Understand integration as the reverse process — Paper 2 · IGCSE Mathematics - Additional 0606

5 (i) Show that [ 0. 4x 5 ( 0. 2 - ln 5 x)] = kx 4 ln 5x , where k is an integer to be found. [2] d x (ii) Express ln 125x 3 in terms of ln 5x . [1] ( x 4 ln 125 x 3) d x . [2] (iii) Hence find y

5 marks

Mark scheme: their 2 x 5(i) ( 4 5  − 5  M1 clearly applies correct form of product ) (0.2 − ln5 x ) + 0.4 x  their  oe or rule  5 x  4  4 5  5   oe their 0.4 x −  ( their 2 x ) ln 5 x + 0 .4 x  their     5 x   − 2 x 4 ln5 x isw A1 nfww 5(ii) 3ln5x or ln5 x + ln5 x + ln5 x B1 −2 x ln5 x ) dx oe 3 45(iii) −∫32 ( 4 M1 FT k = 2 from (i) allow for ( 2 x ln5 x ) dx 2 ∫ or, when k = −2, for 5 (0.2 − ln5 x ) x 4 ln5 x ) dx = −0.2 x ∫ ( 2 4 5 or − 3 x ln5 x ) dx = 0.4 x (0.2 − ln5 x ) oe 3 ∫ ( or, when FT k = 2, for 5 (0.2 − ln5 x ) x 4 ln5 x ) dx = 0.2 x ∫ ( 2 4 5 or 3 x ln5 x ) dx = 0.4 x (0.2 − ln5 x ) oe 3 ∫ ( 3 5 A1 nfww; implies M1 − ( 0.4 x (0.2 − ln5 x ) )[ + c ] oe isw cao 5 2 An answer of 0.6 x (0.2 − ln5 x ) following k = 2 from (i) implies M1 A0

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

9 (i) Show that 3 = 4 . [3] dx x x ln x (ii) Find the exact coordinates of the stationary point of the curve y = 3 . [3] x J ln x N(iii) Use the result from part (i) to find dx . [4] y KK 4 OO x L P

10 marks

Mark scheme: 9(i) d 1 B1 seen (ln x ) = and dx x d 3 2 d −3 −4 x = 3 x or x = −3 x d x dx Substitution of their derivatives into quotient rule M1 3 1 2 A1 correct completion x × − 3 x ln x d  ln x x oe  3 = 6 d x  x  x 9(ii) d y 1 M1 dy = 0 →−1 3ln x = 0 lnx = equate given to zero and solve d x 3 dx for lnx or x 1 A1 seen x = e 3 1 A1 seen y = 3e 9(iii) lnx 1 − 3lnx M1 use given statement in (i) dx oe 4 x 3 =∫ x 1 −1 B1 seen anywhere ∫ x dx = 4 3 x 3 ln x 1 ln x A2 A1 for each term ∫ x d x = − − (+C) oe 4 9 x 3 3 x 3

This question in 0606/23 Oct/Nov 2017

4 2 = 2x + 4 dx (x + 1) dy dy (i) Find , given that = 1 when x = 1. [3] dx dx (ii) Find y in terms of x, given that y = 3 when x = 1. [3]

6 marks

Mark scheme: 4(i) integrate: increase in powers of at least one term M1 * dy 2 1 A1 = x − + ( C ) 3 dx ( x + 1) 1 A1 C = 8 4(ii) integrate their (i): increase in powers of at least one M1 Dep* term 1 3 1 1 A1 two correct terms in x y = x + + x + ( D ) 2 3 2 ( x + 1) 8 29 A1 D = 12

This question in 0606/23 Oct/Nov 2018

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

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

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

7 It is given that 2 2 x = e + 2 for x 2- 1. d x ( x + 1 ) d y d y (a) Find an expression for given that = 2 when x = 0 . [3] d x d x (b) Find an expression for y given that y = 4 when x = 0 . [3]

6 marks

Mark scheme: 7(a)  dy  1 2 x −1 5 B3 1 2 x −1 = e − ( x + 1) + oe M2 for e − ( x + 1) + c oe    dx  2 2 2 or M1 for any two terms correct 1 2 x −1 from e , − ( x + 1) , + c 2 7(b) 1 2 x M1 [ y = ] e − ln ( x + 1) 4 5 M1 FT their c from (a), providing + their × x + d c ≠ 0 2 1 2 x 5 15 A1 [ y = ] e − ln ( x + 1) + x + oe 4 2 4

This question in 0606/22 Oct/Nov 2021

10 (a) Differentiate x ln x - 2x with respect to x. Simplify your answer. [2] d 2 y x + 1 2 dy e 2 e 3 2 (b) A curve is such that 2 = . It is given that = + 2e at the point e, + e . dx e x o dx 2 e 6 o Using your answer to part (a), find the exact equation of the curve. [8] Question 11 is printed on the next page.

