tangent: = 5.5 → y = 5.5 x − 20 oe x − 4 accept 2 y = 11x − 40

This question in 0606/22 Oct/Nov 2017

11 y y = mx + 8 B O A x y = 4 + 3x – x2 The diagram shows the curve y = 4 + 3 x - x 2 intersecting the positive x-axis at the point A. The line y = mx + 8 is a tangent to the curve at the point B. Find (i) the coordinates of A, [2] (ii) the value of m, [3] (iii) the coordinates of B, [2] (iv) the area of the shaded region, showing all your working. [5]

12 marks

Mark scheme: 11(i) y = 0 → ( x − 4 )( x + 1) = 0 M1 Solve → A is ( 4,0 ) nfww A1 Indication somewhere that x = 4 when y = 0 11(ii) 4 + 3 x − x 2 = mx + 8 M1 Eliminate y . x 2 + ( m − 3 ) x + 4 = 0 2 2 M1 M1dep b − 4 ac ( = 0 ) → ( m − 3 ) = 16 Use of discriminant m = − 1 A1 Do not award if m = 7 is not discarded 11(iii) Obtain quadratic x 2 + ( m − 3 ) x + 4 = 0 using M1 Working must be seen for any marks to be awarded. their m and attempt to solve. Must not be awarded if m is not obtained correctly Point B (2, 6) A1 11(iv) 4 2 M1 Area under curve = ∫2 ( 4 + 3 x − x ) d x Integrate powers increased in at least 2 terms 4 A1  3 2 1 3  = 4 x + x − x    2 3  2  64   8  M1 M1dep = 16 + 24 − − 8 + 6 −     Insert limits of their 2 and 4 and  3   3  subtract in correct order. May be 1 = 7 2 3 implied by 18 −… 3 6 × 6 M1 Area of triangle using Intercept is (8,0) so area of triangle = = 18 2 ( their 8 − x B ) their B = × y B 2 or Attempt to find other suitable areas to result in a complete method. 1 2 A1 Accept 10.7. Must not be awarded Shaded area = 18 − 7 = 10 if point B is not obtained correctly. 3 3

This question in 0606/22 Oct/Nov 2017

2 The variables x and y are such that y = ln 3x - 1 for x 2 . 3 dy (i) Find . [2] dx (ii) Hence find the approximate change in x when y increases from ln 1.2 to ln 1.2 + 0. 125 . [3] ^ h ^ h

5 marks

Mark scheme: 2(i) 1 M1 k × 3 x − 1 1 A1 3 × 3 x − 1 2(ii) 11 B1 x = soi 15 d y M1 0.125 ≈ their × δx oe d x x = their 11 15 0.05 nfww A1

This question in 0606/21 May/June 2018

8 y y = x + e 5 - 2x A B 0 5 x The diagram shows part of the curve y = x + e 5 - 2x , the normal to the curve at the point A and the line x = 5 . The normal to the curve at A meets the y-axis at the point B. The x-coordinate of A is 2.5. (i) Find the equation of the normal AB. [4] (ii) Showing all your working, find the area of the shaded region. [6]

10 marks

Mark scheme: 8(i) dy 2 − 5 x B1 = 1 − 2e dx dy B1 x = 2.5 → = −1 and y = 3.5 dx −1 M1 Grad of normal = dy dx y = x + 1 A1 Equation of normal 8(ii) 1 M1 Area of trapezium = × 2.5 × 4.5 2 5.625 sq units A1 5 M1 Area under curve x + e dx ∫ ( 5 − 2 x ) 2.5 5 A1 2  x 1 ( 5 − 2 x )  = − e    2 2  2.5 M1 insert limits and subtract (= 9.87) Shaded area = 15.5 A1 5.625 + 9.87

This question in 0606/21 Oct/Nov 2018

10 The line y = 12 - 2x is a tangent to two curves. Each curve has an equation of the form y = k + 6 + kx - x 2 , where k is a constant. (i) Find the two values of k. [5] The line y = 12 - 2x is a tangent to one curve at the point A and the other curve at the point B. (ii) Find the coordinates of A and of B. [3] (iii) Find the equation of the perpendicular bisector of AB. [3]

11 marks

Mark scheme: 10(i) 12 − 2 x = k + 6 + kx − x 2 M1 * Equate and collect terms → x 2 − ( 2 + k ) x + 6 − k = 0 b 2 − 4ac = 0 M1 Dep* → ( 2 + k ) 2 = 4 ( 6 − k ) k 2 + 8k − 20 = 0 A1 ( k + 10 )( k − 2 ) = 0 M1 k = −10 or 2 A1 M1 Insert values of k in equations10(ii) ( − 4, 20 ) and ( 2, 8 ) 3 and solve for x A1 x 2 + 8 x + 16 = 0 → x = −4 → y = 20 A1 x 2 − 4 x + 4 = 0 → x = 2 → y = 8 10(iii) 1 B1 Grad of perpendicular = 2 Midpoint ( − 1 ,1 4 ) B1 FT y − 14 1 1 B1 FT Eqn = → y = x + 14.5 x + 1 2 2

This question in 0606/21 Oct/Nov 2018

3 A curve has equation y = . Find sin 2x dy (i) , [3] dx (ii) the equation of the tangent to the curve at the point where x = r . [3] 4

