Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 1 · Variant 3

1 (a) Expand b2 - 1 xl in ascending powers of x up to and including the term in x3. [3] 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 6 (b) Hence find the coefficient of x3 in the expansion of `3 - x + 2x 3bj2 - 1 xl . [2] 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 6  6 6 5  x  6 4  x  2 6 3  x  3 M1 OE   2   2     2     2   A correct unsimplified expansion of at least four terms. Condone 0  1  2  2  2  3  2   sign errors for this mark. Condone omission of brackets if   x  6.5  x  2 6.5.4  x  3  recovered. 6 1  6   or 2           This mark can be implied by sight of 64 + 96 x + 60 x 2 + 20 x 3 . 2! 3!  4   4   4    2 3 A1 Three correct simplified terms. 64 − 96 x + 60 x − 20 x A1 Four correct simplified terms. Accept these terms given in a list. 3 1(b) ( 3  their ( −20 ) ) + ( −1 their 60 ) + ( 2  their 64 ) M1 Selecting the correct products using their expansion. 8 A1 Accept 8 x 3 . 2

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2 (a) Solve the equation tan -1 ( 5x - 3) =- 1 r . [2] 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Solve the equation 5 cos 2i = 4 sin i + 4 for 1 r G i G 2 r . [4] 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a)  1   1  M1  1  5 x − 3 = tan  − π  or − tan  π   = −1 Condone 5 x − 3 = tan  π  and 5x −=3 −0.0137  4   4   4  2 A1 x = or 0.4 5 2 2(b) 5 − 5sin 2 = 4sin+ 4 *M1 Replacing cos 2  with 1 − sin 2. 5sin 2 + 4sin− 1 = 0  ( 5sin− 1)( sin+ 1) = 0  DM1 Collecting terms to form a three-term quadratic. Condone only  errors. 1 A1 sin= and − 1 5 = 2.94, 4.71 A1 AWRT  3   337  Accept  π  in place of 4.71and 0.936π or  π  in place  2   360  of 2.94. 3 Condone repetition of π but no other values within the range. 2 π Ignore answers outside the range  2π. 2 4

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1 23 23 The equation of a curve is y = f ( x) , where f ( x) = x ( x - 2) . The following points lie on the curve. 2 Non-exact values of the y-coordinates are given correct to 6 decimal places. A(8, 72), B(8.001, k), C(8.01, 72.300388), D(8.1, 75.038882) (a) Find the value of k. Give your answer correct to 6 decimal places. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The table below shows the gradients of the chords AB and AC, given correct to 4 decimal places. Chord AB AC AD Gradient of chord 30.0039 30.0388 (b) Find the gradient of the chord AD. Give your answer correct to 4 decimal places. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) State what the values in the table suggest about the value of fl ( 8 ) . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 72.030004 B1 CAO. Not AWRT. 1 3(b) 30.3888 B1 CAO. Not AWRT. 1 3(c) 30[.0] B1 CAO. 30 may be accompanied by ‘around’, ‘approximately’ etc. 1

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4 The first, second and third terms of a progression are 20, k and k - 5 respectively. (a) Given that the progression is arithmetic, find the 30th term. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given instead that the progression is geometric, find the sum to infinity. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 20 + ( 30 − 1)  ( their − 5 ) M1 Correct expression for u30 using their calculated value of d. −125 A1 2 4(b) k k − 5 B1 OE r = and r = 20 k k k − 5 For two correct expressions for r. Can be implied by = 20 k OE. k 2 − 20k + 100  = 0 M1 Forming a 3-term quadratic. Condone only  errors.  r = 12 A1 May be seen in the S∞ formula.  20  A1 S = 1 =  40  1 − 2  Alternative Method for Question 4(b) 20 r = k and 20r 2 = k − 5 B1 Stating two correct relationships between k and r. This can be implied by 20 r 2 = 20 r − 5 OE. 20r 2 − 20r + 5 = 0 M1 Forming a three-term quadratic in r. Condone only  errors.  r = 12 A1 May be seen in the S∞ formula.  20  A1 S = 1 =  40  1 − 2  4

