Evaluate definite integrals and apply — Paper 2 · IGCSE Mathematics - Additional 0606

11 y y = x3 + 4x2 – 5x + 5 A B C y = 5 E O D x The diagram shows part of the curve y = x 3 + 4 x 2 - 5 x + 5 and the line y = 5. The curve and the line intersect at the points A, B and C. The points D and E are on the x-axis and the lines AE and CD are parallel to the y-axis. (i) Find y ( x 3 + 4x 2 - 5x + 5 )d x . [2] (ii) Find the area of each of the rectangles OEAB and OBCD. [4] (iii) Hence calculate the total area of the shaded regions enclosed between the line and the curve. You must show all your working. [4] Question 12 is printed on the next page.

10 marks

Mark scheme: 11(i) x 4 4 x 3 5 x 2 B2 B1 for any 3 correct terms + − + 5 x [ + c ] isw 4 3 2 11(ii) x 3 + 4 x 2 − 5 x + 5 = 5 and rearrange to B1 y x ( x 2 + 4 x − 5 ) = 0 oe soi A(−5, 5) B C(1, 5) E (−5, 0) O D(1, 0) x Solves their x 2 + 4 x − 5[ = 0] soi M1 x = –5, x = 1 soi A1 OEAB = 25, OBCD = 5 A1 11(iii) Correct or correct FT substitution of 0, their −5 M1 dependent on at least B1 in (i)  x 0 4 4 x 3 5 x 2  seen in  + − + 5 x   4 3 2  their − 5 Correct or correct FT substitution of their 1, 0 seen M1 dependent on at least B1 in (i)  x their 1 4 4 x 3 5 x 2  in  + − + 5 x   4 3 2  0 1175 49 M1 for the strategy needed to combine the their − theirOEAB + theirOBCD − their oe areas; may be in steps; 12 12 97.916& − 25 + 5 − 4.083& 886 5 A1 all method steps must be seen; not from oe or 73 oe or 73.83& rot to 3 or more sig wrong working 12 6 figs If M0 then allow SC3 for 0 3 2 1 3 2 ∫−5 ( x + 4 x − 5 x ) dx − ∫0 ( x + 4 x − 5 x ) d x oe  x 4 4 x 3 5 x 2  0  x 4 4 x 3 5 x 2  1 =  + −  −  + −   4 3 2  − 5  4 3 2  0   625 500 125     1 4 5   =  0 −  − −   −   + −  − 0    4 3 2     4 3 2   443 = oe 6 or SC2 for 0 3 2 their 1 3 2 ∫their ( − 5) ( x + 4 x − 5 x ) d x − ∫0 ( x + 4 x − 5 x ) dx oe  x 4 4 x 3 5 x 2  0  x 4 4 x 3 5 x 2  their 1 =  + −  −  + −   4 3 2  their ( −5)  4 3 2  0 = [ F (0) − F (their ( −5)) ] − [ F (their1) − F (0) ]

This question in 0606/22 May/June 2017

9 (i) Find x ln x . [2] dx (ii) Hence find lnx d x . [2] y 2k (iii) Hence, given that k 2 0 , show that ln x d x = k ln 4 k - 1 . [4] y k ^ h

8 marks

Mark scheme: 9(i) d 1 M1A1 Product rule. One correct term + ( xlnx ) = x × + lnx isw another term. Allow unsimplified. dx x 9(ii) ∫ 1+ln x d x = x ln x M1 Correct use of (i) and must be dealing with 2 terms. soi ∫ ln xdx = xln x − x + ( C ) A1 Correct answer with no working is fine. 9(iii) 2 k M1 Insert limits and subtract correctly ∫k ln x d x = [ 2 k ln2 k − 2 k ] − [ k ln k − k ] using their result from (ii) which = k (2ln2 k − l n k − 1) must contain an ln function = k ln ( 2 k ) 2 − ln k − 1 M1 Uses n ln a = ln a n somewhere oe ( )   4 k 2   M1  a  = k  ln   − 1  Uses lna − lnb = ln   or  k   b      ln a + ln b = ln ab somewhere = k ( ln4 k − 1) A1 Answer given Correct completion.

