Cambridge A Level Mathematics 9709 — 2015 May/June Paper 2 · Variant 1
9709/21/M/J/15 · 4 questions · 50 marks · ≈56 min
The question paper and its mark scheme, free to read here and free to download. This is Cambridge’s own paper, exactly as it was sat.
Question paper4 pages
Mark scheme5 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q1 · Solve the equation 3x 4 3x
1 (i) Solve the equation 3x 4 3x . [3] + = −11 (ii) Hence, using logarithms, solve the equation 3 2y 4 3 2y , giving the answer correct × + = × −11 to 3 significant figures. [2]
Mark scheme: 1 (i) State or imply equation (3 x + 4) 2 = (3 x − 11) 2 or 3 x + 4 = − (3 x − 11) B1 Attempt solution of ‘quadratic’ equation or linear equation M1 7 Obtain x = or equivalent (and no other solutions) A1 [3] 6 (ii) Use logarithms to solve equation of form 2y = their answer to (i) ( must be + ve) M1 Obtain 0.222 (and no other solutions) A1 [2]
Q3 · The equation of a curve is y 6 sin x cos 2x
3 The equation of a curve is y 6 sin x cos 2x. = −2 Find the equation of the tangent to the curve at the point 1 2 . Give the answer in the form 60, y mx c, where the values of m and c are correct to 3 significant figures. [5] = +
Mark scheme: 3 Differentiate to obtain form p cos x + q sin 2 x or equivalent M1 Obtain correct 6cos x + 4sin 2 x or equivalent A1 Substitute 16π to obtain derivative equal to 5 3 or 8.66 A1 Form equation of tangent (not normal) using numerical value of gradient obtained by differentiation M1 Obtain y = 8.66 x − 2.53 cao A1 [5]
Q6 · Prove that 2 cosec tan sec2 [3] 21 1 1
6 (i) Prove that 2 cosec tan sec2 [3] 21 1 1. (ii) Hence (a) solve the equation 2 cosec tan 5 for 0 [3] 21 1 = < 1 < 0, 160 (b) find the exact value of 2 cosec 4x tan2x dx. [4] Ó 0
Mark scheme: 1 6 (i) State or imply cosec2θ = B1 sin2θ Express left-hand side in terms of sinθ and cosθ M1 Obtain given answer sec 2 θ correctly A1 [3] 1 (ii) (a) State or imply cos θ = or tan θ = 2 at least B1 5 Obtain 1.11 or awrt 1.11, allow 0.353π B1 Obtain 2.03 or awrt 2.03 , allow 0.648π and no other values between 0 and π B1 [3] (b) State integrand as sec 2 2x B1 Integrate to obtain expression of form k tan mx M1 Obtain correct 12 tan 2x A1 Obtain 12 3 or exact equivalent A1 [4] 2 dy 3
Q7 · The equation of a curve is y3 4xy 16
7 The equation of a curve is y3 4xy 16. + = dy 4y (i) Show that [4] dx = − 3y2 4x. + (ii) Show that the curve has no stationary points. [2] (iii) Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. [4]
Mark scheme: 2 dy 37 (i) Obtain 3 y as derivative of y B1 dx dy Obtain 4 y + 4 x as derivative of 4xy B1 dx dy Equate derivative of left-hand side to zero and solve for , must be from dx implicit differentiation M1 dy 4 y Confirm given answer = − correctly A1 [4] dx 3 y 2 + 4 x (ii) State or imply y = 0 B1 Substitute in equation of curve and show contradiction B1 [2] (iii) State or imply 3 y 2 + 4 x = 0 B1 Eliminate one variable from equation of curve using 3 y 2 + 4 x = 0 M1 Obtain y = − 2 A1 Obtain x = − 3 A1 [4]
What was in this paper
The subtopics covered by these 4 questions, and how many questions each got. Open one in a new tab to see every Cambridge question on it.
What you needed in this session
Cambridge’s own grade thresholds for 2015 May/June, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.