Use differentiation to find stationary points — Paper 2 · IGCSE Mathematics - Additional 0606

12 The diagram shows a shape made by cutting an equilateral triangle out of a rectangle of width x cm. x cm The perimeter of the shape is 20 cm. (i) Show that the area, A cm2, of the shape is given by J N 6 + 3 A = 10x - KK OO x 2 . [3] 4 L P (ii) Given that x can vary, find the value of x which produces the maximum area and calculate this maximum area. Give your answers to 2 significant figures. [4]

7 marks

This question in 0606/22 Feb/March 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

9 In this question, all lengths are in metres. y 2r The diagram shows a window formed by a semi-circle of radius r on top of a rectangle with dimensions 2r by y. The total perimeter of the window is 5. (i) Find y in terms of r. [2] 2 r r 2 (ii) Show that the total area of the window is A = 5r - - 2r . [2] 2 (iii) Given that r can vary, find the value of r which gives a maximum area of the window and find this area. (You are not required to show that this area is a maximum.) [5]

9 marks

Mark scheme: 9(i) 2 y + 2r + πr = 5 B1 5 − 2 r −πr B1 Dep y = 2 9(ii) πr 2 M1 A = 2 yr + 2 πr 2 A1 = r ( 5 − 2 r −πr ) + 2 2 πr 2 = 5 r − 2 r − 2 9(iii) M1 differentiate dA A1 = 5 −πr − 4 r dr dA M1 set to zero and attempt to solve = 0 dr 5 A1 r = = 0.7 π+ 4 A = 1.75 A1

This question in 0606/21 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

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

6 A curve has equation y = 3x - 5 - 2x . dy d2 y (i) Find and 2 . [4] dx dx (ii) Find the exact value of the x-coordinate of each of the stationary points of the curve. [2] (iii) Use the second derivative test to determine the nature of each of the stationary points. [2]

8 marks

Mark scheme: 6(i) 9(3 x − 5) 2 − 2 isw B2 B1 for k (3 x − 5) 2 k ≠ 9 seen 54(3 x − 5)[1] isw B2 B1 for k (3 x − 5)[1] k ≠ 54 seen 6(ii) Solves their 9(3 x − 5) 2 − 2 = 0 M1 5 2 A1 [ x = ] ± or exact equivalent 3 9 6(iii) 5 2 M1 Substitutes their + 3 9 5 2 or their − into 3 9 their 54(3 x − 5)[1] and considers sign of result 5 2 d 2 y A1 When x = + > 0 3 9 d x 2 so minimum 5 2 d 2 y and when x = − < 0 3 9 d x 2 so maximum

This question in 0606/23 May/June 2019

11 A particle travelling in a straight line passes through a fixed point O. The displacement, x metres, of the particle, t seconds after it passes through O, is given by x = 5t + sin t . (i) Show that the particle is never at rest. [2] (ii) Find the distance travelled by the particle between t = r and t = r . [2] 3 2 (iii) Find the acceleration of the particle when t = 4. [2] (iv) Find the value of t when the velocity of the particle is first at its minimum. [2] Question 12 is printed on the next page.

8 marks

Mark scheme: 11(i)  d x  B1 v = = 5 + cos t    d t  5 + cos t ≠ 0 (and so never at rest) oe B1 11(ii)  π   π  M1 x = 5   + sin   or  3   3   π   π  x = 5   + sin   seen  2   2  2.75 to 2.752 A1 11(iii)  d v  M1 FT their v provided of the form k ± cos t a = = − sin t    d t  [ t = 4, a = − sin 4 = ] 0.757 or A1 0.7568[024…] 11(iv) Valid method soi e.g. M1 their ( − sin t ) = 0 or cos t = −1 sketch of v = 5 + cos t t = π A1

This question in 0606/23 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

7 x = 2 y 6 y = x + 2 (3 x + 2) x O 6 The diagram shows part of the curve y = x + 2 and the line x = 2 . (3x + 2) (i) Find, correct to 2 decimal places, the coordinates of the stationary point. [6]

10 marks

Mark scheme: 7(i) evidence of differentiation (3x + 2)–3 M1 –12(3x + 2)–3 × 3 A1 may use PR or QR on fraction part +1 B1 d y M1 1 –36(3x + 2)–3 = 0 set their = 0 d x x = 0.43 nfww A1 y = 0.98 only A1 7(ii) − 2 B1 oe 3 x + 2 1 2 B1 x 2  −2   −2  M1 insert correct limits into their two term + 2 −     integral and subtract two non-zero  6 + 2   2  terms in correct order 2.75 nfww A1 2.75 following B1 B1implies M1

