Cambridge A Level Mathematics 9709 — 2024 Feb/March Paper 1 · Variant 2

3 21 Find the exact value of dx . [3] 2 3y x .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Integrate to obtain − 2x−1 B1 OE Substitute limits correctly with clear indication seen that upper limit gives 0 M1 For integral of form − k x− n , where k  0 , n  0 . Obtain 2 A1 WWW 3 Accept 0.667. 3

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2 y O x A The diagram shows part of the curve with equation y = k sin 12 x , where k is a positive constant and x is measured in radians. The curve has a minimum point A. (a) State the coordinates of A. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) A sequence of transformations is applied to the curve in the following order. Translation of 2 units in the negative y-direction Reflection in the x-axis Find the equation of the new curve and determine the coordinates of the point on the new curve corresponding to A. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) State (3π, − k ) B1 1 2(b) 1 M1 Any non-zero c. Obtain equation of form  y =  c  k sin x 2 1 A1 OE Obtain correct equation  y =  2 − k sin x 2 State (3π, 2 + k ) B1 FT Following part (a), i.e. (their x, 2 – their y). 3

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1 dy 23 A curve is such that = 3 ( 4x + 5) . It is given that the points (1, 9) and (5, a) lie on the curve. d x Find the value of a. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 32 *M1 Integrate to obtain form k (4 x + 5) 1 32 A1 Or (unsimplified) equivalent. Obtain correct 2 (4 x + 5) Condone missing ... + c so far. Substitute x = 1, y = 9 to form an equation in c DM1 1 3 9 A1 1 3 9 2 − . Obtain or imply  y =  ( 4 x + 5 ) 2 − May be implied by  a =  ( 4 (1) + 5 ) 2 2 2 2 Substitute x = 5 to obtain a = 58 A1 5

More questions on Differentiation

( sin i + cos i) 2 - 1 i . [3]4 (a) Prove that 2 / 2 tan cos i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ( sin i + cos i) 2 - 1 3 i for - 90c 1 i 1 90c . [3] (b) Hence solve the equation 2 = 5 tan cos i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) Expand bracket to obtain 3 terms and use correct identity M1  may be missing or another symbol used. sin M1 Does not require any further explanation. Use identity tan  may be missing or another symbol used. cos= Conclude with 2tan A1 WWW AG 3 4(b) Attempt solution of 5tan 3 = 2tan to obtain at least one value of tan M1 SOI Can be awarded if tanis cancelled and ignored. Obtain at least two of 0,  32.3 A1 Or greater accuracy. SC B1 if no method shown. Obtain all three values A1 Or greater accuracy; and no others in −90  90 range. Other units SC B1 only for all 3 angles. SC B1 if no method shown. 3

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35 A curve has the equation y = 2 . 2x - 5 Find the equation of the normal to the curve at the point (2, 1), giving your answer in the form ax + by + c = 0 , where a, b and c are integers. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 Differentiate to obtain form kx (2 x 2 − 5) −2 M1 Obtain correct − 12 x (2 x 2 − 5) −2 A1 OE Substitute (2, 1) to obtain gradient − 249 A1 8 OE e.g. − . Allow −2.67. 3 Apply negative reciprocal to their numerical gradient to obtain gradient of *M1 3 Must have been some attempt at differentiation. Expect normal 8 Attempt equation of normal using their gradient of the normal and (2, 1) DM1 3 Expect y −=1 ( x − 2 ) . 8 Obtain 3 x − 8 y + 2 = 0 (allow multiples) A1 Or equivalent of requested form e.g. 8 y − 3 x − 2 = 0 . 6

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6 It is given that the coefficient of x3 in the expansion of ( 2 + ax) 4 ( 5 - ax) is 432. Find the value of the constant a. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 4 2 2 4 1 3 B1 B1 OE 2( ax )  2 ( ax ) ,  Expect 24 a 2 x 2 ,8a 3 x 3 (may be seen in an expansion). 2 3 Multiply terms involving x 2 and 3x by 5 − ax to obtain 3x term *M1 Must find two products only (may be seen in an expansion). DM1 Equate coefficient of 3x to 432 and solve for a Ignore inclusion of 3x at this stage. Obtain a = 3 only A1 5