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

7 A particle moving in a straight line passes through a fixed point O. Its velocity, v ms -1 , t s after passing through O, is given by v = 3 cos 2t - 1 for t H 0 . (i) Find the value of t when the particle is first at rest. [2] r (ii) Find the displacement from O of the particle when t = . [3] 4 (iii) Find the acceleration of the particle when it is first at rest. [3]

8 marks

Mark scheme: 7(i) 1 M1 set v = 0 and solve for cos2t v = 0 → cos2t = 3 →=t 0.615 or 0.616 A1 7(ii) 3 M1A1 M1 for sin2t and ± t s = sin2t − t ( + c ) 2 π π A1 t = → s = 1.5 − ( = 0.715 ) 4 4 7(iii) a = −6sin2t M1A1 M1 for −sin2t t = 0.615 → a = –5.66 or –5.65 or −2 8 A1 condone substitution of degrees

This question in 0606/23 Oct/Nov 2017

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

dy r 8 r(b) A curve is such that = 3 - 2 cos 5x . The curve passes through the point , dx b 5 5 l. (i) Find the equation of the curve. [4] ydx and hence evaluate (ii) Find r ydx . [5] r y 2y

13 marks

Mark scheme: 12(a) 2e x B1 seen 2 2e a 1 M1 Uses limits correctly for their integral − = 50 and sets = 50 2 2 Rearranges and takes logs to base e: M1 Using their integral 2 a = ln101 oe 1 A1 Allow any exact equivalent a = ln101 or ln 101 final answer 2 12(b)(i) 2 B2 B1 for −k sin5 x where k > 0 [ y = ]3 x − sin5 x [ + c ] 5 8π 3π 2  π  M1 = − sin 5 × + c   5 5 5  5  2 A1 y = 3 x − sin5 x + π 5 12(b)(ii)   2   B3 2 y d x = 3 x − sin5 x + π d x B2 for cos5 x oe nfww     ∫ ∫  5    25 2 2 3 x 3 x 2 and B1FT for + … + πx [ + c ] = + cos5 x + πx [ + c ] 2 2 25  π  M1 their F(π) – their F    2  16[.0] or 15.95 to 15.96 or A1 13π 2 2 − 8 25

This question in 0606/21 May/June 2019

9 y y = 16 - x 2 y = 7 O x The diagram shows the curve y = 16 - x2 and the straight line y = 7. Find the area of the shaded region. You must show all your working. [6]

6 marks

Mark scheme: 9 Roots of curve: (4, 0) or (−4, 0) oe B1 Intersections: (−3, 7) or (3, 7) oe B1 Correct strategy for finding area B1 2 x 3 M1 (16 − x )dx = 16 x − ∫ 3 2 x 3 (9 − x )dx = 9 x − or ∫ 3 F(b) – F(a) M1 148 1 A1 or 49 or 49.3[33…] 3 3

This question in 0606/23 May/June 2019

12 A particle P moves in a straight line such that, t seconds after passing through a fixed point O, its acceleration, a ms -2 , is given by a =- 6. When t = 0, the velocity of P is 18 ms -1 . (a) Find the time at which P comes to instantaneous rest. [3] (b) Find the distance travelled by P in the 3rd second. [3]

6 marks

Mark scheme: 12(a) v = −6t + c soi B1 v = −6t + 18 M1 −6t + 18 = 0 , t = 3 A1 12(b) − 6t 2 B1 s = + 18t soi 2 ( −3(3) 2 + 18(3) ) −−( 3(2) 2 + 18(2) ) M1 FT their s provided it is from an attempt to integrate 3 (metres) A1 Not from wrong working

This question in 0606/22 Feb/March 2020

7 Giving your answer in its simplest form, find the exact value of 4 10 (a) dx, [4] y 0 5 x + 2 nl 2 2 4 x + 2 (b) e d.x [5] y 0 ` j

9 marks

Mark scheme: 7(a) 2ln(5x + 2) B2 B1 for kln (5x + 2) 2 ( ln(22) − ln(2) ) oe soi M1 2 A1 2ln11 or ln121 or ln11 e 8 x + 4 d x M17(b)  ln 2 M1  1 8 x+ 4  e    8  0 1 ln2 8 4 4 M2 1 ln2 ( e × e − e ) oe M1 for ( e 8 + 4 − e 4 ) 8 8 A1 255e 4 or exact equivalent 8