This question in 0606/23 Oct/Nov 2019

12 (a) Find the x-coordinates of the stationary points of the curve y = e 3 x ( 2x + 3) 6 . [6] (b) A curve has equation y = f( x) and has exactly two stationary points. Given that f ll ( x) = 4 x - 7 , f l ( 0.5) = 0 and f l ( 3) = 0, use the second derivative test to determine the nature of each of the stationary points of this curve. [2] (c) In this question all lengths are in centimetres. h x 4x The diagram shows a solid cuboid with height h and a rectangular base measuring 4x by x. The volume of the cuboid is 40 cm3. Given that x and h can vary and that the surface area of the cuboid has a minimum value, find this value. [5]

13 marks

Mark scheme: 12(a) 3 x B1 d e ( ) 3 x = 3e d x 6 M1 d ( 2 x + 3) 5 = k ( 2 x + 3) dx their (3e3x) (2x + 3)6 + M1 (e3x) (their 12 (2x + 3)5) (3e3x) (2x + 3)6 + (e3x) (12 (2x + 3)5) A1 (3e3x) (2x + 3)5 (2x + 7) = 0 M1 x = –1.5, –3.5 A1 12(b) x = 0.5 f ′′ ( 0.5 )[ = −5 ] < 0  max B2 B1 for either one correct x = 3 f ′′ ( 3 )[ = 5 ] > 0  min 12(c) 10 B1 h = x 2 2  10  B1 S = 8 x + 10 x  their 2   x  d S − 2 M1 = 16 x − 100 x oe d x 2 25 A1 d S 16 x − 100 x− = 0, x = 3 oe FT their = 0 if possible 4 d x 81.4 or 81.4325… rot to four or more A1 figs

This question in 0606/21 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

9 x m B C D 300 m A E 400 m The rectangle ABCDE represents a ploughed field where AB = 300 m and AE = 400 m . Joseph needs to walk from A to D in the least possible time. He can walk at 0.9 ms -1 on the ploughed field and at 1.5 ms -1 on any part of the path BCD along the edge of the field. He walks from A to C and then from C to D. The distance BC = x m . (a) Find, in terms of x, the total time, T s, Joseph takes for the journey. [3] (b) Given that x can vary, find the value of x for which T is a minimum and hence find the minimum value of T. [6]

9 marks

Mark scheme: 9(a) 2 2 B1 ( AC = ) 300 + x seen isw 2 2 M1 using clearly indicated value for their 300 + x time for AC = oe AC or their CD 0.9 400 − x or time for CD = oe 1.5 2 2 A1 300 + x 400 − x T = + oe seen 0.9 1.5 isw 9(b) − 1 B2 accept unsimplified; 2 2 if incorrect allow B1 for correct dT 1 ( 300 2 + x 2 ) = × 2 x − oe 1 dx 2 0.9 3 2 2 ± 2 differentiation of 300 + x ( ) dT M1 d T set their = 0 must be a function of x dx d x 2 2 2 A1 equation in x2 with square root removed 25 x = 9 300 + x oe ( ) x = 225 (m) A1 T = 533 (s) or 1600/3 (exact value) A1 or 8 min 53 s

This question in 0606/23 Oct/Nov 2020

11 In this question all lengths are in centimetres. 4 3 2 The volume and surface area of a sphere of radius r are rr and 4rr respectively. 3 x y The diagram shows a solid object made from a hemisphere of radius x and a cylinder of radius x and height y. The volume of the object is 500 cm3. (a) Find an expression for y in terms of x and show that the surface area, S, of the object is given by 5 2 1000 S = r x + . [4] 3 x (b) Given that x can vary and that S has a minimum value, find the value of x for which S is a minimum. [4]

8 marks

Mark scheme: 11(a) 4 3 2 M1 500 = πx + πx y oe 6 1  4 3  A1 if first M0, SC1 for oe, isw 500 − y =  πx  1  4 3  6   πx 2 y = 2  500 − πx  oe seen πx  3  2 2  500 2  M1 dep on first M1 S = 2πx + πx + 2πx  2 − x   πx 3  Correct completion to given answer: A1 5 2 1000 S = πx + 3 x 11(b) 10 1000 B2 B1 for each term Differentiates S: πx − oe 3 x 2 10 1000 M1 d S πx − = 0 and attempt to solve FT their providing at least B1 awarded 2 3 x d x 300 A1 x = 3 isw or 4.57[07...] nfww π

This question in 0606/22 May/June 2021

11 The volume, V, of a cone with base radius r and vertical height h is given by rr 2 h . 3 The curved surface area of a cone with base radius r and slant height l is given by rrl . A cone has base radius r cm, vertical height h cm and volume V cm3. The curved surface area of the cone is 4r cm2. 2 16 2 (a) Show that h = 2 - r . [4] r (b) Show that V = r 16r 2 - r 6 . [2] 3 (c) Given that r can vary and that V has a maximum value, find the value of r that gives the maximum volume. [5]

