Differentiate products and quotients of — Paper 2 · IGCSE Mathematics - Additional 0606

9 (a) Find e 2x + 1 dx . [2] y x d (b) (i) Given that y = , find y. [3] ln x d x J 1 1 1 N - + (ii) Hence find 2 2 OO d x . [3] y KK ln x ( ln x) x L P

8 marks

This question in 0606/22 Feb/March 2017

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

7 Differentiate with respect to x, (i) 1 + 4x 10 cos x , [4] ^ h e 4x - 5 (ii) . [4] tanx

8 marks

Mark scheme: 7(i) 9 M1 k (1 + 4 x ) 9 A1 4 × 10 (1 + 4 x ) or better (1 + 4 x )10 ( their − sin x ) + M1 clearly applies product rule cos x their 4 × 10 × (1 + 4 x ) 9 ( ( ) ) 10 9 A1 all correct (1 + 4 x ) ( − sin x ) + cos x 4 × 10 × (1 + 4 x ) ( ) 7(ii) d 4 x − 5 4 x − 5 B1 e = 4e soi ( ) dx d 2 B1 ( tan x ) = sec x soi dx clearly applies correct form of quotient rule M1 or correct form of product rule to 4 x − 5 4 x − 5 e 4 x − 5 (tan x ) −1 tan x ( their 4e their sec 2 x ) − e ( ) ( tan x ) 2 4e 4 x − 5 (tan x ) −1 + e 4 x − 5 (tan x ) −2 × sec 2 x 4 x − 5 4 x − 5 A1 all correct tan x ( 4e sec 2 x ) − e ( ) isw ( tan x ) 2

This question in 0606/23 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

7 Differentiate with respect to x (i) 4x tan x , [2] e 3x + 1 (ii) 2 . [3] x - 1

5 marks

Mark scheme: 7(i) 4tan x + 4 x sec 2 x isw B2 Fully correct B1 for one correct term as part of e.g. a sum of 2 terms 3 x +1 3 x +1 B17(ii) d ( e ) = 3e dx 3 x +1 3 x +1 M1 ( x 2 − 1)( their 3e ) − their (2 x )e 2 2 ( x − 1) 3 x +1 3 x +1 A1 ( x 2 − 1)( 3e ) − 2 xe oe isw 2 2 ( x − 1)

This question in 0606/21 May/June 2018

7 Differentiate with respect to x (i) 4x tan x , [2] e 3x + 1 (ii) 2 . [3] x - 1

5 marks

Mark scheme: 7(i) 4tan x + 4 x sec 2 x isw B2 Fully correct B1 for one correct term as part of e.g. a sum of 2 terms 3 x +1 3 x +1 B17(ii) d ( e ) = 3e dx 3 x +1 3 x +1 M1 ( x 2 − 1)( their 3e ) − their (2 x )e 2 2 ( x − 1) 3 x +1 3 x +1 A1 ( x 2 − 1)( 3e ) − 2 xe oe isw 2 2 ( x − 1)

This question in 0606/23 May/June 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

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

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

7 (i) Given that y = x x 2 + 1 , show that = p , where a, b and p are positive constants. [4] dx 2 x + 1 ` j (ii) Explain why the graph of y = x x 2 + 1 has no stationary points. [2]

6 marks

Mark scheme: 7(i) B2 2 = x 2 + 1 2 × 2 x = kx x 2 + 1 B1 for d ( x d ( x 2 + 1 ) 2 + 1 ) 1 ( ) − 1 ( ) − 1 dx 2 dx where k ≠ 1 2 M1 x + 1 −   1 2 1 2 x + 1 + x × their  × 2 x  ( )   2    d y  2 x 2 + 1 A1 =   1  d x  2 2 x + 1 ( ) 1 or a = 2, b = 1, p = nfww 2 7(ii) Complete argument B2 FT their positive a and b d y e.g. For stationary points = 0 and when a and B1 FT for a partially correct argument d x 2 d y b are positive, ax + b cannot be 0 e.g. Because cannot be 0. d x or 2x2 cannot be −1

