Cambridge A Level Mathematics 9709 — 2021 May/June Paper 2 · Variant 2

2 The solutions of the equation 5 x 5 are x a and x b, where a b. = −2x = = < Find the value of 3a 7b . [5] −1 + −1 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Solve 5 5 2 = − x x to obtain 5 7 = x Attempt solution of linear equation where signs of 5x and 2x are the same M1 Obtain 5 3 = − x A1 Allow AWRT –1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 Alternative method for Question 2 State or imply non-modulus equation 2 2 25 (5 2 ) = − x x B1 Attempt solution of 3-term quadratic equation M1 Obtain 5 3 − and 5 7 A1 Allow AWRT 0.714 and AWRT -1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 5

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5 (a) Find the quotient when x4 55 is divided by x 2 and show that the remainder is 7. −32x + −2 [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Factorise x4 48. [2] −32x + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 48 0, giving your answer in an exact form. [2](c) Hence solve the equation e−12y −32e−3y + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Carry out division at least as far as x kx or equivalent … M1 OE, e.g. comparing coefficients with coefficient of 2 x equal to 1 and attempt at a second coefficient. Obtain quotient 2 4 12 + + x x A1 Confirm remainder is 7 A1 AG 3 5(b) Include 2 ( 2) − x as a factor M1 Must be a product of factors only SC B1 for ( )( ) 2 2 4 4 4 12 − + + + x x x x Conclude 2 2 ( 2) ( 4 12) − + + x x x A1 isw any attempt to factorise the quotient. 2 5(c) Apply logarithms and use power law for 3 e− = y k where 0 > k M1 Obtain 1 1 1 ln2, ln 3 3 2 = − y A1 Or exact equivalent Must be simplified e.g. not lne or 6 3 ISW extra solutions but A0 if undefined solutions are included. 2

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6 y A B x O M The diagram shows the curve with equation y ln x 2 ln x. = −2 The curve crosses the x-axis at the points A and B, and has a minimum point M. (a) Find the exact value of the gradient of the curve at each of the points A and B. 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(b) Find the exact x-coordinate of M. 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Mark scheme: 6(a) Obtain 1 = x 0e . Must come from correct work, e.g. ln 0 = x . Obtain 2 e = x B1 Differentiate to obtain at least one correct term *M1 Obtain correct first derivative 2ln 2 − x x x A1 Allow ln ln 2 + − x x x x x Allow 2 2 −x x Substitute at least one of their x-values corresponding to 0 = y to find gradient DM1 Allow unsimplified. Obtain gradient 2 − [at A] and gradient 2 2e− at [B] A1 Must be simplified. 6 6(b) Equate first derivative to zero M1 Their derivative must have at least 2 terms. Obtain e = x A1 Allow 1e 2

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7 y P x O The diagram shows the curve with parametric equations x 4t e2t, y 6t sin 2t, = + = for 0 The point P on the curve has parameter p and y-coordinate 3. ≤t ≤1. 1 (a) Show that p [1] = 2 sin 2p. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Show by calculation that the value of p lies between 0.5 and 0.6. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Use an iterative formula, based on the equation in part (a), to find the value of p correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (d) Find the gradient of the curve at P. 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Mark scheme: 7(a) Equate y to 3 and confirm 1 2sin 2 = p p 1 7(b) Consider sign of 1 2sin 2 − p p or equivalent for 0.5 and 0.6 M1 Obtain 0.09... − and 0.06... or equivalents and justify conclusion A1 AG 2 7(c) Use iteration process correctly at least once M1 Need to see 0.55494… Obtain final answer 0.557 only A1 Allow recovery. Allow if iterations are to 4sf Allow if insufficient iterations seen. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [0.5565, 0.5575] A1 If not starting at 0.55 then max marks M1A1A0 3 Question Answer Marks Guidance 7(d) Obtain 2 d 4 2e d = + t x t B1 Use product rule to find d d y t M1 Must be of the form sin 2 cos2 + p t qt t Obtain 6sin 2 12 cos2 + t t t A1 Allow unsimplified. Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have either B1 or previous M1. Obtain 0.826 A1 AWRT 5

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