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7 The straight line y = x + 5 meets the curve 2x 2 + 3y 2 = k at a single point P. (a) Find the value of the constant k. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the coordinates of P. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) Attempt substitution for y in quadratic equation *M1 Or substitution for x … k (all terms gathered together). Obtain 5 x 2 + 30 x + 75 − k  = 0 or 5 y 2 − 20 y + 50 − k  = 0 A1 OE e.g. x 2 + 6 x + 15 − 5 Use b 2 − 4 ac = 0 with their a, b and c DM1 ‘ = 0’ may be implied in subsequent working or the answer. Obtain 900 − 20(75 − k ) = 0 or equivalent and hence k = 30 A1 … obtaining 400 − 20(50 − k ) = 0 and k = 30 . 4 7(b) Substitute their value of k in equation from part (a) and attempt solution M1 2 2 Expect 5 x + 30 x + 45 = 0 or 5 y − 20 y + 20  = 0 . Obtain coordinates ( −3, 2) A1 SC B1 only ( −3, 2) without attempt at quadratic solution. 2

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8 (a) An arithmetic progression is such that its first term is 6 and its tenth term is 19.5 . Find the sum of the first 100 terms of this arithmetic progression. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) A geometric progression a1, a2, a3, ... is such that a 1 = 24 and the common ratio is 1.2 The sum to infinity of this geometric progression is denoted by S. The sum to infinity of the even-numbered terms (i.e. a2, a4, a6, ...) is denoted by SE. Find the values of S and SE. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) Substitute n = 10 and a = 6 into u n = a + ( n − 1) d *M1 Expect 6 + 9d = 19.5 or equivalent.  d = 1.5 A1 Substitute a = 6 and their d into correct formula for the sum of 100 terms DM1 Obtain 8025 A1 4 8(b) Obtain S = 48 B1 Identify for S E first term 12 and common ratio 14 B1 Attempt sum to infinity, S E , with at least one of first term and common ratio M1 Only awarded if |r| < 1. correct Obtain S E = 16 A1 4

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9 The functions f and g are defined for all real values of x by f ( x) = ( 3 x - 2) 2 + k and g ( x) = 5 x - 1, where k is a constant. (a) Given that the range of the function g f is gf ( x) H 39 , find the value of k. 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(b) For this value of k, determine the range of the function f g. 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(c) The function h is defined for all real values of x and is such that gh ( x) = 35x + 19 . Find an expression for g -1 ( x) and hence, or otherwise, find an expression for h ( x) . 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Mark scheme: − 1; fg(x) is M0. ( 3 x − 2 ) 2 + k9(a) Attempt to form expression for gf ( x ) *M1 Expect 5 ( ) Do not allow algebraic errors. Obtain 5 ( 3x − 2 ) 2 + 5k − 1 A1 OE e.g. 45 x 2 − 60 x + 5k + 19 . Their 5k − 1 = 39 or 5k − 1 ⩾ 39 DM1 Or use b 2 − 4 ac = 0 (must be ‘= 0’, could be implied later) on 45 x 2 − 60 x + 5 k + 19 − 39 0 OE. Obtain k = 8 A1 Do not accept k 8 . 4 9(b) 2 M1 May simplify and/or use k at this stage; k may have come Obtaining ( 3 ( 5 x − 1) − 2 ) + their k from an inequality in (a). A1 FT OE Conclude  8 allow y 8  fg ( x )  Following their value of k; must be ⩾, not >. Allow an accurate written description. 2 9(c) −1 1 B1 OE State g ( x) = 5 ( x + 1) 1 ( x + 1) must be indicated as the inverse. 5  7 x + 4 B1B1 If 7 x + 4 only, it must be clear that this is h ( x ) .  h ( x ) = 3

