Cambridge A Level Mathematics 9709 — 2025 May/June Paper 1 · Variant 3

12 1 A curve has equation y = 2 x + . x 2 Find the equation of the tangent to the curve at the point (-2, -1). Give your answer in the form y = mx + c . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 2 + (12 )( −2 ) x− 3 B1 Correct differential but can be unsimplified. −3 *M1 Substitute x = −2 into their differential, which must contain 2 + (12 )( −2 )( −2 )  = 5  x−3. y −−( 1) DM1 Attempt to find equation of tangent through ( −2, − 1) with Either ( their 5 ) = x −−( 2 ) their numerical gradient obtained as described above. or −=1 ( their 5 ) −( 2 ) + c  c = y = 5x + 9 A1 4

More questions on Differentiation

2 The first two terms of a geometric progression are 4 sin 2i, 8 sin 3i, where i is an angle such that 0 1 i 1 1 r . 6 Given that the sum to infinity of the progression is 1, find the value of i. Give your answer in the form 2 sin -1 k , where k is a rational number. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 3 B1 A correct unsimplified expression for r, this may only be seen 8sin  r = 2  = 2sin  in a correct S∞ 4sin  2 *M1 OE 4sin  1 = Correct S∞ with their expression for r. 1 − 2sin  2 Or 3 2 B1 8sin  4sin 1 r = 2 and = 4sin  1 − r 2 3 *M1 2 8sin  1 − 8sin  = 2 4sin  Then Expect 8sin 2 + 2sin − 1  = 0  DM1 OE and attempt to solve. Attempt to solve a three-term quadratic to find at least one value of sin. This can be implied by a final answer of = 0.253 c AWRT −1 1 A1 OE  =  sin ,0.25 Clearly identify one answer only. 4 1 1 Accept sin =  k = etc. 4 4 4

More questions on Series

3 a 3 Given that e 2 + 2o d x = 12 , find the value of the constant a. [4] y 1 ( 4x - 3) .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3  1  −1 B1 OE a (4 x − 3) + 2 x   Do not accept ( −+2 1) as equivalent to −1.  −1 4  Apply correct limits, x = 3 & 1, to their integral *M1 Their integral must contain ( 4 x − 3 ) −1 . Condone using x = 1 and 3. − a  − a   8a  DM1 OE + 6 − + 2 = 12  + 4 = 12     Equate their linear unsimplified expression in a to 12. 36  4   36  a = 36 A1 4

More questions on Integration

4 (a) Find the first three terms in the expansion of b2 - 3 xl in ascending powers of x. [3] 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use your answer to part (a), with a suitable value of x, to find an approximation to .19855. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 0 2 B1B1B1 Coefficients must be simplified. 32  −120 x +180 x    x  Ignore further terms in the expansion. 5  15 45 2  SC B3: 2  1 − x + x .  can be awarded full marks,  4 8  ISW. 3 4(b) 1 B1 Identify 0.01 as the only value to substitute, condone any x = 0.01 or method. 100 2 x = ( their 0.01) M1 Clear use of their x-value from their solution of ( Their 32 − 120 x + 180 x ) with 3 2 − x = 1.985 and their answer to (a). 2 Expect 32 – 120(0.01) + 180(0.01)2 = 32 – 1.2+ 0.018. Condone substitution into further terms in the expansion. 30.8[18] A1 Correct answer to at least 3sf coming from substitution into three or more terms. ‘Correct’ answers without appropriate working get A0. 3

More questions on Series

5 Solve the equation 4 sin i tan i = 1 + 5 cos i for - 180° 1 i 1 180° . [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 sin  M1 Replace tan  with cos Replace sin 2  with 1 − cos 2  leading to a 3-term quadratic *M1 To obtain a 3-term quadratic in cos. Accept a cubic that has a common factor of cos. Condone +/–sign errors during simplification. 9cos 2 + cos− 4  = 0  A1 OE Attempt to solve (their quadratic in cos ) using a valid method DM1 Only available for solution of a three-term quadratic. Correct values of cos are 0.6134 and –0.7245. Any two of 52.2,  136.4 A1 AWRT as final answers. Any two correct answers between −180 and 180 as their final answers. All four values A1 Condone other answers outside the range but no others between −180 and 180. SC after M1 *M1 A1 DM0, correct answers can score B1 B1. 6