6 marks

Mark scheme: 3(i) 3 x 2 sin2 x − x 3 × 2cos2 x 3 M1 Quotient rule 2 A2/1/0 minus one each error ( sin2 x ) isw 3(ii) π3 B1 y = [ = 0.48…] 64 dy 3π 2 B1 = [=1.85] oe dx 16 3π 2 π 3 B1 cao y = x − 16 32 [ y = 1.85 x − 0.97 ]

This question in 0606/22 Oct/Nov 2018

9 y y = 2√x A (4, 4) O B x The diagram shows part of the curve y = 2 x . The normal to the curve at the point A (4, 4) meets the x-axis at the point B. (i) Find the equation of the line AB. [4] (ii) Find the coordinates of B. [1]

9 marks

Mark scheme: 9(i) dy − 12 B1 = x dx dy 1 B1 x = 4 → = dx 2 grad of normal = −2 M1 y − 4 A1 = −→2 [ y = −2 x + 12 ] x − 4 9(ii) (6, 0) B1 FT 9(iii) 1 B1 FT Area of triangle = × 2 × 4 = 4 2 1 M1 Area under curve 2 x 2 d x =∫ 3 A1 4 2 = x 3 2 A1 FT Total area = 14 [14.7 ] 3 OR Area of trapezium OBAP B1 FT 1 = ( 6 + 4 ) × 4 = 20 2 Area between curve and y- axis M1 y 2 = dy ∫ 4 y 3 A1 = 12 2 A1 FT Total area = 14 [14.7 ] 3

This question in 0606/22 Oct/Nov 2018

10 Two lines are tangents to the curve y = 12 - 4x - x 2 . The equation of each tangent is of the form y = 2 k + 1 - kx , where k is a constant. (i) Find the two possible values of k. [5]

9 marks

Mark scheme: 10(i) 2 k + 1 − kx = 12 − 4 x − x 2 M1 * x 2 + 4 x − kx + 2 k − 12 + 1 b 2 − 4ac M1 Dep* → ( 4 − k ) 2 − 4 ( 2 k − 11) k 2 − 16k + 60 A1 ( k − 6 )( k − 10 ) M1 k = 6 or1 0 A1 OR k = 4 + 2 x M1 * −4 x − 2 x 2 + 8 + 4 x + 1 = 12 − 4 x − x 2 M1 Dep*  k − 4   k − 4  2 or 2 k + 1 − k   = 12 − 2 ( k − 4 ) −    2   2  x 2 − 4 x + 3 A1 or k 2 − 16 k + 60 ( x − 1)( x − 3 ) M1 or ( k − 6 )( k − 10 ) x = 1 or x = 3 → k = 6 or 10 A1 10(ii) k = 6 → [ y ] = 13 − 6 x B1 FT k = 10 → [ y ] = 21 − 10 x B1 FT M1 solve x = 2 , y = 1. 2 cao

This question in 0606/22 Oct/Nov 2018

10 The equation of a curve is y = x 2 3 + x for x H- 3 . dy (i) Find . [3] dx (ii) Find the equation of the tangent to the curve y = x 2 3 + x at the point where x = 1. [3] (iii) Find the coordinates of the turning points of the curve y = x 2 3 + x . [4]

10 marks

Mark scheme: 10(i) d 1 − 1 B1 2 3 + x = ( 3 + x ) d x 2 1 − 1 M1 2 correctly substitute their ( 3 + x ) 2 and their 2x into product rule d y 2 1 − 1 1 A1 2 = x × ( 3 + x ) 2 + 2 x ( 3 + x ) d x 2 10(ii) y = 2 B1 d y 17 B1 = d x 4 y − 2 17 17 9 B1 17 = ( y = x − ) oe FT on their 2 and their from x − 1 4 4 4 4 or use y = mx + c and find c d y their d x 10(iii) d y M1 set their = 0 d x obtain correct quadratic equation A1 5x2 + 12x [= 0] soi (0, 0) and (–2.4, 4.46) A2 A1 for one point or two correct values of x

This question in 0606/23 Oct/Nov 2018

11 A line with equation y =- 5x + k + 5 is a tangent to a curve with equation y = 7 - kx - x 2 . (i) Find the two possible values of k. [5] (ii) Find, for each of your values of k, • the equation of the tangent • the equation of the curve • the coordinates of the point of contact of the tangent and the curve. [5] (iii) Find the distance between the two points of contact. [2]

12 marks

Mark scheme: 11(i) −5 x + k + 5 = 7 − kx − x 2 M1 * 2 2 M1 Dep* b − 4 ac ( = 0 ) → ( k − 5 ) − 4 ( k − 2 ) ( = 0 ) k 2 − 14 k + 33 ( = 0 ) A1 ( k − 11)( k − 3 ) ( = 0 ) M1 Dep dep * solve quadratic in k k = 11 and k = 3 A1 11(ii) y = –5x + 16 and y = 7 – 11x – x2 B2 FT their k B1 for any two correct y = –5x + 8 and y = 7 – 3x – x2 solve one tangent/curve pair for one variable from M1 quadratic equation with repeated root (–3, 31) and (1, 3) A2 A1 for one correct point or two correct x values 11(iii) find distance between any two points found in (ii) M1 800 oe A1

This question in 0606/23 Oct/Nov 2018

2 Variables x and y are related by the equation y = . ex d y l - x ln x (i) Show that = x . [4] d x xe (ii) Hence find the approximate change in y as x increases from 2 to 2 + h, where h is small. [2]

