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.

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Question paper4 pages

Cambridge A Level Mathematics 9709 2011 May/June Paper 2 · Variant 2 question paper, page 1 of 4
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Cambridge A Level Mathematics 9709 2011 May/June Paper 2 · Variant 2 question paper, page 2 of 4
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Cambridge A Level Mathematics 9709 2011 May/June Paper 2 · Variant 2 question paper, page 3 of 4
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Mark scheme5 pages

Answers below. Sit the paper first if you are practising.

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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]

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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]

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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

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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

More questions on Differentiation

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]

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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

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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]

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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 2011 May/June, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.

A41/50
B38/50
E20/50