More questions on Trigonometry

6 An arithmetic progression has first term a and common difference 2. The N th term is 55 and the sum of the first 3N terms is 5760. Find the values of N and a. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 a + 2 ( N − 1) = 55 B1 SOI 1 B1 SOI  3 N ( 2 a + 2 ( 3 N − 1) ) = 5760 2 Eliminate either a or N M1 Using a valid substitution, expect either ( 57 − a ) a = 57 − 2 N or N = . 2 Condone +/–sign errors in their substitution or during simplification. N 2 + 56 N − 1920 or a 2 − 226 a + 1953  = 0  A1 All terms need not be collected on one side. N = 24 , a = 9 A1 A1 Ignore presence of extra solutions for N and a. 6

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dy 2 7 A curve is such that = 3 x + 10 x - 8 . dx (a) Find the set of values of x for which y decreases as x increases. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is given that the maximum point of the curve has y-coordinate 27. Find the equation of the curve. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 2 *B1 OE 2  5  49 3x + 10 x −8 0 or 3  x +  −  0 Condone = or ⩽ 0, but use of > or ⩾ can only score B0B1B0.  3  3 If there is no < , = or ⩽ 0, then a correct final answer implies < 0 here. 2 B1 − 4 and 3 2 DB1 FT OE −4 x  3 2 Use of ⩽ sign(s) gets DB0. Condone x −4, x  , 3 2 2 x −4 and x  but not x −4 or x  . 3 3 FT on their critical values. 3 7(b) Identify x = −4 as the x-value of the maximum point. B1 FT SOI FT on their lower critical value from part (a). 3 3 10 2 B1 Correct integration of the three terms given.  y =  x + x − 8 x  + c  x ) = . 3 2 Condone f (  2  M1  2  Use of y = 27, x = their  −4 or  in their integral. Their integral must be a cubic, their  −4 or  must come  3   3  from (a) or a restart. c = −21  y = x 3 + 5 x 2 − 8 x − 21 A1 Do not ISW if they continue to find a straight line. Condone 3 + 5 x 2 − 8 x + c omission of final statement if y or f ( x ) = x has been seen earlier. 4

More questions on Differentiation

8 A 12 cm B E 12 cm D F C The diagram shows a square ABCD where each side has length 12 cm. Points E and F lie on the sides BC and CD respectively and are such that BE = 1 BC and DF = 1 DC . The arc EF is part of a circle 3 3 with centre A. The shaded region is bounded by the arc EF and the line segments EC and FC. (a) Show that the size of angle EAF is 0.9273 radians, correct to 4 significant figures. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the perimeter of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the area of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) 1 −1 4 −1 4 −1 12  M1 Attempt a complete valid method for finding angle EAF. π −2 tan or sin or cos   2  12 160 160  160 may be replaced by 12.65 AWRT and 32 by 5.657 Or AWRT. ˆ  1  Working in degrees should give EAF = 53.13° which EF 32 2   converts to the correct value in radians. −1  2  2sin   = =   or other valid method.  AE  160 10     ˆ [ EAF = 0.927295…=] 0.9273 A1 AG If decimals are used, at least one of 0.32175.., 0.6435 or 0.92729.. must be seen in the first method and 5.6569 and 12.649 or 0.46364 seen in the second method for the correct level of accuracy. Note: EAFˆ = 53.14° does not convert to the correct answer, so A0. Alternative Method for Question 8(a) 2 1 M1 Attempt a complete valid method for finding angle EAF. ˆ 12 − ( 24 + 24 + 32 ) = ( theirAE  their AF ) si n EAF Area of ABCD – (Area ∆s ADF, ABE &CEF) = Area ∆AEF. 2  4  ˆ = sin −1 2 12 2 − ( 24 + 24 + 32 ) EAF  their  . ˆ sin EAF =  5  ( theirAE  their AF ) −1 3 2 Using the cosine rule should lead to cos . 2 12 − ( 24 + 24 + 32 ) 5 ˆ −1 EAF = sin ( theirAE  their AF ) 1 A1 AG EF −1 4  2  ˆ EAF = sin =    5  AE 2 8(b) 4 B1 OE [AE or AF =] 160 or ˆ Expect AWRT 12.65. sin EAB their 160 .  r= ( their 12.65 )  0.9273 M1 Use of r with ( ) Note: using r = 12 scores M0. 11.729... + 8 + 8 =  27.7 AWRT A1 3 8(c) 1 2 M1 1 2 ( their 12.65 )  0.9273 [= 74.184] Use r  for area of sector with ( their 12.65 ) . 2 2 Note: using r = 12 scores M0. 1 2 M1 Attempt a complete method for finding area of shaded region. 144 − 24 − 24 − ( their12.65 )  0.9273 Condone use of r = 12.  2  [Area =] 21.8 A1 AWRT Alternative Method for Question 8(c) 1 2 M1 Area of the segment with ( their 12.65 ) . ( their 12.65 ) ( 0.9273 − sin 0.9273 ) [=10.183] 2 Note: using r =12 scores M0. 1 2 M1 Attempt a complete method for finding area of shaded region, 32 − ( their12.65 ) ( 0.9273 − sin 0.9273 ) condone use of r = 12. 2 [Area =] 21.8 A1 AWRT 3