6 marks

Mark scheme: 2(i) x B2 B1 for each e d ( ln x ) 1 d ( ) x = , = e soi dx x dx x 1 x M1 e × their − ( ln x ) × their e d y x = dx x 2 e ( ) correct completion to given answer, A1 d y 1 − x ln x = d x xe x 2(ii)  1 − 2ln 2  M1 δy = × h soi  2   2e  −0.0261[…]h isw A1

This question in 0606/22 Feb/March 2019

8 The equation of a curve is given by y = xe -2x . dy (i) Find . [3] dx (ii) Find the exact coordinates of the stationary point on the curve y = xe -2x . [2] -2x 1(iii) Find, in terms of e, the equation of the tangent to the curve y = xe at the point ,1 [2] e e2 o. (iv) Using your answer to part (i), find xe -2x d x . [3] y

10 marks

Mark scheme: 8(i) –2e–2x seen B1 Product rule M1 Clear attempt e −2 x (1 − 2 x ) A1 8(ii) dy M1 Must have two terms Set = 0 and attempt to solve dx  1 1  A1  ,   2 2e  8(iii) d y M1 Attempt to find at x = 1 d x 1 −1 1 2 A1 y − = x + 2 2 ( x − 1) or y = − 2 e e e e 2 8(iv) Integrate part(i) M1 xe −2 x = ∫− 2 xe −2 x + e −2 x d x ( ) Integrate e −2 x and make ∫ xe −2 x dx the M1 subject − xe − 2 x e −2 x A1 − + c 2 4

This question in 0606/21 Oct/Nov 2019

8 The roots of the equation x 3 + ax 2 + bx + 24 = 0 are 2, 3 and p, where p is an integer. (i) Find the value of p. [1] (ii) Show that a =-1 and find the value of b. [4] Given that a curve has equation y = x 3 - x 2 + bx + 24 find, using your value of b, dy (iii) , [1] dx (iv) the integer value of x for which the gradient of the curve is 2 and the corresponding value of y. [3] The coordinates of the point P on the curve are given by the values of x and y found in part (iv). (v) Find the equation of the tangent to the curve at P. [1]

10 marks

Mark scheme: 8(i) p = –4 B1 8(ii) (x – 2) (x – 3) (x + 4) M1 FT (x – 2) (x –3) (x – p) (x2 – 5x + 6) (x + 4) A1 FT (x2 – 5x + 6) (x – p) multiply out two factors correctly obtain a = –1 A1 answer given x3 – x2 – 14x + 24 b = –14 stated B1 8(iii) d y 2 B1 FT their numerical b 3x2 – 2x + b = 3 x − 2 x − 14 d x 8(iv) d y M1 FT their numerical b set their equal to 2 d x x = 2 A1 y = 40 only A1 no additional answers 8(v) y – 40 = 2(x + 2) (y = 2x + 44) B1

This question in 0606/23 Oct/Nov 2019

6 (a) Find the equation of the tangent to the curve 2y = tan 2x + 7 at the point where x = r . 8 Give your answer in the form ax - y = r + c , where a, b and c are integers. [5] b (b) This tangent intersects the x-axis at P and the y-axis at Q. Find the length of PQ. [2]

7 marks

Mark scheme: 6(a) dy 2 B1 = sec 2 x dx dy B1 d y their = their 2 FT their dx x = π d x 8 π B1 x = , y = 4 8  π  M1 y − their 4 = ( their 2 )  x −  oe  8  π A1 2 x − y = − 4 4 6(b) 2 2 M1  π   π   − 2  +  4 −  oe  8   4  3.59 or 3.59[03…] rot to four or more A1 figs

This question in 0606/22 May/June 2020

10 The gradient of the normal to a curve at the point (x, y) is given by . x + 1 (a) Given that the curve passes through the point (1, 4), show that its equation is y = 5 - ln x - x . [5] (b) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 3 . [3]

8 marks

Mark scheme: 10(a) d y (1 + x )  1  2 M1 for using m1 × m2 = −1 = − = −  + 1  dx x  x  y = − lnx − x + C 2 1 M1 for integrating x A1 for all correct including C 4 = −ln1 −+1 C A1 Insert (1, 4) and arrive at correct C = 5 → y = 5 − lnx − x answer. AG 10(b) x = 3 → y = 2 − ln3 B1 d y 1 4 and = − −=1 − d x 3 3 y − ( 2 − ln3 ) 4 M1 = − x − 3 3 4 A1 y = − x + 6 − ln3 3 or y = −1.33 x + 4.90

This question in 0606/21 Oct/Nov 2020

5 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Find the equation of the tangent to the curve y = x 3 - 6x 2 + 3x + 10 at the point where x = 1. [4] (b) Find the coordinates of the point where this tangent meets the curve again. [5]

9 marks

Mark scheme: 5(a) x = 1 → y = 8 B1 dy M1 Attempt to differentiate. Powers reduced = 3x2 – 12x + 3 by 1 in all four terms. dx dy A1 x = 1 → = –6 dx y − 8 A1 Either form. = −→6 y = −6 x + 14 isw x − 1 5(b) x3 – 6x2 + 9x – 4 = 0 2 M1 for equating their tangent to curve and (x – 1)(x2 – 5x + 4) = 0 simplifying to 4 term cubic. or (x – 4)(x2 – 2x + 1) = 0 M1Dep for finding a factor or stating that (x – 1) is a factor or makes at least 3 attempts to find a factor. (x – 1)(x – 1)(x – 4) = 0 2 A1 for (x – 1) or x = 1 can be implied. nfww A1 for (x – 4) or x = 4 not repeated. nfww x = 4 → y = –10 only A1 nfww