This question in 0606/22 May/June 2020

12 (a) (i) Given that f( x) = , show that f l ( x) = tan x sec x. [3] cos x (ii) Hence find y `3 tan x sec x - 4 e 3 xj dx. [3] 5 p (b) Given that dx = ln 2, find the value of the positive constant p. [5] y 2 px + 10

11 marks

Mark scheme: 12(a)(i) −−( sin x ) B2 − sin x oe B1 for oe cos 2 x cos 2 x Correct completion to given answer: B1 dep on all previous marks having been tanxsecx awarded 12(a)(ii) 3 x B1 4 3e x = e 4 oe 3 x 3 x M1 3 4 3 4 e d x = − ke oe cos x − cos x 3 x A1 3 4 4 − e + c oe cos x 3 = ln 2 12(b) [ ln( px + 10) ]52 M1 ln(5 p + 10) − ln(2 p + 10) = ln 2 M1 ln  5 p + 10 = ln 2 M1    2 p + 10  5 p + 10 = 2(2 p + 10) M1 p = 10 A1

This question in 0606/23 May/June 2020

11 The equation of a curve is y = x 16 - x 2 for 0 G x G 4 . (a) Find the exact coordinates of the stationary point of the curve. [6] d 2 23 2(b) Find 16 - x and hence evaluate the area enclosed by the curve y = x 16 - x and the d x ` j lines y = 0, x = 1 and x = 3 . [5]

11 marks

Mark scheme: 11(a) d y 1 2 − 12 2 12 3 d 2 12 B1 for = x × (16 − x ) × ( − 2 x ) + (16 − x ) (16 − x ) d x 2 d x − 1 2 1 = 2 × ( − 2 x ) (16 − x ) 2 M1 for product rule A1 for all correct 1 2 3 dy d y 2 2 x M1 for setting = 0 and attempt to = 0 → 16 − x = ( ) 1 dx dx 2 2 16 − x ( ) solve 2 2 M1 for obtaining x = k x = 8 2 2, 8 ( ) A1 11(b) 1 2 M1 for attempt at chain rule 3 2 2 × ( − 2 x ) A1 for all correct unsimplified (16 − x ) 2 3 1 3 3 3 3   1 2 2 M1 for obtaining k − 2 dx = ( 16 − x 2 ) Area =   ( 16 − x 2 ) ( 16 − x 2 ) x 3 1  1 1  32 32  A1 for obtaining k = − 1 = −  7 − 15  = 13.2 3 3   A1 for 13.2

This question in 0606/21 Oct/Nov 2020

11 (a) (i) Find 6 dx . [2] 10x - 1 e dd` j 2 c 2 x 3 + 5 (ii) Find ` j d x . [3] dd x e (b) (i) Differentiate y = tan ( 3 x + 1) with respect to x. [2] c 10r sec 2 ( 3x + 1) (ii) Hence find dd - sin x d x . [4] r e 2 o e12 Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a)(i) (10 x − 1) −5 B2 (10 x − 1) −5 1 ( + c ) isw B1 for k ( + c ) , where k ≠ −×5 10 −5 10 11(a)(ii)  5 2 25  B1  4 x + 20 x +  d x   x  4 6 20 3 B2 B1 for any 3 terms correct x + x + 25ln x + c 6 3 11(b)(i) 3sec 2 (3 x + 1) B2 B1 for k sec 2 (3 x + 1) where k ≠ 3 11(b)(ii) sec 2 (3 x + 1) tan(3 x + 1) B1 dx =  2 6 oe, soi B1 − sin x d x = cos x oe   π   π  M1 F   − F   where  10   12  F(x) = k1 tan(3 x + 1) + k 2 cos x oe 0.322 or 0.3222[32...] rot to 4 figs A1

This question in 0606/22 Feb/March 2021

12 A particle P travels in a straight line so that, t seconds after passing through a fixed point O, its velocity, v ms -1 , is given by t v = for 0 G t G 2 , 2e t 2 v = e - for t 2 2 . Given that, after leaving O, particle P is never at rest, find the distance it travels between t = 1 and t = 3. [6]