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5 X O 4 cm 4 cm θ rad P Q 4 3 cm The diagram shows part of a circle with centre O and radius 4 cm. The chord PQ is of length 4 3cm and angle POQ = i radians. The point X lies on the circle. (a) Find the exact value of i. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact area of the segment PXQ. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a)  1 2 3 B1  Use of a correct expression for either or . sin  =  2  4 2  1 2 or cos =    2  4 2 4 2 + 4 2 − 4 3 ( )  1  or cos= = −   2 4 4  2  B1 Correct answer can imply B1B1. = 2π Condone inconsistent use of . 3 2 5(b) 1 2 2 1 2 4 1 2 2 *M1 Use of a correct formula with theiror 2− their,  0 . Triangle  4  sin or  4  or  4  2 3 2 3 2 3 and/or minor sector may be seen in the formula for finding the minor segment. 1 2 4π 1 2 2π DM1 Addition of relevant sector area with 2− their, where  0,  4  +  4  sin 2 3 2 3 and triangle area with their, where  0.  1 2 2π 1 2 2π  Subtraction of triangle area, with their, 0 , from relevant Or 16π −   4  −  4  sin   2 3 2 3  sector area, with their, 0, and subtract from circle area. 32π A1 OE. Exact equivalents only. + 4 3 SC B2 following M0DM0 for 40.4 AWRT. 3 3

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6 y 6 4 y = x3 2 – 1 0 1 2 3 4 5 x y = f(x) – 2 – 4 – 6 – 8 The diagram shows the graphs of y = x3 and y = f ( x) . The graph of y = x3 is transformed to the graph of y = f ( x) by a sequence of transformations. (a) Describe fully a suitable sequence of transformations. Make clear the order in which the transformations are applied. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) You are given that f ( x) = a ( x + b ) 3 + c . State the values of the constants a, b and c. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) {Stretch} {[scale] factor 2} {[parallel to] in/on y[-axis] or x-axis B2,1,0 B2 for all three {} elements correct. B1 for two {} elements invariant} correct.  3  B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation}   or {[+]3 in/on [the] x[direction]} and correct.   − 4  Do not condone use of ‘Right’ and ‘Down’ in place of x and y {– 4 in/on [the] y [direction]} directions. 3  0  Note: The vectors can be seen as  and   . 0  −4  Correct transformations in the correct order without any extra B1 Combined translations (condone transformations accompanied by transformations. Down and right can be condoned for this mark if  0  a vector) must come after the stretch or, if separate, the the candidate’s intention is clear.    −4  must come after the stretch. Alternative Method for Question 6(a)  3  B2,1,0 B2 for all three {} elements correct. B1 for two {} elements {Translation}   or {[+]3 in/on [the] x [direction]} and correct.   − 2  Do not condone use of ‘Right’ and ‘Down’ in place of x and y {-2 in/on [the] y [direction]} directions. 3  0  Note: The vectors can be seen as  and   . 0  −2  {Stretch} {[scale]factor 2} {[parallel to] in/on y-[axis] or x axis B2,1,0 invariant OE} Numerically correct transformations in the correct order without B1 Combined translations (condone transformations accompanied by any extra transformations. Down and right can be condoned for  0  this mark if the candidate’s intention is clear. a vector) must come before the stretch or, if separate, the    −2  must come before the stretch. 5 6(b) a = 2 B1 Individual values are considered the final answer and NOT those in a ( x + b )3 + c, unless individual values are not stated. b = −3 B1 c = − 4 B1 3

More questions on Functions

2 1 3 7 The function g is defined by g ( )x = + for x 2 , where a is a positive constant. ax - 3 2 a Find g -1 ( )x and hence verify that if a = 6 then g -1 ( x) / g ( x) . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 7  2 1   1  2 *M1 OE y = +  ax − 3 ) = 2 or ax − 3 = For setting ‘y =’ and rearranging.    y −  (  ax − 3 2   2  1 y − Condone only  errors. 2 4 DM1 For making ax the subject. ax = + 3 2 y − 1 Condone only  errors. −1 4 3 2 3 A1 OE [ y = or g ( x ) = ] + or + a ( 2 x − 1) a  1  a Note: x and y may be interchanged from the start. a  x −   2  Clear replacement of a with 6 in g −1 and, if necessary, followed by A1 This mark is dependent on getting the correct inverse. simplification to arrive at exactly the same expressions. Alternative Method for Question 7  ax + 1  *M1 OE  y =   2 y ( ax − 3 ) = ax + 1 For setting ‘y =’ and rearranging.  2 ( ax − 3 )  Condone only  errors. ax ( 2 y − 1) = 6 y + 1 DM1 For making ax ( 2 y − 1) the subject, condone only  errors. 1 A1 OE 3 x + 6 x + 1 2 Note: x and y may be interchanged from the start. or  y = or g −1 ( x ) =    a ( 2 x − 1)  1  a  x −   2  Clear replacement of a with 6 in g −1 and, if necessary, followed by A1 This mark is dependent on getting the correct inverse. simplification to arrive at exactly the same expressions. 4