This question in 0606/22 Oct/Nov 2020

4 It is given that y = ln ( 1 + sin x) for 0 1 x 1 r. d y. [2] (a) Find d x d y r 1(b) Find the value of when x = , giving your answer in the form , where a is an integer. d x 6 a [2] d y (c) Find the values of x for which = tan x . [5] d x

9 marks

Mark scheme: 4(a) dy 1 M1 = dx 1 + sinx cosx A1 × cosx = 1 + sinx 4(b) π dy M1 insert into their 6 dx 1 A1 3 not 3 3 4(c) cos x sin x M1 sinx their = replace tan x with 1 + sin x cos x cos x use cos 2 x = 1 − sin 2 x M1 earned when equation reduced to a 2 quadratic in sinx 2sin x + sin x −=1 0 ( ) ( 2sin x − 1)( sin x + 1) = 0 M1 solve three term quadratic in sinx π A1 or 0.524 or better radians only x = 6 if M0 M0 M0 and (a) and (b) correct, allow SC2 for 1 π tanx = , x = 3 6 5 π A1 or 2.62 or better radians only x = A0 if extra solution(s) in range 6

This question in 0606/23 Oct/Nov 2020

7 A curve has equation y = x cos x . d y. [2] (a) Find d x (b) Find the equation of the normal to the curve at the point where x = r , giving your answer in the form y = mx + c . [4] r (c) Using your answer to part (a), find the exact value of x sin x d x . [5] 6y0

11 marks

Mark scheme: 7(a) − x sin x + cos x isw B2 accept unsimplified if incorrect allow B1 for d ( cos x ) = − sin x clearly seen d x 7(b) x = π, y = −π B1 or –3.14 or better d y B1 d y x = π, = − 1 from correct d x d x gradient of normal =1 M1 use m1m2 = –1 with their grad of tangent y = x −π2 cso A1 or y = x − 6.28 or better fully correct solution their ( a ) = x cos x7(c)  M1 * − sin x + cos xd x = x cos x (  ) B1 clearly seen anywhere cos xdx = sin x  − x cos x + sin x A1 implies previous marks if (a) is correct π M1 * dep insert into their integral 6 1 3 A1 reject decimals −π 2 12

This question in 0606/23 Oct/Nov 2020

7 A curve has equation y = p ( x) , where p ( )x = x 3 - 4x 2 + 6x - 1. (a) Find the equation of the tangent to the curve at the point (3, 8). Give your answer in the form y = mx + c . [5] (b) (i) Given that p -1 exists, write down the gradient of the tangent to the curve y = p -1 ( x) at the point (8, 3). [1] (ii) Find the coordinates of the point of intersection of these two tangents. [2]

8 marks

Mark scheme: 7(a) dy 2 B1 = 3 x − 8 x + 6 dx d y M1 condone one slip Finds their d x x = 3 mT1 = 9 A1 y – 8 = their 9(x – 3) M1 or y = 9x + c and 8 = 9(3) + c y = 9x – 19 cao A1 7(b)(i) 1 B1 FT their 9 mT2 = their 9 7(b)(ii) [Uses y = x in their ( y = 9x – 19) M1 to form] their ( x = 9x – 19) or their ( y = 9y – 19) oe and solves for x or y or solves e.g. x + 19 their (9x – 19) = their 9  19 19  A1 FT equal x and y coordinates providing at  ,  oe least 3 marks earned in (a)  8 8 

This question in 0606/22 Feb/March 2021

3 A curve has equation y = . x + 1 dy r k (a) Show that the exact value of at the point where x = can be written as 2 , where k dx 6 r is an integer. + 1 [5] b 6 l (b) Find the equation of the normal to the curve at the point where x = 0 . [4]

9 marks

Mark scheme: 3(a) d B1 ( sin3 x ) = 3cos3 x soi d x Applies the correct form of the quotient rule M1 dy ( x + 1)(3cos3 x) − (2 + sin3 x) [1] A1 d FT their ( sin3 x ) = 2 dx dx ( x + 1)  π  3π   3π  M1  + 1  3cos  −  2 + sin  [1] dy  6  6   6  = dx  π  2  + 1   6   dy  −3 A1 not from wrong working  =  2  d x   π   + 1   6  3(b) [When x = 0 ] y = 2 B1 dy B1 dy [When x = 0 ] = 1 FT their dx dx [ m⊥=] = −1 M1 FT −1 their1 y – 2 = −x oe A1 FT their m⊥

This question in 0606/21 Oct/Nov 2021

10 y 5 2 y = + x - x x O x 5 2 The diagram shows part of the curve y = + x - x . x (a) Find, in the form y = mx + c , the equation of the tangent to the curve at the point where x = 1. [5]

9 marks

Mark scheme: 10(a) dy −2 M2 M1 for any two correct terms = −5 x + 2 x − 1 oe dx d y A1 [When x = 1 ] = − 4 and y = 5 d x y – 5 = −4(x – 1) oe M1 d y FT their and y; dep on at d x x =1 least M1 y = −4x + 9 A1 FT 10(b) x 3 x 2 B2 B1 for 5lnx and one other term F( x ) = 5ln x + − (+c) correct 3 2 F(3) − F(1) M1 dep on at least B1 for integration 14 A1 5ln 3 + 3

