Cambridge A Level Mathematics 9709 — 2012 May/June Paper 2 · Variant 3
9709/23/M/J/12 · 7 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 13, showing all your working
1 Solve the equation 13, showing all your working. [4] |x3 −14| =
Mark scheme: 1 Either: Obtain value x3 = 27 from inspection, equation, … B1 Obtain value x3 = 1 similarly B2 Obtain x = 1 and x = 3 B1 Or: Attempt to square both sides obtaining 3 terms on LHS M1 Attempt solution for x3 of 3-term quadratic DM1 Obtain x3 = 1 and x3 = 27 A1 Obtain x = 1 and x = 3 A1 [4]
Q2 · Ln y (5, 4.49) (0, 2.14) x O The variables x and y satisfy the equation y where A and b…
2 ln y (5, 4.49) (0, 2.14) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y = A(bx), against x is a straight line passing through the points and as shown in the diagram. (0, 2.14) (5, 4.49), Find the values of A and b, correct to 1 decimal place. [5]
Mark scheme: 2 State or imply that ln y = ln A + x ln b B1 Equate intercept on y-axis to ln A M1 Obtain ln A = 2.14 and hence A = 8.5 A1 Attempt gradient of line or equivalent (or use of correct substitution) M1 Obtain 0.47 = ln b or equivalent and hence b = 1.6 A1 [5]
Q3 · The polynomial is defined by p(x) ax3 a 4, p(x) = −3x2 −5x + + where a is a constant
3 The polynomial is defined by p(x) ax3 a 4, p(x) = −3x2 −5x + + where a is a constant. (i) Given that is a factor of find the value of a. [2] (x −2) p(x), (ii) When a has this value, (a) factorise completely, [3] p(x) (b) find the remainder when is divided by [2] p(x) (x + 1).
Mark scheme: 3 (i) Substitute 2 and equate to zero or divide and equate remainder to zero M1 Obtain a = 2 A1 [2] (ii) (a) Attempt to find quadratic factor by division, inspection or identity M1 Obtain 2x2 + x – 3 A1 Conclude (x – 2)(2x + 3)(x – 1) A1 [3] (b) Attempt substitution of –1 or attempt complete division by x + 1 M1 Obtain 6 A1 [2] 2 2
Q4 · Given that 35 sec2θ 12 tan θ, find the value of tan θ
4 (i) Given that 35 sec2θ 12 tan θ, find the value of tan θ. [3] + = (ii) Hence, showing the use of an appropriate formula in each case, find the exact value of (a) [2] tan(θ −45◦), (b) tan 2θ. [2]
Mark scheme: 4 (i) Use sec2 θ = 1 + tan2 θ B1 Attempt solution of quadratic equation in tan θ M1 Obtain tan2 θ – 12 tanθ + 36 = 0 or equivalent and hence tan θ = 6 A1 [3] (ii) (a) Attempt use of tan(A – B) formula M1 Obtain 75 following their value of tan θ A1√ [2] (b) Attempt use of tan 2θ formula M1 Obtain − 1235 A1 [2] 1 x
Q5 · Y M x O 12x The diagram shows the curve y 4e 3 and its minimum point M
5 y M x O 12x The diagram shows the curve y 4e 3 and its minimum point M. = −6x + (i) Show that the x-coordinate of M can be written in the form ln a, where the value of a is to be stated. [5] (ii) Find the exact value of the area of the region enclosed by the curve and the lines x 0, x 2 = = and y 0. [4] =
Mark scheme: 2 x5 (i) Differentiate to obtain expression of form ke + m M1 1 2 x Obtain correct 2e − 6 A1 Equate attempt at first derivative to zero and attempt solution DM1 Obtain 12 x = ln 3 or equivalent A1 Conclude x = ln 9 or a = 9 A1 [5] 1 2 x 2 (ii) Integrate to obtain expression of form ae + bx + cx M1 1 2 x 2 Obtain correct 8e − 3 x + 3 x A1 Substitute correct limits and attempt simplification DM1 Obtain 8e – 14 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 23 3
Q6 · A curve has parametric equations 1 x , y = = √(t + 2)
6 A curve has parametric equations 1 x , y = = √(t + 2). (2t + 1)2 The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1. 1 (i) Show that p 6 2. [6] = (p + 2) −1 (ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places. [3]
Mark scheme: 6 (i) Obtain derivative of form k(2t + 1)–3 M1 Obtain –4(2t + 1)–3 or equivalent as derivative of x A1 1 − 12 Obtain 2 (t + 2) or equivalent as derivative of y B1 dy Equate attempt at to –1 M1 dx 3 12 Obtain ( 2 p + )1 = 8( p + 2) or equivalent A1 1 Confirm given answer p = ( p + 2) 6 − 12 A1 [6] (ii) Use iteration process correctly at least once M1 Obtain final answer 0.678 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6775, 0.6785) A1 [3] [0.7 → 0.68003 → 0.67857 → 0.67847 → 0.67846] 2 2
Q7 · Show that sin x cos can be written in the form 5 2 sin 2x cos 2x
7 (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos dx. [4] ã 0 (2 + x)2
Mark scheme: 7 (i) Expand to obtain 4 sin2 x + 4 sin x cos x + cos2 x B1 Use 2 sin x cos x = sin 2x B1 Attempt to express sin2 x or cos2 x (or both) in terms of cos 2x M1 Obtain correct 12 k 1( − cos 2 x ) for their k sin2 x or equivalent A1√ Confirm given answer 52 + 2 sin 2 x − 32 cos 2 x A1 [5] (ii) Integrate to obtain form px + q cos 2x + r sin 2x M1 Obtain 52 x − cos 2 x − 34 sin 2 x A1 Substitute limits in integral of form px + q cos 2x + r sin 2x and attempt simplification DM1 Obtain 85 π + 14 or exact equivalent A1 [4]
What was in this paper
The subtopics covered by these 7 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 2012 May/June, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.