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8 The points (6, 1) and ( -2, 7) lie at the opposite ends of a diameter of a circle. (a) Find the equation of the circle. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) There are two tangents to the circle which have gradient - 1 . 2 Find the exact values of the x-coordinates of the points at which these tangents touch the circle. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) Centre is ( 2, 4 ) B1 2 2 M1 2 [(Radius)2 =] ( 6 − 2 ) + (1 − 4 ) Use of a correct method to find ( radius ) or radius.  1 2 2  2 This can be implied by r = 5 . = 25 or  ( 6 −−( 2 ) ) + (1 − 7 )    2  ( x − 2 ) 2 + ( y − 4 ) 2 = ( 5 ) 2 A1 OE Alternative Method for Question 8(a) y − 1 y − 7 B1 Correct expression using either ( 6,1) or ( −2,7 ) and ( x , y ) . Either or x − 6 x + 2 y − 1 y − 7 M1 Product of their gradients = −1.  = −1 x − 6 x + 2 x 2 + y 2 − 4 x − 8 y − 5 = 0 A1 OE Simplification to get the correct five terms. 3 8(b) [Gradient of normal through centre =] 2 B1 SOI y − 4 = ( their gradient )( x − 2 ) *M1 Correct form of a line equation with their normal gradient i.e. not 1 Or 4 = ( their 2 )( 2 ) + c  c = [0] − 2 ( x − 2 ) 2 + ( 2 x − 4 ) 2 = 25 DM1 OE Using their 2 x and their circle to form an equation in x.  5 ( x − 2 ) 2 = 25   ( x − 2 ) 2 = 5 A1 OE   2 Accept x − 4 x − 1  = 0  .    x = 2  5 A1 OE Exact values only. Alternative Method for Question 8(b) dy dy B1 OE 2 x + 2 y − 4 − 8 = 0 Implicit differentiation of the correct circle equation. dx dx 1   1  *M1 d y 1 2 x + 2 y − − 4 − 8 − = 0 Replacing with − in their differential which must contain      2   2  d x 2 d y two terms in . Expect y = 2 x . d x ( x − 2 ) 2 + ( 2 x − 4 ) 2 = 25 DM1 OE Using their 2 x and their circle to form an equation in x.  5 ( x − 2 ) 2 = 25   ( x − 2 ) 2 = 5 A1 OE   2 Accept x − 4 x − 1 = 0.  x = 2  5 A1 OE Exact values only. 8(b) Alternative Method 2 for Question 8(b) 1 B1 This mark can be implied by a correct differential. − x 2 + 4 x + 21 2 + 4 y = 25 − ( x − 2 ) 2 + 4 or ( ) 1 *M1 Differentiating their expression for y. Use of the chain rule must − dy 1 2 2 = 25 − ( x − 2 )  ( 4 − 2 x ) be clear. ( ) dx 2 1 DM1 dy 1 − 1 1 . Equating their to − − = 25 − ( x − 2 ) 2 2  ( 4 − 2 x )   25 − ( x − 2 ) 2 = ( 2 x − 4 ) 2  ( ) dx 2   2 2 5 x 2 − 20 x − 5 = 0 A1 OE  x =2  5 A1 OE Exact values only. Alternative Method 3 for Question 8(b) B1 Replacing y with − 12 x + c in a correct circle equation. = 25 ( x − 2 ) 2 + ( ( − 12 x + c ) − 4 ) 2 5 2 2 5 2 2 M1 Forming a three-term quadratic in x. x − cx + c − 8c − 5 = 0 x − cx + ( c − 4 ) − 21 = 0 or ( ) Condone only  errors. 4 4 Using " b 2 − 4 ac " = 0 with their values from the quadratic in x DM1 Expect c = 0. ( c − 4 ) 2 − 21 2 − 4 54    1600 + 400  10  5 5  A1 OE  c = 40  =  8  2   x =2  5 A1 OE Exact values only. 5