This question in 0606/21 Oct/Nov 2021

(b) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. Find the exact x-coordinate of each of the two points where the normal cuts the curve again. [5]

10 marks

Mark scheme: 9(a) dy 2 M2 M1 for any two terms correct = 3 x + 2 x − 4 dx dy A1 x = 1 → = 1 dx [ m⊥= ] −1 M1 −1 FT their1 y – 4 = −1(x – 1) oe isw A1 FT their m⊥ 9(b) x 3 + x 2 − 4 x + 6 = their ( −+x 5 ) M1 FT their linear equation of the form y = mx + c where m ≠ 0 and 3 2 c ≠ 0 → x + x − 3 x + 1 [ = 0 ] from (a) Correct quadratic factor: x2 + 2x – 1 B2 B1 for any two out of three terms correct Must be from the correct cubic Solves their (x2 + 2x – 1) = 0 using the formula or by M1 dep on M1 and valid attempt at completing the square finding quadratic factor M0 if their quadratic factor does not have real roots −±2 8 −±2 2 2 A1 isw or isw 2 2

This question in 0606/22 Oct/Nov 2021

6 It is given that x = 2 + sec i and y = 5 + tan 2i . (a) Express y in terms of x. [2] dy (b) Find in terms of x. [1] dx (c) A curve has the equation found in part (a). Find the equation of the tangent to the curve when r i = . [4] 3

7 marks

Mark scheme: 6(a) y = ( x − 2) 2 + 4 oe, isw B2 B1 for a correct expression in x and y only, that is not of the form y = f(x) 6(b)  dy  B1 dep on B2 in (a) = 2( x − 2) oe    dx  6(c) π B1 [When θ = ] x = 4 soi 3 π B1 [When θ = ] y = 8 soi 3 π dy M1 d y [When x = 4 or θ = ] = 4 FT their providing non-zero 3 dx dx x = 4 y − 8 = 4( x − 4) oe isw A1 d y FT their providing non-zero dx x = 4

This question in 0606/23 Oct/Nov 2021

10 (a) It is given that f ( )x = 4x 3 - 4x 2 - 15x + 18 . Find the equation of the normal to the curve y = f ( x) at the point where x = 1. [5] (b) DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. It is also given that x + a , where a is an integer, is a factor of f ( )x . Find a and hence solve the equation f ( )x = 0 . [6]

11 marks

Mark scheme: 10(a) [ f ′( x ) = ] 12 x 2 − 8 x − 15 M2 M1 for any two terms correct or 12 x 2 − 8 x − 15 + c y = 3 and f ′(1) = − 11 A1 1 M1 −1 [ m⊥= ] soi FT 11 their f ′(1) 1 A1 FT their m⊥ and their 3, provided y − 3 = ( x − 1) oe, isw 11 their 3 ≠ 1 or 0 or −11 10(b) [f(−2) =] −32 – 16 + 30 + 18 = 0 M1 Method must be seen and be fully or [f(−a) = ] −4a3 – 4a2 + 15a + 18 and shows this to correct with no clear evidence of be 0 when a = 2 calculator use or uses algebraic long division or synthetic division to show that x + 2 is a factor of f(x) or that a – 2 is a factor of f(−a) a = 2 A1 as the only value of a Uses (x + 2) is a factor to find the correct quadratic B2 B1 for any two out of three terms factor 4x2 – 12x + 9 correct Correctly solves their( 4x2 – 12x + 9) (x + 2) = 0 or M1 dep on using a quadratic factor that correctly factorises their( 4x2 – 12x + 9) (x + 2) has earned at least B1; method must be seen; M0 if their quadratic factor does not have real roots x = −2 or 1.5 A1 dep on M1 B2 M1

This question in 0606/23 Oct/Nov 2021

3 (a) Find the coordinates of the point on the curve y = 1 + 3 x where the gradient of the normal 8 is - . [5] 3 (b) Find the equation of the normal to the curve y = 1 + 3 x at the point (8, 5) in the form y = mx + c . [3]

8 marks

Mark scheme: 13(a) − dy 1 − 12 B2 dy 1 = (1 + 3 x )  3 oe, isw B1 for = (1 + 3 x ) 2  ... dx 2 dx 2 dy 1 − 12 or = (...)  3 d x 2  1 1 − dy 1  their 2  or = their (1 + 3 x )  3 d x 2 dy − 12 or = k (1 + 3 x )  3, k is a constant, dx 1 k ≠ 2 3 B1 d y mtangent = or 0.375 FT their if necessary providing at 8 d x −2 least B1 previously awarded or mnormal = 1 oe − 3(1 + 3 x ) 2  dy 3   dx   8 M1 d y FT their if necessary providing at their − = −  = or   their   d x dx   8 dy 3     least B1 previously awarded (5, 4) A1 3(b) 10 B1 mnormal = − 3 10 M1 FT their mnormal y – 5 = − (x – 8) 3 −10  −10  or y = x + c and 5 =   (8) + c 3  3  oe soi 10 95 A1 FT their mnormal y = − x + 3 3

This question in 0606/21 Oct/Nov 2022

8 The equation of a curve is y = x sin x . d y (a) Find . [2] d x (b) Find the equation of the tangent to the curve at x = r in the form y = mx + c . [3] 2 (c) Use your answer to part (a) to find x cos x dx . [3] y r (d) Evaluate x cos x dx , giving your answer correct to 2 significant figures. [2] 4y0

