Cambridge A Level Mathematics 9709 — 2011 May/June Paper 2 · Variant 2
9709/22/M/J/11 · 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 · Use logarithms to solve the equation 3x giving your answer correct to 3 significant figures
1 Use logarithms to solve the equation 3x giving your answer correct to 3 significant figures. = 2x+2, [4]
Mark scheme: 1 Attempt use of power law for logarithms M1* Obtain xlog3 = xlog2 + 2log2 or equivalent A1 Attempt solution for x of linear equation M1 dep* Obtain 3.42 A1 [4]
Q2 · Y 3 B A x O 2 The diagram shows the curve y Region A is bounded by the curve and the…
2 y 3 B A x O 2 The diagram shows the curve y Region A is bounded by the curve and the lines x 0, = √(1 + x3). = x 2 and y 0. Region B is bounded by the curve and the lines x 0 and y 3. = = = = (i) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give your answer correct to 2 decimal places. [3] (ii) Deduce an approximation to the area of region B and explain why this approximation under- estimates the true area of region B. [2]
Mark scheme: 2 (i) Show or imply correct ordinates 1, 2 or 1.414, 3 B1 Use correct formula, or equivalent, with h = 1 M1 Obtain 3.41 A1 [3] (ii) Obtain 6 – 3.41 and hence 2.59, following their answer to (i) provided less than 6 B1√ Refer, in some form, to two line segments replacing curve and conclude with clear justification of given result that answer is an under-estimate. B1 [2]
Q3 · The sequence x1, x2, x3,
3 The sequence x1, x2, x3, . . . defined by x2n 6 x1 1, 12 3p = xn+1 = + converges to the value α. (i) Find the value of α correct to 3 decimal places. Show your working, giving each calculated value of the sequence to 5 decimal places. [3] (ii) Find, in the form ax3 bx2 c 0, an equation of which α is a root. [2] + + = 2
Mark scheme: 3 (i) Use the iteration process correctly at least once M1 Obtain at least two correct iterates to 5 decimal places A1 Conclude α = 0.952 A1 [3] [1 → 0.95647 → 0.95257 → 0.95223 → 0.95220] 1 3 2 (ii) State or imply equation is x = x + 6 B1 2 Obtain 8x3 – x2 – 6 = 0 B1 [2] 1
Q5 · A curve has equation x2 2y2 5x 6y 10
5 A curve has equation x2 2y2 5x 6y 10. Find the equation of the tangent to the curve at the + + + = point Give your answer in the form ax by c 0, where a, b and c are integers. [6] (2, −1). + + =
Mark scheme: dy 5 Obtain 4 y as derivative of 2y2 B1 dx dy Differentiate LHS term by term to obtain expression including at least one M1 dx dy dy Obtain 2 x + 4 y + 5 + 6 A1 dx dx dy Substitute 2 and –1 to attempt value of M1 dx 9 Obtain − A1 2 Obtain equation 9x + 2y –16 = 0 or equivalent of required form A1 [6] GCE AS/A LEVEL – May/June 2011 9709 22
Q6 · The curve y 4x2 ln x has one stationary point
6 The curve y 4x2 ln x has one stationary point. = (i) Find the coordinates of this stationary point, giving your answers correct to 3 decimal places. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]
Mark scheme: 6 (i) Attempt differentiation using product rule M1 Obtain 8 x ln x + 4 x (a.c.f.) A1 Equate first derivative to zero and attempt solution M1 Obtain 0.607 A1 Obtain –0.736 following their x-coordinate A1√ [5] (ii) Use an appropriate method for determining nature of stationary point M1 Conclude point is a minimum (with no errors seen, second derivative = 8) A1 [2]
Q7 · The cubic polynomial is defined by p(x) 6x3 ax2 bx 10, p(x) = + + + where a and b are…
7 The cubic polynomial is defined by p(x) 6x3 ax2 bx 10, p(x) = + + + where a and b are constants. It is given that is a factor of and that, when is divided (x + 2) p(x) p(x) by the remainder is 24. (x + 1), (i) Find the values of a and b. [5] (ii) When a and b have these values, factorise completely. [3] p(x)
Mark scheme: 7 (i) Substitute x = –2 and equate to zero M1 Substitute x = –1 and equate to 24 M1 Obtain 4a – 2b = 38 and a – b = 20 or equivalents A1 Attempt solution of two linear simultaneous equations (dependent on M1 M1) M1 Obtain a = –1 and b = –21 A1 [5] (ii) Attempt to find quadratic factor by division, inspection or use of identity M1 Obtain 6x2 – 13x + 5 A1√ Conclude ( x + 2 )(2 x − 1)(3 x − 5 ) A1 [3] 1 1
Q8 · Prove that cos 2θ
8 (i) Prove that cos 2θ. [3] sin22θ(cosec2θ −sec2θ) ≡4 (ii) Hence (a) solve for equation 3, [4] 0◦≤θ ≤180◦the sin22θ(cosec2θ −sec2θ) = (b) find the exact value of [2] cosec215◦−sec215◦.
Mark scheme: 1 1 8 (i) Use cosecθ = and secθ = B1 sin θ cos θ Attempt to simplify left-hand side M1 Confirm given right-hand side 4cos2θ with no errors seen A1 [3] 3 (ii) (a) State or imply cos2θ = B1 4 Attempt correct process to find at least one angle M1 Obtain 20.7° A1 Obtain 159.3° and no others in range A1 [4] 4 cos 30 o (b) Recognise as 2 o B1 sin 30 Obtain 8 3 B1 [2]
What was in this paper
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What you needed in this session
Cambridge’s own grade thresholds for 2011 May/June, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.