This question in 0606/22 Feb/March 2019

2 Two variables x and y are such that y = for x 2 0. x3 dy 1 - 3 ln x (i) Show that = 4 . [3] dx x (ii) Hence find the approximate change in y as x increases from e to e + h, where h is small. [2]

5 marks

Mark scheme: 2(i) d 1 B1 (ln x ) = soi dx x 3  1 2 M1 x  − 3 x ln x d y  x  = d x 3 2 x ( ) −3  1 −4 or x + −3 x ln x   ( )  x  Completion to given answer: A1 dy 1 − 3ln x = dx x 4 2(ii)  1 − 3lne  M1  4  h  e  −h2 oe or −0.0366h awrt A1 e 4

This question in 0606/21 May/June 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

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

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

10 (a) Find ( e x + 1 ) 3 d x . [2] (b) (i) Differentiate, with respect to x, y = x sin 4x . [2] r 3 1 r 3 (ii) Hence show that 4x cos 4xdx = - . [4] yr 8 6 4

8 marks

Mark scheme: 10(a) 1 3 x + 3 1 3 3 x B2 3 x + 3 3 3 x 1 e + c or e × e + c nfww B1 for k e or ke × e where k ≠ or 0 3 3 3 10(b)(i) d(sin4x ) B1 = 4cos4 x soi dx Applies correct form of product rule: B1 FT their 4 cos 4x if possible 4x cos 4x + [1] sin 4x isw M1 FT use of their mx cos 4x + n sin 4x where m10(b)(ii)  (4 x cos4 x )d x =  x sin 4 x − sin 4 xd x     and n are constants 1 A1 x sin4 x + cos4 x [ + c ] soi 4 π  π  1  π  A1 sin  4 ×  + cos  4 ×  − 3  3  4  3   π  π  1  π    sin  4 ×  + cos  4 ×    4  4  4  4   Correct completion to given answer A1 1 π 3 − 8 6

This question in 0606/22 May/June 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

5 It is given that y = 3 tan 2 x for 0° 1 x 1 360° . dy 2 (a) Show that = m tan x sec x where m is an integer to be found. [2] dx dy (b) Find all values of x such that = 3 sec x cosec x . [5] dx

7 marks

Mark scheme: 5(a) dy 2 B2 B1 for = 6tan x sec x dx d 2 1 2 (tan x ) = 2(tan x ) sec x dx 5(b) 6 tan x sec 2 x − 3sec x cosec x = 0 B1 NB division by secx is B0 3sec x (2 tan x sec x − cosec x ) = 0 oe 2 tan 2 x = 1 oe B1 1 M1 FT tan 2 x = k where k > 0 tan x = [ ± ] or [±] 0.707[1…] 2 35.3 or 35.2643… rot to 2 or more dp A2 no extras in range 215.3 or 215.2643… rot to 2 or more dp 144.7 or 144.7356… rot to 2 or more dp A1 for any two correct answers 324.7 or 324.7356… rot to 2 or more dp

This question in 0606/21 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

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

5 You are given that y = . cos 2x d y k sin 2x (a) Show that = 2 where k is a constant to be found. [2] d x cos 2x d y 5 r (b) Find the values of x such that = for 0 1 x 1 . [4] d x sin 2x 2

6 marks

Mark scheme: 5(a) dy −2 2sin2 x B2 B1 for − (cos2 x ) −2  m sin 2 x = −(cos2 x ) −2sin2 x = dx cos 2 2 x 0[cos2 x ] − ( m sin2 x ) or where m = 2 dy 0[cos2 x ] −−( 2sin2 x ) 2sin2 x 2 or = = cos 2 x dx cos 2 2 x cos 2 2 x or m < 0 5(b) 2tan22x = 5 M1 FT their k or 7cos22x = 2 or 7sin22x = 5 5 A1 tan2 x =  2 2 or cos2 x =  7 5 or sin2 x =  7 0.503 or 0.5034[26…] rot to 4 or more sf A2 A1 for either, ignoring extras 1.07 or 1.067[36…] rot to 4 or more sf and no extras in range

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

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

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