10 marks

Mark scheme: 8(a) Product rule attempted M1 at most one error  dy  A1 = sin x + x cos x oe    dx  8(b)  π  π B1 When x = y =    2  2  π  dy M1 FT their derivative providing at least M1 When x = = 1   awarded in (a)  2  dx y = x A1 8(c) xsinx + cosx + c B3 B2 for xsinx + cosx or B1 for  x cos xdx =  sin xdx x sin x −     8(d) π π π M1 sin + cos − ( 0 + cos0 ) 4 4 4 0.26 A1

This question in 0606/21 Oct/Nov 2022

3 In this question a and b are constants. a 1 The normal to the curve y = + 3x - 2 at the point where x = 1 has equation y =- x + b . x 4 Find the values of a and b. [6]

6 marks

Mark scheme: 3  −2 B1  mtangent =  − ax + 3 oe [When x = 1, mnormal = ] B1 −1 FT if appropriate −1 dy oe or gradient of tangent = 4 soi their 3 −a dx x =1 −1 1 M1 −1 = − oe FT their (3 − a ) 4 dy their dx x =1 or their (3 – a) = 4 oe d y or their and their evaluation of d x x =1 − 1 1 − 4 a = −1 nfww A1 1 M1 FT y = (their a) + 1 providing at least [When x = 1] their 0 = − [1] + b oe 2 of the first 3 marks awarded 4 1 A1 b = nfww 4

This question in 0606/22 Oct/Nov 2022

2 The tangent to the curve y = ax 2 - 5x + 2 at the point where x = 2 has equation y = 7x + b . Find the values of the constants a and b. [5]

5 marks

Mark scheme: 2 dy B1 = 2 ax − 5 dx 2a −=2 5 7 oe M1  dy FT their  = 7  dx x = 2  a = 3 A1 7 +2 b = their 4 M1 dep on previous M1 or b = 2 −4 theira where their 4 is an attempt to evaluate y = ax 2 − 5 x + 2 using x = 2 and their a b = −10 A1 Alternative ( −12 )2 − 4a ( 2 − b ) = 0 oe (B1) for use of discriminant on ax 2 − 12 x + 2 − b = 0 144 − 8 a + 4 a ( 4 a − 22 ) = 0 oe (M1) Condone one sign or arithmetic error or 144 − ( b + 22 )( 2 − b ) = 0 oe a 2 − 6a + 9  = 0 oe (A1) for correct 3-term quadratic in solvable form or b2 + 20b + 100  = 0 oe a = 3 (A2) A1 for a = 3 or b = −10 and b = −10

This question in 0606/23 Oct/Nov 2022

7 (a) Variables x and y are such that y = . Use differentiation to find the approximate tan x r r change in y as x increases from to + h , where h is small. [5] 4 4 2 1 d y 1 d y ( x + 1)( x - 4)(b) Given that y = 3 show that y - - 2 can be written as 5 . [4] ( x - 3) dx 3 f d x p ( x - 3)

9 marks

Mark scheme: 7(a) d 2 B1 (cos x ) = −2cos x sin x soi dx Attempts the quotient rule M1 d 2 2 2 FT their (cos x ) dy −2cos x sin x tan x − (1 + cos x )sec x dx = dx tan 2 x Fully correct isw A1 d 2 FT their (cos x ) only dx δy dy M1  their h dx x = π 4 δy − 4 h cao A1 7(a) Alternative method 1 d 3 2 (B1) (cos x ) = −3cos x sin x soi dx Attempts the quotient rule: (M1) d 3 2 3 FT their (cos x ) dy (sin x )( − sin x − 3cos x sin x ) − (cos x + cos x )cos x dx = dx sin 2 x Fully correct isw (A1) d 3 FT their (cos x ) only dx δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) Alternative method 2 (B1) d  −2 sec 2 x  2 = −2(tan x ) dx  tan x  Attempts the product rule (M1) d  2  FT their   dy d x  tan x  = −2(tan x ) −2 sec 2 x − (sin x ( − sin x ) + cos x (cos x )) dx Fully correct isw (A1) d  2  FT their   only d x  tan x  δy dy (M1)  their h dx x = π 4 δy − 4 h cao (A1) 7(b) dy −4 B1 = −3( x − 3) oe, soi dx d 2 y −5 B1 = −−4 3( x − 3) oe, soi dx 2 ( x − 3) 2 + 3( x − 3) − 4 M1 dy −4 FT = k ( x − 3) ( x − 3) 5 dx d 2 y −5  x −+3 3 4  x ( x − 3) − 4 and = m ( x − 3) where k 2 or  4 − 5 =  5 dx  ( x − 3) ( x − 3)  ( x − 3) and m are constants Correct completion to given answer: A1 x 2 − 3 x − 4 ( x + 1)( x − 4) = ( x − 3)5 ( x − 3)5

This question in 0606/22 Feb/March 2023

11 The normal to the curve y = sin ( 4 x - r ) at the point A(a, 0), where r 1 a 1 r , meets the y-axis at 2 the point B. Find the exact area of triangle OAB, where O is the origin. [9]

9 marks

Mark scheme: 11 Solves sin(4 x − π) = 0 oe M1 3π A1 a = 4 dy B2 dy = 4cos(4 x − π) B1 for = k cos(4 x − π) , dx dx where k > 0 or k = −4  −1  1 B2 nπ = − FT their a = , n is a  3π  4 4  4cos(4  their − π)   4  positive integer B1 for −1 3π theirk cos(4  their − π) 4 1  3π  1  3π  M1 FT their perpendicular y − 0 = −  x −  or 0 = −   + coe gradient and their a 4  4  4  4   3π  B1 B  0,  soi  16  9π 2 B1 [Exact area =] 128