More questions on Coordinate geometry

9 y x O 1 4 The diagram shows part of the curve with equation y = x + and the line y = 4.5. 2 x Find the exact volume of the solid formed when the shaded region is rotated through 360° about the x-axis. 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Mark scheme: 9 1 4 2 M1 Equating the curve and the line to form a 3-term quadratic. x + = 4.5  x − 9 x + 8 = 0 2 x  x =1, 8 A1 SOI   2  567  A1 FT FT their limits from the quadratic, both limits must be positive. π or 445.3 π  4.5  ( their 8 − their 1)  =   4  2 *M1  1 4  For use of y 2dx . x + π   [dx ]    2 x   1 2 16   1 3 16  A1 OE x + 4 + x + 4 x −   dx = π   First A1 for two correct terms, second A1 for all three correct. π  x x 2  4   12  May be unsimplified. May include +C. A1  218 143  1015 DM1 Clear use of their limits in their integrated expression. At least one π  +  = π correct power required. Both limits must be positive, or if correct  3 12  12  218 143  limits are used, then   +  is the minimum required.  3 12  343 A1 Condone a negative answer changed to a positive. Volume of solid = π A0 for omission of π. 6 SC B1: following DM0. A0 for 179.594… Alternative Method for Question 9 1 4 2 M1 Equating the curve and the line to form a 3-term quadratic x + = 4.5  x − 9 x + 8 = 0 2 x  x =1, 8 A1 SOI   *M1  2  . For use of 1  4  y12 − y2 2 )  ( x +  4.5 2 − [Volume =] π   dx       2  2 x  y1 − y2 ) for this *M1.   Condone  (  2  1 3  16   A1 OE  − x  +   A1 for each element of the integral.  π  4.5 x − 4 x    12  x   A1 May be unsimplified. May include +C. A1 2 ‘Correct’ terms from y1 − y2 ) score A0A0A0 XP.  ( 268π  − 193  DM1 Clearcorrectusepowerof theiris required.limits in their integrated expression. At least one  3 6  Both limits must be positive, or if correct limits are used, then  268 193    −  is the minimum required.  3 6  343 A1 Condone a negative answer changed to a positive. Volume of solid = π A0 for omission of π. 6 SC B1: following DM0. A0 for 179.594… 8

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10 A function f is defined by f ( )x = px 2 + 4 x + q for x ! R , where p and q are constants. (a) It is given that p = 2 and q = 10 . (i) Express f ( )x in the form a ( x + b ) 2 + c , where a, b and c are constants. [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) State the range of f. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (b) It is given instead that q =-5 and the roots of f ( )x = 0 are 5m and -9m , where m is a constant. Find the values of p and m. 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Mark scheme: 10(a)(i) (2 x + 1) 2 +8 B1 B1 for each {} element.   B1 B1 3 10(a)(ii)  Range or f  or y  ( x )    8 B1 FT OE. FT their value of c. x 8 scores B0. 1 10(b) 2 2 *M1 Allow one error for this mark if the intention is clear. p ( 5m ) + 4 ( 5m ) − 5 = 0 and p ( −9m ) + 4 ( −9m ) − 5  = 0  5 − 20m  2 *DM1 Eliminate p by substitution or equating two expressions for p. 81  m + 4 ( −9 m ) − 5  = 0   2   (5m)    36 m + 5 2 or 25   m + 4 ( 5 m ) − 5  = 0  2    ( − 9 m )   5 − 20 m   36 m + 5  or  2  =  2   (5 m )   ( −9 m )  2520m = 280 DM1 OE Simplify as far as a linear equation or a quadratic where m = 0 is ignored or dismissed. m = 19 , p = 9 A1, A1 Alternative Method for Question 10(b) p ( 5m )2 + 4 ( 5m ) − 5 = 0 and p ( −9m )2 + 4 ( −9m ) − 5  = 0 *M1 Allow one error for this mark if the intention is clear. 2 2 *DM1 Equating their two expressions. p ( 5m ) + 4 ( 5m ) −=5 p ( −9m ) + 4 ( −9m ) − 5 DM1 Simplify as far as a quadratic where m = 0 is ignored or dismissed  2 1  45 56 pm − 56m = 0  m =   5 = or 45m = 5 and substitute for m or p.  p  p p = 9, m = 19 A1, A1 10(b) Alternative Method 2 for Question 10(b) 2 2 *M1 Equating 5m and − 9m to the two solutions from the quadratic −+4 4 − 4 ( p )( −5 ) −−4 4 − 4 ( p )( −5 ) = 5m and = −9 m formula. 2 p 2 p Allow one error for this mark if the intention is clear. 2 2 *DM1 Attempt to solve simultaneously by equating their expressions for −+4 4 − 4 ( p )( −5 ) −−4 4 − 4 ( p )( −5 ) 5 = m, or dividing one by the other and equating to −.9 ( 5  2 ) p ( −9 2 ) p 16 + 20 p = 196 DM1 OE Simplify as far as a linear expression – no square root. p = 9, m = 19 A1, A1 Alternative Method 3 for Question 10(b)  x 2 + 4mx − 45m2  = 0 *M1 Recognising the relationship between 5m and − 9m and the ( x − 5m )( x + 9m )  quadratic equation. Or +  = −4 m and  = −45 m 2 4 2 5 *DM1 2 4 5 4 m = , −45m = − Using x + x −  = 0  . Comparing coefficients or using p p p p b c +  = − and = . a a 1 2 DM1 Forming an equation in one unknown. m =  − 45 m = − 5 m p 1 2 or p =  5 p = 45 p m m = 19 , p = 9 A1, A1 5