6 marks

Mark scheme: 12 t t 2 B1 For 0 ≤≤t 2 : dt =  2e 4e t t t − − − t 3 B2 − 2 2 2 2 For t > 2: e d t = −2e + oe B1 for e dt = −2e ( + c ) oe  e  3 M2 M1 for − 3 1 3 2 + − −2e e 4e 1 − 2 3 [s(1) =] and [ s (3) = ] − 2e + their 4e e 3 or at least one term correct in the difference: −   3 1 1 1   3 2 + − − −2e OR   +   −  3  1 e e e 4e     2e 2 + their  −  −  e  4e   or for one bracket correct in:  − 3   1 1  2 3 1  −2e + their −  +  −   e e   e 4e  0.565 or 0.5654 to 0.56541 nfww A1 12 Alternative method (using def int): 2   t 2 M1* for =    4e 1  4 1  M1 for  −  (dep*)  4e 4e oe 3  − t  M1** for  − 2e 2    2  − 3  2 2 M1 for  −2e +  (dep**)  e oe  − 3   4 1  2 2 M1 for  −2e +  +  −   e   4e 4e  oe A1 for 0.565 or 0.5654 to 0.56541 nfww

This question in 0606/22 Feb/March 2021

12 DO NOT USE A CALCULATOR IN THIS QUESTION. y A 1 y = 2x + 1 5y = x - 1 C O B x 1 The diagram shows part of the curve y = and part of the line 5y = x - 1. 2x + 1 The curve meets the y‑axis at point A. The line meets the x‑axis at point B. The line and curve intersect at point C. (a) (i) Find the coordinates of A and B. [1] (ii) Verify that the x‑coordinate of C is 2. [2] (b) Find the exact area of the shaded region. [5] Question 13 is printed on the next page.

8 marks

Mark scheme: 12(a)(i) A(0,1) and B(1, 0) B1 12(a)(ii) 1 2 − 1 B2 1 [ y = ] and [ y = ] B1 for [ y = ] and 5 y = 2 − 1 oe 2(2) + 1 5 2(2) + 1 and 1 evaluates both expressions as 5 Alternative 1 (B2) 1 1 2 − 1 1 1 1 1 [ y = ] = or[ y = ] = B1 for = and 5 × = x − 1 oe 2(2) + 1 5 5 5 2(2) + 1 5 5 and 2 1 1 1 1 or −= and = oe 1 5 5 5 2 x + 1 solves 5 × = x − 1 oe to get x = 2 5 1 1 or = oe to get x = 2 5 2 x + 1 Alternative 2 (B2) 2x2− x – 6 = 0 B1 for (2x + 1)(x – 1) = 5 or 2x2− x – 6 = 0 and solves or factorises to get (2x + 3)(x – 2) and states x = 2 OR shows 2(22) – 2 – 6 = 0 oe Alternative 3 (B2) (2x + 1)(x – 1) = 5 oe B1 for (2x + 1)(x – 1) = 5 and shows (2 × 2 + 1)(2 – 1) = 5 12(b) 1 B1 × 1 × 0.2 oe 2 2 2 2  12 1  or − −  −  oe 5 × 2 5  5 × 2 5  1 B2 [ F( x ) = ] ln(2 x + 1) [+c] oe 1 1 2 B1 for ln 2 x + 1 or ln x + 0.5 2 2 1 or ln( x + 0.5) [+c] oe or k ln(2 x + 1) or k ln( x + 0.5) , k ≠ 0.5 or 0 2 F(2) – F(0) – their 0.1 M1 FT their F(x) providing at least B1 for integration of curve awarded 0.5ln5 − 0.1 or exact equivalent A1

This question in 0606/22 May/June 2021

8 y 5 y = + 2 x x - 1 2y = 9x x = 4 x 0 5 The diagram shows part of the curve y = + 2x , and the straight lines x = 4 and 2y = 9x . x - 1 5 (a) Find the coordinates of the stationary point on the curve y = + 2x . [5] x - 1 (b) Given that the curve and the line 2y = 9x intersect at the point (2, 9), find the area of the shaded region. [5]