This question in 0606/22 Feb/March 2023

8 A curve has equation y = cos where x is in radians. The normal to the curve at the point where 4 4 r x = cuts the x-axis at the point P. Find the exact coordinates of P. [7] 3

7 marks

Mark scheme: 8 dy 1 x M2 dy x 1  sin M1 for  k sin , k < 0 or k = dx 4 4 dx 4 4 y = 0.5 B1 dy 1 3 3 M1 3   or  FT (their k)  providing at least M1 dx x  4π 4 2 8 2 3 awarded 8 M1 1 FT  mnormal   soi 3 their d y dx x  4π 3 8  4π  A1 dep on previous M1; y  0.5   x   oe must have exact values 3  3  FT their mnormal and their 0.5 providing both are non-zero, exact values  4π 3  A1   ,0  or exact equivalent;    3 16  mark final answer

This question in 0606/22 May/June 2023

6 y A x O y = 5e 2 x - 3 The diagram shows the curve y = 5e 2 x - 3 . The curve meets the y-axis at the point A. The tangent to the curve at A meets the x-axis at the point B. Find the length of AB. [6]

6 marks

Mark scheme: 6 dy 2 x B1  10e dx [At A, m = ] 10 B1 [At A, y = ] 2 B1 [Equation tangent is] y = 10x + 2 oe B1 2 M1 providing their 10 is derived using   their 2 2  their 2  oe differentiation AB2 =     their10  [AB = ] 2.01 or 2.009[9…] nfww, isw A1

This question in 0606/23 May/June 2023

5 In this question p and q are constants. p The normal to the curve y = 2 + 5x - 2 , at the point where x = 1, has equation y =- x + q . x Find the values of p and q. [6]

6 marks

Mark scheme: 5 [ mtangent =] − 2 px−3 + 5 oe B1 −1 B1 −1 [When x = 1, mnormal = ] FT if appropriate −2 p + 5 dy their or gradient of tangent = 1 nfww dx x =1 −1 = −1 oe M1 −1 their ( −2 p + 5) FT dy or their( –2p + 5) = 1 their dx x =1 d y − 1 or their and their evaluation of d x x =1 −1 p = 2 nfww A1 [When x = 1] their5 = −+1 q or y = –x + 6 M1 FT y = (their p) + 3 providing at least 2 of the first 3 marks awarded q = 6 nfww A1

This question in 0606/21 Oct/Nov 2023

6 Find the value of the constant a for which the line y = ( 2a + 1 ) x - 10 is a tangent to the curve y = ax 2 - 5 x + 2 . [6]

6 marks

Mark scheme: 6 ax 2 − 5 x + 2 = 2 ax + x − 10 M1 ax 2 − ( 2a + 6 ) x + 12  = 0 oe A1 Correct use of b2 – 4ac [*0]: M1 where * is any inequality sign or =; (– (2a + 6))2 – 4(a)(12) [*0] oe FT their 3-term quadratic in x and a 4 a 2 − 24 a + 36[*0] A1 Factorises or solves their 3-term quadratic M1 FT their 3-term quadratic in a in a a = 3 A1 6 Alternative method 3 a + 3 (2) M1 for 2ax – 5 a = oe or x = x − 1 a  6   3  2 (M1) k  + 1  x − 10 =   x − 5 x + 2 oe FT their a of the form where k, b, c are  x − 1   x − 1  bx + c or non-zero constants 2 da + e  a + 3   a + 3   a + 3  or their x of the form where d, e, f are (2 a + 1)   − 10 = a   − 5   + 2 fa  a   a   a  oe non-zero constants 3x2 – 12x + 12 [= 0] oe or a 2 − 6 a + 9[ = 0] (A1) Solves their 3-term quadratic in x as far as (M1) x = ... or factorises or solves their 3-term quadratic in a a = 3 (A1)

This question in 0606/21 Oct/Nov 2023

5 (a) Find the equation of the normal to the curve y = x 3 - 7x 2 + 12x - 5 at the point (1, 1). [5] (b) Find the x-coordinates of the two points where the normal cuts the curve again. Give your answers in the form x = a ! b where a and b are integers. [5]

10 marks

Mark scheme: 5(a) Correct first derivative: M2 M1 for two terms of x 3 − 7 x 2 + 12 x − 5 3 x 2 − 14 x + 12 differentiated correctly [At x = 1] gradient of tangent: 1 A1 y – 1 = their(–1)(x – 1) oe M1 −1 FT or y = –x + c and 1 = –1 + c soi dy their dx x =1 y – 1 = –1(x – 1) or y = −+x 2 oe, isw A1 5(b) x 3 − 7 x 2 + 12 x − 5 = their ( −+x 2) B1 FT their y = ax + b where a is a non-zero constant Uses the correct linear factor x – 1 and the M1 correct cubic x3 − 7 x 2 + 13x − 7  = 0 to find a quadratic factor with at least two terms correct x 2 − 6 x + 7 A1 Correct use of formula or completing the M1 FT their 3-term quadratic providing it is square on their 3-term quadratic, e.g., from an attempt at finding a quadratic factor 2 and the discriminant is not negative −−( 6 )  ( −6) − 4[1](7) x = 2[1] 6  36 − 4[1](7) or x = 2 x = 3  2 A1