11 marks

Mark scheme: 11(a) 4 B1 l = r h 2 = l 2 − r 2 or l 2 = r 2 + h 2 M1 2 2 M1 FT their l ; dep on previous M1 2 4 2 4 2 2 = h =  − r or  r + h r r 2 16 2 2 2 or l = and h = l − r r 2 2 16 2 A1 Correct, convincing completion to h = − r r 2 Alternative method 2 2 (B1) l = r + h 2 2 (M1) πr r + h = 4π 2 (M1) 2 4 2 2 ( r + h ) = r 2 16 2 (A1) Correct, convincing completion to h = − r r 2 11(b) π 2 16 2 M1 r 2 − r 3 r A1 π 4  16 2  r  2 − r  3  r  π 2 6 and correct completion to 16r − r 3 111(c) − d V π  1 2 6  5 B3 B2 for 2 1 =  (16 r 32 r − 6 r ) oe − r )  ( 2 6 − d r 3  2 16 r − r 32 r − 6 r   k ( 2 5 ) ( ) where k is a constant and k ≠ 0 or B1 for 1 2 6 − 2 k 16 r − r × ( f ( r ) ) where ( ) f(r) ≠ 32 r − 6 r 5 d V M1 FT their f(r) = ar +br5 Equates their to 0 and solves as far as for a, b ≠ 0 d r r4 = … r = 1.52 or 1.519[67…] rot to 4 or more sf or A1 2 oe 4 3

This question in 0606/21 Oct/Nov 2021

4 (a) Find the x-coordinates of the stationary points on the curve y = 3 ln x + x 2 - 7 x , where x 2 0 . [5] (b) Determine the nature of each of these stationary points. [3]

8 marks

Mark scheme: 4(a) dy 3 B2 B1 for the first term correct and = + 2 x − 7 one other term correct dx x or for all terms correct with extra terms seen dy M1 Equates their to zero and rearranges to 3-term dx quadratic in x Solves their 3-term quadratic M1 Dep on previous M1 x = 0.5 , 3 nfww isw A1 no extra solutions 4(b) d 2 y 3 M1 dy = − + 2 FT their providing B1 dx 2 x 2 dx earned in (a) d 2 y d 2 y A1 x = 0.5 , < 0 → max or = −10 → max dx 2 dx 2 d 2 y d 2 y 5 A1 x = 3 , > 0 → min or = → min dx 2 dx 2 3 Alternative method Considers gradient at x ‒ h and x + h for x = 0.5 or (M1) dy FT their providing B1 x = 3 [where h is small] dx earned in (a) or Considers y-values at x ‒ h and x + h for x = 0.5 or x = 3 [where h is small] Correct conclusion for one turning point (A1) max at x = 0.5 or min at x = 3 Correct method and conclusion for second turning point (A1)

This question in 0606/22 Oct/Nov 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

12 In this question all lengths are in centimetres. O R 12 P Q C h A B 8 The diagram shows a right triangular prism of height h inside a right pyramid. The pyramid has a height of 12 and a base that is an equilateral triangle, ABC, of side 8. The base of the prism sits on the base of the pyramid. Points P, Q and R lie on the edges OA, OB and OC, respectively, of the pyramid OABC. Pyramids OABC and OPQR are similar. 3 3 2 (a) Show that the volume, V, of the triangular prism is given by V = ( ah + bh + ch ) where a, 9 b and c are integers to be found. [4] (b) It is given that, as h varies, V has a maximum value. Find the value of h that gives this maximum value of V. [3]

7 marks

Mark scheme: 12(a) PQ 12 − h M1 = oe 8 12 2  or  12 − h = Area Δ PQR oe  12  16 3 8(12 − h ) A1 PQ = oe 12  12 − h  2 or Area ΔPQR = 16 3    12 oe 1  8(12 − h )  2 π M1 FT their PQ or Area ΔPQR providing of correct V = ×   × sin × h oe structure 2  12  3 2  or 16 3  12 − h × h  12  3 3 2 A1 V = ( h − 24 h + 144 h ) 9 12(b) d V 3 2 B1 3 3 2 = ( 3h − 48 h + 144 ) FT their V = ( h − 24 h + 144 h ) if of the d h 9 9 same structure 3 2 M1 their ( 3h − 48 h + 144 ) = 0 and must be a 3-term quadratic; must be an attempt 9 at a derivative factorises/solves 4 oe identified as the only solution; cao A1