10 marks

Mark scheme: 10(a) ln x  1 B2 1 B1 for [1]ln x  x    2  oe x 10(b) 1 B1 x   2 x dy x 2 M2 d y   ln x  2 x   c  M1 for  ...  ln x  ... or dx 2 d x dy x 2 for   ...  2 x dx 2 Substitution to find c: M1 FT their attempt to integrate e 2 e 2 dependent on at least M1  2e   lne  2e  c 2 2 [c = 1]  x 2  A1 dx   2 x  ln x  1  y    2  Integrates and uses (a) M1 Dependent on M2M1 3 2 FT error in c only x 2 x y    x ln x  2 x  C 6 2 e 3 2 e 3 2 M1 Substitution to find C  e   e  elne  2e  C Dependent on previous M1 6 6 x 3 2 A1 y   x  x ln x  2 x  e 6

This question in 0606/21 May/June 2022

12 A curve is such that 2 = 4 . Given that the gradient of the curve is at the point (1, −1), dx x 3 find the equation of the curve. [7]

7 marks

Mark scheme: 1 112 1 B2 1   2 2 4 4 2 x  2 x  1 x  2  x B1 for ( x  x ) or oe seen x 1 1  or for two terms correct in x 2  2  x 2 At least two terms correct in their M1 1  1 3 1 FT their x 2  2  x 2 providing at least  dy  2 2 2  x  2 x  2 x   c  two terms correct and no extra spurious    dx  3 terms 4 2  32   12  M1 dep on previous M1 and having an  1   2[1]  2 1   c arbitrary constant; condone one sign or 3 3 arithmetic slip 3 1 A1  2 10  x 2  2 x  2 x 2    dx   3 3  10 5 3 FT their  4 2 2 4 2 10 3  x  x  x  x  A 15 3 3 5 on previous M1 dep A1; condone one sign or  3  4  4  10 2 1 1   1[2]  12   1  A arithmetic slip 15 3 3 5 3 A1 4 2 2 4 2 10 4 y  x  x  x  x  oe 15 3 3 15

This question in 0606/22 May/June 2022

8 (a) Differentiate y = 2xe 4 x with respect to x. [2] (b) Hence find xe 4 x dx . [4] y

6 marks

Mark scheme: 8(a) d 4 x 4 x B1 (e )  4e d x 8 xe 4 x  2e 4 x isw B1 FT their ke4x 8(b) Use of part (a) B1 FT part (a) providing of form 2e 4 x dx oe kxe 4 x  2e 4 x  8 xe 4 x dx  2 xe 4 x   4 x 1 4 x 1 4 x M1 FT part (a) providing of form e d x oe xe dx  xe   kxe 4 x  2e 4 x 4 4  xe 4 x e 4 x A2 A1 for any 2 terms correct   c oe 4 16

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

3 (a) Differentiate ln ( x 3 + 3x 2 ) with respect to x, simplifying your answer. [2] x + 2 (b) Hence find dx . [2] y x ( x + 3)

4 marks

Mark scheme: 3(a) 3( x  2) 3 x  6 2 mark final answer or or simplified equivalent; x ( x  3) x 2  3 x 3 x 2  6 x B1 for oe x 3  3 x 2 23(b) 1ln( x 3  3 x 2 )  c 3  3 x 2 ) B1 for 1ln( x 3 3

This question in 0606/23 May/June 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

= cos 4x - . Given that = at the point , on the curve, find8 A curve is such that 2 r r r d x 4 dx 4 16 4 the equation of the curve. [7]

7 marks

Mark scheme: 8  π  1  π  B2 B1 for  cos  4 x  4  dx  4 sin  4 x  4    c   π  1 k sin  4 x   where k > 0 or k   4  4 3 1   3π  π  M1 FT their k providing B1 awarded  sin  4      c 4 4   16  4  1  π  1 2 M1 FT for  cos  4 x    x   A  16  4  2  π   1  A  m cos  4 x     their  x    4   2   π  FT their k sin  4 x    their c  4  providing at least B1 M1 awarded 1  π  1 5π 2 M1 FT for y  cos  4 x    x  oe, cao 16  4  2 32 π 1  3π π  1  3π   cos      A   4 16  4 4  2  16  FT  π   1   their m  cos  4 x     their  x  A  4   2  providing previous M1 awarded

This question in 0606/23 May/June 2024