10 marks

Mark scheme: 8(a) dy −2 B2 d −1 −2 = −5( x − 1) + 2 oe B1 for ( −5( x − 1) ) = k ( x − 1) dx dx soi 2 5 2 M1 dep on at least B1 ( x − 1) = or 2 x − 4 x − 3 = 0 2 10 A1 implies M1 x = 1 + oe, isw or 2.58[11…] 2 y = 2 + 2 10 oe, isw or 8.32 to 8.325 A1 8(b) [Area of triangle =] 9 soi B1 4 2 M2 5  2 x  M1 for dx = k ln( x − 1) [Area under curve = F(x) = ]  5ln( x − 1) +  oe  x − 1  2  2 k ≠ 0 soi or for 5ln x – 1 their 9 + F(4) – F(2) M1 dep on at least M1 21 + 5ln3 isw or 26.49 to 26.5 A1

This question in 0606/23 Oct/Nov 2021

11 y A B y = 6 + e 4x − 5 0 2 x The diagram shows part of the graphs of y = 6 + e 4 x - 5 and x = 2 . The line x = 2 meets the curve at the point B(2, b) and the line AB is parallel to the x‑axis. Find the area of the shaded region. [7]

7 marks

Mark scheme: 11 Length of rectangle: 6 + e3 M1 2 or  (6 + e 3 )x  0 Area of rectangle: A1 2(6 + e3) soi Area between curve and x-axis: M2 1 4 x− 5 2 M1 for e 2 4 x − 5  1 4 x − 5  4 (6 + e )d x = 6 x + e  0  4  0 1 4 x − 5 M1 dep on an attempt to integrate that results in Correct use of limits in their 6 x + e : – 5 ax + be4x 4 F(2) – F(0) their area of rectangle – their area between M1 dep on an attempt to integrate using correct curve and x-axis limits 7 3 1 A1 nfww e + or 35.2 4 4e 5 or 35.15 … isw 11 Alternative method Length of rectangle: 6 + e3 soi (M1) 2 3 4 x − 5 (M1) 0 ( their (6 + e ) − (6 + e ) ) dx 2 3 4 x − 5 (A1) 0 ( (6 + e ) − (6 + e ) ) dx oe 2 3 4 x − 5  3 1 4 x − 5  2 (M2) 1 4 x− 5  0 (e − e )d x =  e x − 4 e oe M1 for e  0 4 3 1 4 x − 5 (M1) dep on an attempt to integrate that results in Correct use of limits in their e x − e – ax + be4x 5; may be unsimplified 4 F(2) – F(0) 7 3 1 (A1) nfww e + or 35.2 4 4e 5 or 35.151374 … rot to 4 or more sf

This question in 0606/22 Feb/March 2022

11 y D x 2 y = e r x = 4 A y = cos 5x O B C x x r 2 The diagram shows part of the curves y = e and y = cos 5 x and part of the line x = . The 4 r curves intersect at A. The curve y = cos 5x cuts the x-axis at B. The line x = cuts the x-axis at C and 4 x 2 at D. Find the exact area of the shaded region, ABCD. [7] the curve y = e

7 marks

Mark scheme: 11  π  B1 B  ,0  soi  10  π x π M2 x x 04 e 2 dx  010 cos5 x d x M1 Integrates e 2 to ke 2 , k  0 M1 Integrates cos5x to k sin5 x , 1 k  0 or k  5 π π A2 A1 for each part correct  x  4  sin5 x  10 2  2e  0     5  0 π M1 M1 for correct attempt at    1 5π 1  0 8  2e  2e   sin  sin0 subtraction and for one    5 10 5  correct use of limits dependent on at least M1for integration and B1 π A1 8 11 2e  isw 5

This question in 0606/21 May/June 2022

11 y P Q B y = 1 A y = 1 + cos x O R x The diagram shows part of the line y = 1 and one complete period of the curve y = 1 + cos x , where x is in radians. The line PQ is a tangent to the curve at P and at Q. The line QR is parallel to the y-axis. Area A is enclosed by the line y = 1 and the curve. Area B is enclosed by the line y = 1, the line PQ and the curve. Given that area A : area B is 1 : k find the exact value of k. [9] Continuation of working space for Question 11. Question 12 is printed on the next page.