This question in 0606/22 Oct/Nov 2023

8 A curve has equation y = x sin 2x . dy (a) Find . [2] dx (b) Find the equation of the tangent to the curve at x = r . [3] 4 r (c) Use your answer to part (a) to find the exact value of 2x cos 2xdx . [5] 6y0

10 marks

Mark scheme: 8(a) Derivative of sin2x: 2cos2x soi B1 Product rule: x  2cos2x + [1]sin 2x isw B1 FT their 2cos2x 8(b) π B1 y = soi, isw 4 gradient of tangent: 1 soi B1 dep on correct derivative y = x or y – x = 0 or x – y = 0 B1 dep on correct derivative 8(c) π M3 M2 for x sin2 x + k cos2 x  1  6 x sin 2 x + cos2 x nfww 1    2  0 where k > 0 or k = − ; nfww 2 or M1 for x2 cos2 x dx = x sin2 x − sin2 x dx − cos2 x or + x2 cos2 x dx = x sin2 x 2 π π 1 π 1 A1 sin + cos − cos0 6 3 2 3 2 A1 π 3 1 π 3 − 3 − or 12 4 12

This question in 0606/22 Oct/Nov 2023

9 A curve has equation y = xe 2 x . d y. [2] (a) Find d x (b) Find the equation of the normal to the curve at x = 1. [4] (c) Use your answer to part (a) to find the exact value of 2 xe 2 x d x . [5] 2y0

11 marks

Mark scheme: 9(a) Derivative of e 2 x : 2e 2 x soi B1 2 x 2 x B1 FT their 2e2x x  2e + e isw 9(b)  When x = 1 y = e2 B1 dy B1 FT their derivative which must include  gradient tangent =  their x dx x =1 at least one term in 2e −1 B1 −1 Gradient of normal = 2 FT their (3e ) d y their d x x =1 −1 B1 dep on 2 marks awarded in part (a) and y – e2 = (x – 1) oe, isw 3e 2 all previous marks awarded in this part 9(c)  2 x 1 2 x  2 M3 M2 for xe 2 x + ke 2 x where k < 0 or xe − e    2  0 k = 1 2 e 2 x dx or M1 for  2 xe 2 x dx = xe 2 x −  4 1 4 1 A1 2e − e −− ( 2 ) ( 2 ) 1.5e4 + 0.5 or exact equivalent A1

This question in 0606/23 Oct/Nov 2023

4 (a) (i) Given that y = 3 sin 2 x + cos x , show that y + cot x = k ( 1 + cos 2 x) , where k is an dx integer. [4] (ii) Using your value of k, solve the equation k ( 1 + cos 2 x) = 4 for - r G x G r . [4] (b) (i) Differentiate y = tan x - x with respect to x. [2] ` j 2 x - 1 (ii) Hence find dx . [2] y 2 x cos x - x ` j

12 marks

Mark scheme: 4(a)(i) dy B2 B1 for an attempt to differentiate both = 6sin x cos x − sin x oe, isw terms with one term correct dx 2 cos x M1 d y 3sin x + cos x + ( 6sin x cos x − sin x ) FT their of the form sin x d x k sin x cos x  sin x Correct simplified step e.g. A1 3sin 2 x + cos x + 6cos 2 x − cos x or 3sin 2 x + 6cos 2 x or 3 + 3cos 2 x leading to 3(1 + cos 2 x ) nfww 4(a)(ii) 2 1 M1 FT their k providing 0 < k ≤ 4 cos x = 3 M1 dep on previous M1; FT their k cos x = 1 oe 3 0.955 or 0.9553[1...] rot to 4 or more sf A2 and no other angles in range 2.19 or 2.186[2...] rot to 4 or more sf A1 for any two correct angles, ignoring extras M1 for f( x)sec 2 ( x − x )4(b)(i)  1 − 12  2 2  1 − x  sec ( x − x ) oe, isw  2  4(b)(ii) Correctly writes M1 where k is a non-zero constant;  1 − 12  2 dependent on part (b)(i)  1 − x  sec ( x − x ) =  2  2 x − 1 2 x − 1sec 2 ( x − x ) or 2 x 2 x cos 2 ( x − x ) and states an answer k tan ( x − x ) or 1 2 x − 1 states dx = tan( x − x ) 2  x cos 2 ( x − x ) 2tan ( x − x ) + c nfww A1

This question in 0606/22 Feb/March 2024

5 Variables x and y are related by the equation y = . Use differentiation to find the approximate ln 3x change in y when x increases from 1 to 1+ h , where h is small. [4]

4 marks

Mark scheme: 5 Correct quotient rule: 2 M1 for  1   1  (ln3 x )[1] − x   3  (ln3 x )[1] − x  their   3  dy  3 x  dy  3 x  = oe = OR dx (ln3 x ) 2 dx (ln3 x ) 2 OR correct product rule using y = x(ln3x)–1: for dy  −2 3  −1 dy  −2 3  = x  − (ln3 x )   + 1 (ln3 x ) = x  their  −(ln3 x )   dx  3 x  dx  3 x  +[1](ln3 x ) −1 δ y ln3 − 1 M1 d y = oe, soi FT their providing quotient 2 h (ln3) d x x =1 rule or appropriate product rule attempted ln3 − 1 A1 must have evidence of correct δy = h or δy = 0.0817 h nfww 2 derivative (ln3)

This question in 0606/22 Feb/March 2024