More questions on Trigonometry

9 Three points P, Q and R have coordinates P (-13, 5), Q (5, 1) and R (2, k), where k is a constant. It is given that the angle PRQ is a right angle. (a) Show that one of the possible values of k is 10, and find the other possible value. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is now given that k = 10 . A circle passes through the points P, Q and R. Find the equation of the tangent to the circle at R. Give your answer in the form ax + by + c = 0 , where a, b and c are integers. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 9(a) k − 5 k − 1 B1 Obtain at least one relevant gradient. Gradient of PR = or gradient of RQ = 2 + 13 2 − 5 (their gradient of PR) × (their gradient of RQ) = − 1 and attempt to simplify M1 Clear of fractions and expand brackets. Expect k 2 − 6k + 5 = 45 OE. Condone +/–sign errors during simplification. k 2 − 6 k − 40  = 0  A1 OE k = 10, k = −4 A1 Alternative Method for Question 9(a) 2 2 2 B1 2 PR = ( k − 5) + ( 2 + 13 ) Obtain at least one of PR and RQ 2. 2 2 2 These expressions may be seen under a square root sign. or RQ = ( k − 1) + ( 2 − 5 ) 2 2 2 + 4 2 and attempt to simplify M1 Expect 2k 2 − 12k + 260 = 340 OE. ( their PR ) + ( their RQ ) = 18 Condone +/–sign errors. 2 k 2 − 12 k − 80  = 0  A1 OE k = 10, k = −4 A1 9(a) Alternative Method 2 for Question 9(a) x 2 + y 2 + 8 x − 6 y − 60 = 0 B1 OE 22 + k 2 + 8 −−2 6 k 60 = 0 M1 Substitution of ( 2,k ) into their circle equation. k 2 − 6 k − 40  = 0  A1 OE k = 10, k = −4 A1 4 If none of the above marks are awarded, then SC B1 for using 1 and k = 10 to find the gradients of QR (–3) and PR ( 3 ) showing that their product is –1 (or other similar verification), and then stating that this shows that 10 is a possible value for k. 9(b) B1 SOI [Centre is] ( −4, 3 ) ( their 3 ) − 10  7  *M1 Attempt at finding the gradient of the radius using ( 2, 10 ) Radius gradient = = ( their − 4 ) − 2  6  and their centre, but not P, Q, the mid-point of PR or QR. −1 y − 10 DM1 OE Either = Correct method to find the equation of the tangent using  7  x − 2  their  −1  6  and ( 2, 10 ) . their radius gradient −1 Or 10 =  2 + c  c = ...  7   their   6  6  6  82 A1 OE y − 10 = − ( x − 2 ) or y =  −  x + Correct equation but not stated in the required form. 7  7  7 6 x + 7 y − 82 = 0 or 82 − 6 x − 7 y = 0 A1 All correct terms on one side, but condone them being in the wrong order. Alternative Method for Question 9(b) x 2 + y 2 + 8 x − 6 y − 60 = 0 B1 OE Equation of the circle. dy dy *M1 Differentiate implicitly to arrive at an expression with two 2 x + 2 y + 8 − 6 = 0 d y dx dx terms containing . dy −2 x − 8 2 − 12 d x Or = ( 85 − x − 8 x − 16 ) Or rearrange to make y the subject and differentiate to arrive dx 2 at an expression of the form f  ( x )  f ( x ) . 9(b) dy dy DM1 Substitute (2,10) into their implicit differential. 2 ( 2 ) + 2 (10 ) + 8 − 6 = 0 dx dx 1 d y − Or substitute x = 2 into their expression for . dy −2 ( 2 ) − 8 2 2 Or = d x ( 85 − ( 2 ) − 8 ( 2 ) − 16 ) dx 2 6 6 82 A1 OE y − 10 = − ( x − 2 ) or y = − x + Correct equation but not stated in the required form. 7 7 7 6 x + 7 y − 82 = 0 A1 All correct terms on one side, but condone them being in the wrong order. 5