9 marks

Mark scheme: 11 Full, complete and actioned method to find 5 B1 for F( x )   (1  cos x )dx  x  sin x the first area:  1 1 oe A or A or B or B 2 2 B1 for the area of an appropriate rectangle or rectangles soi M1 for correct use of correct limits to find an appropriate area under the curve A1 for the accurate area under the curve 1 1 B1 for area A or A or B or B 2 2 OR M1 for attempting to integrate sinx or ±cosx M1 dep for using correct limits A1 for a correctly integrated expression with correct limits M1 for correct use of correct limits A1 for exact value A = 2 OR equivalent correct plan Full, complete and actioned method to find 3 1 B1 for Area(A + B) or Area(A + B) oe the second, corresponding area 2 M1 for using Area(A + B) to find A or B 1 1 or for using Area(A + B) to find A 2 2 1 or B oe 2 A1 for exact value OR B1 for  π  2  3π π  4   2 (1  cos x )dx      oe  0   2 2    M1 for correct use of correct limits A1 for exact value B = 2  2 OR equivalent correct plan k = π – 1 cao B1 dep on all previous marks; allow 2π  2 k = 2

This question in 0606/22 May/June 2022

10 y A y = 3 + 2x - x 2 B O C x The diagram shows part of the curve y = 3 + 2 x - x 2 . The point A lies on the curve and has an x‑coordinate of 1.5. The tangent to the curve at A meets the x‑axis at B. The curve meets the x‑axis at C. Find the area of the shaded region. [10]

10 marks

Mark scheme: 10 A(1.5, 3.75) soi B1 Factorises or solves 3  2 x  x 2  0 M1 C(3, 0) soi A1 implies M1 dy d y B1  2  2 x , when x = 1.5  1 dx d x y – 3.75 = (x – 1.5) oe B2 B1 for y – 3.75 = (x – 1.5) oe leading to B(5.25, 0) soi 1 B1 1  (5.25  1.5)  3.75 OR  (5.25  3)  2.25 2 2 2 5.25 2  1.5 2  2 5.25 2  32  5.25    5.25(1.5)   5.25    5.25(3)   2  2  2  2   3 2 3  M1 OR  (3  2 x  x )dx   F( x ) 1.5   3   1.5   3 2  3 x  x )d x   G( x ) 1.5  3  1.5(2.25   2 3  2 x x   3 x     3 x 3 2 x 3   2 3 1.5  2.25 x    oe  2 3 1.5  225 M1 dep on at least 2 correct terms in the  their   F(3)  F(1.5)  integration  32   81  OR  G(3)  G(1.5)   their   32  117 A1 or 3.65625 32

This question in 0606/23 May/June 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

= 0 the10 The acceleration, a ms -2 , of a particle at time t seconds is given by a =- 2 . When t ( t + 1 ) -1 velocity of the particle is 50 ms . (a) Find an expression for the velocity of the particle in terms of t. [4] (b) Find the distance travelled by the particle between t = 1 and t = 10 . [4]

8 marks

Mark scheme: 10(a)  −45  −45(t + 1) −1 B2  −45  −1 dt = k (t + 1) dt = + C or B1 for 2   2   v =  (t + 1) v =  (t + 1)     −1 better their 45 M1 50 = + C 0 + 1 A1  v = 45 + 5 t + 1 10(b)  F(t ) =  45ln(t + 1) + 5t 101 B2 B1 for (their 45)ln(t + 1) F(10) – F(1) M1 dep on at least B1 122 (m) or 121.7[13…] rot to 4 or more sf A1 dep on all previous marks awarded

This question in 0606/21 Oct/Nov 2022

9 The equation of a curve is y = kxe - 2 x , where k is a constant. dy (a) Find . [2] dx (b) Find the coordinates of the stationary point on the curve y = 10xe - 2 x . [3] (c) Use your answer to part (a) to find 4xe - 2 x dx . [3] y 1 (d) Find the exact value of 4xe - 2 x dx . [2] y0