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8 6 11 The equation of a curve is y = - . 3x - 8 x - 1 (a) Find the coordinates of the point at which the tangent to the curve at the point (3, 5) intersects the line y =-8x . 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(b) (i) Find the x-coordinates of each of the stationary points of the curve. 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Mark scheme: 11(a)    B1 B1 for each {} element. d y  −8 3  6  =  2  + 2  d x  ( 3 x − 8 )  ( x − 1)  B1    *M1 Using their differentiated expression, which must contain  −8 3  6  −45  −2 −2  2  + 2  =  ( 3x − 8 ) and ( x − 1) , and x = 3 . This may be seen in their line  ( 3.3 − 8 )  ( 3 − 1)   2  equation.  −45  *DM1 Correct form of a line equation with their gradient, but not the their ( x − 3 ) y − 5 =   negative reciprocal, and ( 3, 5 ) . 2    −45   145  or 5 =  their  ( 3 ) + c  c =    2   2   −45  DM1 Replacing y with −8x and collecting terms. 29 x = 145  −8 x − 5 = ( x − 3 )    2  x = 5, y = −40 A1 Accept ( 5, − 40 ) . 6 11(b)(i) −24 6 2 2 *M1 d y −2 −2 + = 0 ⇒ 24 ( x − 1) = 6( 3 x − 8 ) Equate their , which must contain ( 3x − 8 ) and ( x − 1) to 0, 2 2 ( 3 x − 8 ) ( x − 1) d x and clear of fractions.   4 ( x − 1) 2 = ( 3 x − 8 ) 2  Only condone  errors.   5 x 2 − 40 x + 60 [= 0] or  2 ( x − 1) = 3 x − 8 DM1 OE Forming a three-term quadratic. Condone only  errors. Or taking square roots. For this method, the  must be present. 2, 6 A1 Both values. 3 11(b)(ii) 2     B1FT d y d y  −24 −2 3   6 −2  Correct differentials of the elements of their , which must 2 =  3  +  3  d x B1FT −2 − d x  ( 3 x − 8 )   ( x − 1)  contain ( 3 x − 8 ) and ( x − 1) 2. d 2 y their 144 their 12 M1 d 2 y At x = 2, 2 = 3 − 3 Replacing x with their 2 and their 6 in their 2 , which must dx ( 6 − 8 ) ( 2 − 1) dx −3 −3 contain ( 3 x − 8 ) and ( x − 1) . d 2 y their 144 their 12 and at x = 6, 2 = 3 − 3 Correct final answers. dx (18 − 8 ) ( 6 − 1) d 2 y A1 WWW At x = 2, 2 = −30  0, therefore max[imum] Correct values, or working, and  0 and  0 are required. dx d 2 y 48  6  At x = 6, 2 = 3    0, therefore min[imum] dx 10  125  4

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