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10 y A C i rad B x O The diagram shows the circle with centre C (– 4, 5) and radius 20 units. The circle intersects the y-axis at the points A and B. The size of angle ACB is i radians. (a) Find the equation of the tangent to the circle at the point (–6, 9). 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(b) Find the equation of the circle in the form x 2 + y 2 + ax + by + c = 0 . 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(c) Find the value of i correct to 4 significant figures. 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(d) Find the perimeter and area of the segment shaded in the diagram. 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Mark scheme: 10(a) Obtain gradient of relevant radius is –2 B1 Using m1m2 = −1 obtain the gradient of the tangent and use it to form a M1 m1 must be from an attempt to find the gradient of the straight line equation for a line containing (–6, 9) radius using the centre and the given point. Obtain y = 12 x + 12 A1 1 OE e.g. y − 9 = ( x + 6 ) . 2 3 10(b) State or imply ( x + 4) 2 + ( y − 5) 2 = 20 B1 If x 2 + y 2 − 2 gx − 2 fy + c = 0 is used correctly with ( − g , − f ) = ( −4, 5 ) and c = g 2 + f 2 − r 2 then M1. Obtain x 2 + y 2 + 8 x − 10 y + 21 = 0 B1 A1 if above method used. 2 10(c) Substitute x = 0 in equation of circle to find y-values 3 and 7 B1 May be implied by AB = 4 or use of |x-coordinate of C|. or state C to AB = 4 Attempt value of  either using cosine rule or via 12 using right-angled M1 Using their AB. If /2 used, must be multiplied by 2. triangle Obtain = 0.9273 A1 Or greater accuracy. A correct answer implies the M1. 3 10(d) Attempt arc length using r formula with their  (not their /2) and M1 Expect 4.15. r =  20 Obtain perimeter = 8.15 or greater accuracy A1 Condone missing units or incorrect units. 1 2 M1 If sector – triangle used, both formulae must be correct. Attempt area using 2 r (− sin) formula or equivalent with their and If triangle ACM used, area must be multiplied by 2. r =  20 Obtain area = 1.27 or greater accuracy A1 Condone missing units or incorrect units. 4

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11 y A B O x M 1 3 The diagram shows the curve with equation y = 2x - 2 3 - 3x - + 1 for x 2 0 . The curve crosses the x-axis at points A and B and has a minimum point M. (a) Find the exact coordinates of M. 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(b) Find the area of the region bounded by the curve and the line segment AB. 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Mark scheme: 11(a) 4 − 53 − 34 B1 1 2 1 −  − Differentiate to obtain − 3 x + x  3 3 − 3 x + 1 OE Expect quadratic 2  x  1 1   − 3 3   or rewrite as a quadratic equation in x or x Allow 2 x 2 − 3 x + 1 . 1 1 M1 Substitution SOI if dealt with correctly later − Equate first derivative to zero and reach a solution for x 3 or x 3 with no error in use of indices −  3  2 1 1 or complete square to find minimum point 2  a −  − where a = x 3  4  8 Obtain x = 6427 A1 Or exact equivalent. SC B1 if no working shown. Ignore extra solution x = 0 . y = − 18 seen B1 Or exact equivalent. Allow −0.125 . 4 1 − 1311(b) M1 − or equivalent and attempt solution Recognise equation as quadratic in x 2 2a − 3a + 1 = 0 where a = x 3 . 1 1 A1 OE 3 1 3 Obtain x −= 1 and x −= 2 SC B1 if no M mark awarded. Obtain 1 and 8 A1 SC B1 if no M mark awarded. 1 2 1 2 *M1 9 3 + x or 2 out of 3 correct terms Integrate to obtain form k1 x 3 + k 2 x Expect 6 x 3 − x 3 + x . 2 1 2 A1 No other terms from a second integral. Obtain correct 6x 3 − 9 x 3 + x 2 Apply their limits correctly DM1 Their limits must be from their working. [Obtain –0.5 and conclude area is] 0.5 A1 7

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