10 marks

Mark scheme: 9(a) d −2 x −2 x B1 e = −2e soi ( ) dx dy −2 x −2 x B1 FT for use of product rule = ke − 2 kxe oe, isw dx −2 x  d −2 x  k .e + kx. their e  ( )   dx  Alternative d 2 x 2 x (B1) e = 2e soi ( ) dx d y k e 2 x − 2 kx e 2 x (B1) FT for use of quotient rule = oe, isw 2 x 2 x 2 k .e − kx. their 2e d x ( ) e 2 x ( ) 2 e 2 x ( ) 9(b) dy M1 FT their (a), provided of the form Equates = 0 and finds 10 – 20x = 0 oe −2 x −2 x 2 x 2 x dx me + nxe or me + nxe  1 5  A2 For both values:  ,  oe only 1  2 e  x = 0.5 and y = 5e− or 1.84 or 1.839[39...] rot to 4 or more sf 1 A1 for x = only 2 9(c) −2 xe −2 x − e −2 x + c B3 For fully correct answer or B2 for −2 xe −2 x − e −2 x or   4 xe −2 x dx  = −2 xe −2 x +  2e −2 x dx   or B1 for kxe −2 x =  ke −2 x − 2 kxe −2 x dx ( ) or better 9(d) −2e −2 − e −2 − 0 − e 0 oe M1 Correct substitution of limits into ( ) correct expression 3 2 A1 1 − or 1 − 3e− e 2

This question in 0606/23 Oct/Nov 2022

4 y = cosec 5x r Show that y = a sin bx , where a and b are integers, and hence find the value of y d x . [4] 5y0

4 marks

Mark scheme: 4 1 B1 [y =] = sin5x nfww cosec5 x π B1 FT their asin5x π 5  cos5 x  5  y0 dx =  −  5  0 1  π   1  M1 FT their a(kcosbx) where − cos  5   −− cos ( 5  0 )  1 5  5   5  k < 0 or k = 5 2 A1 5

This question in 0606/22 Feb/March 2023

9 A particle travels in a straight line so that, t seconds after passing a fixed point, its velocity, v ms -1 , is given by t v = e 4 for 0 G t G 4 , 16e v = 2 for 4 G t G k . t The total distance travelled by the particle between t = 0 and t = k is 13.4 metres. Find the value of k. [6]

6 marks

Mark scheme: t9 t 16e 16e B3 B2 for either correct 4 4 t t e dt  4e (  c ) and 2 dt  (  c )   t t 4 4 or B1 for e dt  ae (  c )  where a is a constant, a > 0 16e b or 2 dt  (  c )  t t where b is a constant b > 0 Correct plan: M1 4 t k 16e 4 e dt  2 dt  13.4 soi 0 4 t Correct equation: A1 dep on B3; implies M1 1 0  16e 16e  4e  4e    13.4    k 4  OR [When t = 4 s = 4e – 4  16e When t = k s =  8e  4 and] k  16e 13.4 =  8e  4 k k = 10 or awrt 10.0 A1 dep on all previous marks awarded

This question in 0606/22 May/June 2023

9 (a) Show that 3 dx = 36.6 . [3] 1 x (b) y 10y = 7 - 3x 1 y = 3x + 4 A O x 1 The diagram shows part of the line 10y = 7 - 3 x and part of the curve y = . 3x + 4 The line and curve intersect at the point A. Verify that the y-coordinate of A is 0.1 and calculate the area of the shaded region. [8]

11 marks

Mark scheme: 19(a) B1  x  4 23 3  x  4 x 3 x 8 5 2 M1 FT providing one term is correct in  3 3 3  2 1  x  6 x    5 1 x 3  4 x 3 3 53 23  3 53 23  A1 (8)  6(8)   (1)  6(1)  = 36.6 5  5    9(b) 1 M2 1 10(0.1) = 7 – 3x and 0.1 = M1 for 10(0.1) = 7 – 3x and 0.1 = 3 x  4 3 x  4 and oe evaluates both expressions as x = 2 oe 1 B1 [Area trapezium =] (0.1  0.7) their 2 oe or 2 7(their 2) 3(their 2) 2   [0] oe or 0.8 10 20  1  1 B2 1 d x  ln(3 x  4) [  c ] B1 for k ln(3 x  4) k  or for     3 x  4  3 3 1ln3 x  4 3 1 1 M1 dep on at least previous B1 ln(3(2)  4)  ln(3(0)  4) 3 3 their 0.8 – 0.3054…oe M1 dep previous M1; FT their 0.8 providing the difference results in a positive value 0.495 or 0.4945[69…] rot to 4 or more sf A1

This question in 0606/23 May/June 2023

3 (a) Find 4x + 5 - dx . [3] 2x + 3 3 1 (b) Hence find the exact value of 4x + 5 - dx , simplifying your answer. [3] + 3 o y1 e 2x