This question in 0606/22 Feb/March 2022

8 The function f is defined by f ( x) = 3 sin 2 x - 2 cos x for 2 G x G 4 , where x is in radians. (a) Find the x-coordinate of the stationary point on the curve y = f ( x) . [5]

10 marks

Mark scheme: 8(a) 3(2sin x cos x ) ( 2sin x )   0  B2 B1 for the correct derivative for either term; may be unsimplified or better Factors out sin x and equates to 0: B1 FT an expression of the form 2 (sin x )(3cos x  1)  0 oe asinxcosx +bsinx for non-zero constants a and b sin x  0  theira cos x theirb  M1 FT an expression of the form asinxcosx +bsinx for non-zero constants a and b; dep on previous B1 x = π as only solution A1 dep on all previous marks awarded If B2 B0 M0 then SC1 for dividing by 1 1  sinx and using cos    to find  3  x = 1.91 and 4.37 and clearly reject them. 8(b) For use of sin2x = 1 cos2x to write in terms M1 of cosx only e.g. 3(1 cos2x ) – 2cosx = 1  3cosx Collects terms A1 e.g. 3cos2x – cosx 2 = 0 Factorises the left-hand side or solves: M1 (3cosx + 2)(cosx – 1) = 0 x  2.3[0] , x = 3.98 and no extras A2 not from wrong working A1 for either; not from wrong working

This question in 0606/22 May/June 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

! 0 .2 A curve has equation y = 32x 2 + 2 where x 8x (a) Find the coordinates of the stationary points of the curve. [5] (b) These stationary points have the same nature. Use the second derivative test to determine whether they are maximum points or minimum points. [3]

8 marks

Mark scheme: 2(a) dy 2 x 3 B2  64 x  oe, isw dy 3 dx 8 B1 for  64 x  kx or dx dy 2 x 3  kx  where k is a non-zero d x 8 constant or dy 2 3 SC1 for  64 x  x  c dx 8 d y M1 FT their derivative providing it has two their = 0 and attempt to solve terms and at least one term is a correct dx power of x (0.25, 4), (0.25, 4) nfww, isw A2 A1 for either stationary point correct or for x =  0.25 nfww dy 3 or, if  64 x  16 x , then award dx  1 65  SC2 for  , oe or    2 4  SC1 for either of these stationary points 1 or x  oe 2 2(b) Correct second derivative: M1 d y 3 2 FT their = mx  nx where m  0 d y 3 4 dx 2  64  x oe, isw dx 4 and n  0 seen in part (a)  4  4 A2 dep on x = 0.25 nfww in part (a) 3  1  3  1  64  = 256 or 64  > 0     4  4  4  4  or A1 dep on x = 0.25 or x = –0.25 nfww in d 2 y d 2 y part (a) for correctly showing or stating when x = 0.25 2  256 or 2  0 oe 2 dx dx d y  3  is positive  64    and minimum [points] oe dx 2 4 x 4   OR 4  4 3  1  3  1  64    256 or 64    0     4  4  4  4  or d 2 y d 2 y when x = 0.25 2  256 or 2  0 oe dx dx and minimum [points] oe OR d 2 y 3 2  64  4 and this is positive for any dx 4 x value of x and minimum [points]

This question in 0606/22 May/June 2023

6 DO NOT USE A CALCULATOR IN THIS QUESTION. (a) Given that x - 3 and x + 1 are both factors of 2x 3 - 3x 2 - 8x - 3 , solve the equation 2x 3 - 3 x 2 - 8x - 3 = 0 . [2] (b) The polynomial p ( x) = x 3 + ax 2 + bx + c , where a, b and c are constants, has remainder - 5 4 when divided by x - 1. The curve y = p ( x) has stationary points at x = and x = 2 . 3 (i) Find the values of a, b and c. [7] (ii) Hence use the second derivative test to show that the stationary point at x = 2 is a minimum. [2]

11 marks

Mark scheme: 6(a) (2x + 1)(x – 3)(x + 1) nfww M1 Correct method leading to A1 1 x =  , x = 3, x = 1 2 6(b)(i)  p ( x )   3 x 2  2 ax  b   0  B1 2 B1 OR forms the product (3x – 4)(x – 2) = 0  4   4  3    2 a    b  0  3   3  3 2 2  2 a 2  b  0 B1 OR multiplies out to find 3 x 2  10 x  8  0 Solves to find the value of one unknown M1 FT their linear equations in a and b oe OR compares coefficients to state a value of a or b a = 5, b = 8 A1 [p(1)=] 1 + a + b + c = –5 oe, soi M1 [1 – 5 + 8 + c = 5] A1 c = 9 6(b)(ii)  p( x )   6 x  2(their a ) soi M1 FT their a 6(2) – 10 = 2 > 0 [therefore minimum] A1

This question in 0606/23 May/June 2024