More questions on Coordinate geometry

9 10 A curve C has equation y = + 2 x - 5 . 2x - 5 (a) Find the coordinates of the two stationary points. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find and hence determine the nature of each stationary point. [3] dx 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ - 3 (c) The curve C is transformed to the curve C1 using a translation of e o followed by reflection in 7 the x-axis. (i) State the coordinates of the maximum point of C1. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... a (ii) Find the equation of C1 in the form y = + dx + e , where a, b, c, d and e are integers. bx + c [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: 10(a) −9 2(2 x − 5) −2 + 2 B1 Correct differential. M1 d y ( their − 18(2 x − 5) −2 + 2 ) = 0 and rearrange to form a quadratic. Equating a two term to 0 and dealing correctly with the 2 2 d x  ( 2 x − 5 ) = 9 or 8 x − 40 x + 32 = 0    negative power. d y −2 Their two term must contain (2 x − 5) . d x (1, − 6 ) and ( 4, 6 ) A1, A1 A1 for two correct x-values or one correct point, second A1 for all correct. 4 10(b)   B1 FT Following through on their first derivative which must −3 −3 144 x − 360 −2 −18 −2 2(2 x − 5)  = 72(2 x − 5) or 4  contain (2 x − 5) .   ( 2 x − 5 )  M1 Substitute x-coordinate of each stationary point and determine 2 x − 5) −3 Use ( their x = 1 and x = 4 ) in (their 72 ( ) their nature. Nature of the turning points must correctly To determine the nature of both turning points. 2 d y follow from their values of 2 . d x  d 2 y  72 A1 CWO For x = 1 ,  2 =  − or  0 ⟹ maximum  d x  27  d 2 y  72 For x = 4 ,  2 =  or  0 ⟹ minimum  d x  27 3 10(c)(i) (1, − 13 ) B1 1 10(c)(ii) 9 M1  − 3   y = + 2 ( x  3 ) − 5  7 Application of to the original expression for C but   (2 x  3 ) − 5  7  condone +/−sign errors.  9  M1  9  y =  −  + 2 x − 5  .  y =  −  + 2 ( x  3 ) −5 7  SC B1 for   2 ( x  3 ) − 5   2 x − 5  9 A1 Answer must be in this format; the ' y = ' can be implied by y = − − 2 x − 8 2 x + 1 earlier inclusion. 3

More questions on Differentiation

11 The function f is defined by f ( )x = x 2 + 4ax + a for x ! R , where a is a constant. The function g is such that g -1 ( )x = 3 2 x - 4 for x ! R . (a) Given that the range of f is f ( )x H - 33 , find the possible values of a. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given instead that fgg ( 0 ) = 96 , find the value of a. 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Mark scheme: 11(a)  ( x + 2 a ) 2 − 4 a 2 + a B1  f ( x ) =  2 M1 Condone or  . their −4a + a = −33 ( ( ) ) 4a 2 −−a 33 [= 0] B1 Condone or . A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. Alternative Method for Question 11(a) x = −2 a , y = −4 a 2 + a B1 The co-ordinates of the minimum point (likely to be either from differentiation or completing the square). 2 M1 Their y-coordinate equated to −33. Condone or  . their − 4 a + a = −33 ( ) −4a 2 + a + 33  = 0 B1 OE Condone or  . A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. 11(a) Alternative Method 2 for Question 11(a) x 2 + 4ax + a + 33 [=0] B1 Condone or  . ( 4 a ) 2 − 4 (1 a + 33 ) [ = 0] M1 Condone or  . 16a 2 − 4a − 132 [=0] B1 OE Condone or  . Accept 4a 2 − a − 33 or multiples thereof. A1 OE − 11, 3 Do not ISW if their final answer is given as a range or if one 4 of the answers is rejected. 4 11(b) 1 3 B1 Expression for g ( x ) . SOI  g ( x ) =  ( x + 4 ) 2 Either 1 3 M1 Replacing x with 0 in their g ( x ) . g ( 0 ) = 2   g ( 0 ) =  ( ( 0 )  4 )  2 gg ( 0 ) = 6 A1 f ( their 6 )  = 36 + 24a + a  M1 Or 3 2 3 M1 Either composite function using their g ( x ) .  x + 4   x + 4  fg ( x ) =   + 4 a   + a  2   2   x 3 + 4  3 + 4   3   ( x 3 + 4 )  2  or gg ( x ) =  = + 2  2  16    A1 Complete algebraic expression for fgg ( x ) .  3  2  3     ( x 3 + 4 )  ( x 3 + 4 ) fgg ( x ) =  + 2  + 4 a  + 2  + a  16   16      x ) .  ( 0 + 4 ) 3  2  ( 0 + 4 ) 3  M1 Replacing x with 0 in their fgg ( fgg ( 0 ) =  + 2  + 4 a  + 2  + a  16   16  11(b) Then 36 + 24 a + a = 96 A1 OE A1 OE  a = 12 5 6

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