6 marks

Mark scheme: 3(a) 2 1 B3 2 1 2 x + 5 x − ln ( 2 x + 3 ) + c oe B2 for 2 x + 5 x − ln ( 2 x + 3 ) 2 2 2 1 or 2 x + 5 x − ln2 x + 3 + c 2 or 2 x 2 + 5 x + k ln ( 2 x + 3 ) + c with k ≠ 0 or B1 for 2 x 2 + 5 x +…+c 1 or ... − ln2 x + 3 2 or ... + k ln ( 2 x + 3 ) with k ≠ 0 3(b) Substitutes limits and subtracts in correct M1 FT their part (a) providing it includes a term order k ln ( 2 x + 3 ) with k ≠ 0  1   1  A1 18 + 15 − ln9 − 2 + 5 − ln5      2   2  1 9 1 5 A1 26 − ln or 26 + ln oe 2 5 2 9

This question in 0606/21 Oct/Nov 2023

7 A particle moves in a straight line. At time t seconds after passing through a fixed point O, its velocity, v ms -1 , is given by v = 10 sin 2 t - 6 cos 2 t . (a) Find an expression for the acceleration of the particle. [2] (b) Find the acceleration when t = r . [1] 4 (c) Find the first time at which the acceleration is zero. [3] (d) Find the displacement of the particle between t = r and t = r . [4] 4 2

10 marks

Mark scheme: 7(a) 20cos2t + 12sin2t 2 B1 for 20cos2t or 12sin2t 7(b) 12 B1 7(c) 20  M1 FT acos2t + bsin2t where a and b are non-zero tan 2t = their −  oe integers  12  t = 1.06 or 1.055[60…] rot to 3 or more dp A2 A1 for 2t = –1.030[3…] or 2t = 2.111[2…] 7(d) s = −5cos2t −3sin2t (+ c) B2 B1 for −5cos2t or −3sin2t π π M1 FT providing at least B1 previously awarded −5cosπ − 3sinπ − (−5cos − 3sin ) 2 2 or s = −5cos2t −3sin2t + 5 and s π − s π = 10 − 2 2 4 8 A1

This question in 0606/21 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

6 Find the exact area of the region enclosed by the curve y = e 2 - 4 x , the x-axis, the line x =- 0.25 and the line x = 0.5 . [4]

4 marks

Mark scheme: 6  1 2 − 4 x  0.5 B2 B1 for ke 2 −x4 , k −4 − e oe    4  −0.25 Correct use of correct limits: M1 1 2 − 4 x FT their − e providing B1 4 1 0  1 3  awarded − e −− e  oe 4  4  1 1 3 A1 − + e or exact equivalent, isw 4 4

This question in 0606/22 Feb/March 2024

9 y -1 x = 9 y = 4 + 3x - 1 ` j A O B x The diagram shows a sketch of part of the curve y = 4 + ( 3 x - 1 ) -1 and the line x = 9 . The point A has x-coordinate 1. The tangent to the curve at A meets the x-axis at the point B. Find the area of the shaded region. [10] Question 10 is printed on the next page.

10 marks

Mark scheme: 9 d y  2 M2 d y  2  (3 x  1)  3 M1 for  k (3 x  1) where k > 0 d x d x d y 3 9 A1 d y [When x = 1]  and y = FT their providing M1 has been dx 4 2 d x awarded Equation of tangent: M1 d y FT the value of their at x = 1 9 3 d x y   ( x  1) oe isw 2 4 and their y B(7, 0) oe A1 Area of triangle: M1 FT their 7 and their –0.75x + 5.25 of the 1 9 27 form mx + c if needed     their 7   1 or nfww or 2 2 2 3 21  3 21    49   (7)    8 4  8 4  [Area under curve = F(x) = ] B2 B1 for  1  9 1 d x  k ln(3 x  1) or 1 ln3 x  1 4 x  ln(3 x  1)   oe  3 x  1 3  3 1 Correct and actioned plan e.g. M1 dep on at least previous B1 and correct 27 plan or correct FT area of triangle oe F(9) – F(1) – their 2 1 1 A1 18  ln13 2 3 or 19.4 or 19.35[49...]

This question in 0606/23 May/June 2024