Cambridge A Level Mathematics 9709 — 2012 May/June Paper 2 · Variant 2
9709/22/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 inequality [4] |x + 3| < |2x + 1|
1 Solve the inequality [4] |x + 3| < |2x + 1|.
Mark scheme: 1 Either: State or imply non-modular inequality (x + 3)2 < (2x + 1)2 or corresponding equation or pair of linear equations B1 Attempt solution of 3-term quadratic or of 2 linear equations M1 Obtain critical values − 43 and 2 A1 State answer x < − 43 , x > 2 A1 Or: Obtain critical value x = 2 from graphical method, inspection, equation B1 Obtain critical value x = − 43 similarly B2 State answer x < − 43 , x > 2 B1 [4] 2
Q2 · Given that 52x 5x 12, find the value of 5x
2 (i) Given that 52x 5x 12, find the value of 5x. [3] + = (ii) Hence, using logarithms, solve the equation 52x 5x 12, giving the value of x correct to + = 3 significant figures. [2]
Mark scheme: 2 (i) State or imply equation in the form (5x)2 + 5x – 12 = 0 B1 Attempt solution of quadratic equation for 5x M1 Obtain 5x = 3 only A1 [3] (ii) Use logarithms to solve equation of the form 5x = k where k > 0 M1 Obtain 0.683 A1 [2]
Q3 · Find the quotient when the polynomial 8x3 13 −4x2 −18x + is divided by 4x2 4x and show…
3 (i) Find the quotient when the polynomial 8x3 13 −4x2 −18x + is divided by 4x2 4x and show that the remainder is 4. [3] + −3, (ii) Hence, or otherwise, factorise the polynomial 8x3 9. −4x2 −18x + [2]
Mark scheme: 3 (i) Attempt division, or equivalent, at least as far as quotient 2x + k M1 Obtain quotient 2x – 3 A1 Complete process to confirm remainder is 4 A1 [3] (ii) State or imply (4x2 + 4x – 3) is a factor B1 Obtain (2x – 3)(2x – 1)(2x + 3) B1 [2]
Q4 · Α4 (i) Express 9 sin θ 0 and Give the value cos θ in the form R where R −12 sin(θ −α), >…
α4 (i) Express 9 sin θ 0 and Give the value cos θ in the form R where R −12 sin(θ −α), > < 90◦. 0◦< of α correct to 2 decimal places. [3] Hence (ii) solve the equation 9 sin θ cos θ 4 for [4] −12 = 0◦≤θ ≤360◦, (iii) state the largest value of k for which the equation 9 sin θ cos θ k has any solutions. [1] −12 =
Mark scheme: 4 (i) State or imply R = 15 B1 Use appropriate formula to find α M1 Obtain 53.13° A1 [3] (ii) Attempt to find at least one value of θ – α M1 Obtain one correct value 68.6° of θ A1 Carry out correct method to find second answer M1 Obtain 217.7° and no others in range A1 [4] (iii) State 15, following their value of R from part (i) B1√ [1] dx 1
Q5 · The parametric equations of a curve are x y e2t 2t
5 The parametric equations of a curve are x y e2t 2t. = ln(t + 1), = + dy (i) Find an expression for in terms of t. [4] dx (ii) Find the equation of the normal to the curve at the point for which t 0. Give your answer in = the form ax by c 0, where a, b and c are integers. [4] + + =
Mark scheme: d x 1 5 (i) State = B1 d t t + 1 dy 2 t State = 2e + 2 B1 dt dy Attempt expression for M1 dx dy 2 t Obtain = ( 2e + 2)(t + )1 or equivalent A1 [4] dx (ii) Substitute t = 0 and attempt gradient of normal M1 dy 1 Obtain − 4 following their expression for A1√ dx Attempt to find equation of normal through point (0, 1) M1 Obtain x + 4y – 4 = 0 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 22
Q6 · Y M 1 x O a p 2 sin 2x for 0 The diagram shows the curve y The x-coordinate of the…
6 y M 1 x O a p 2 sin 2x for 0 The diagram shows the curve y The x-coordinate of the maximum point M ≤x ≤12π. = x 2 is denoted by α. + dy (i) Find and show that α satisfies the equation tan 2x 2x 4. [4] dx = + (ii) Show by calculation that α lies between 0.6 and 0.7. [2] 2 tan−1(2xn + 4) (iii) Use the iterative formula xn+1 = 1 to find the value of α correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
Mark scheme: 6 (i) Attempt use of quotient rule or equivalent M1 2( x + 2) cos 2 x − sin 2 x Obtain or equivalent A1 ( x + 22) Equate numerator to zero and attempt rearrangement M1 Confirm given result tan 2x = 2x + 4 A1 [4] (ii) Consider sign of tan 2x – 2x – 4 for 0.6 and 0.7 or equivalent M1 Obtain –2.63 and 0.40 or equivalents and justify conclusion A1 [2] (iii) Use iteration process correctly at least once M1 Obtain final answer 0.694 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6935, 0.6945) A1 [3] [0.6 → 0.69040 → 0.69352 → 0.69363 0.65 → 0.69215 → 0.69358 → 0.69363 0.7 → 0.69384 → 0.69364 → 0.69363] 2 2
Q7 · Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx
7 (i) Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx. ã 0 (tan2x + cos2x) [7] (ii) y 1 x O 4p 0 is shown in The region enclosed by the curve y tan x cos x and the lines x 0, x = + = = 14π and y = the diagram. Find the exact volume of the solid produced when this region is rotated completely about the x-axis. [4]
Mark scheme: 7 (i) Replace tan2 x by sec2 x – 1 B1 Express cos2 x in the form ± 12 ± 12 cos 2 x M1 Obtain given answer sec 2 x + 12 cos 2 x − 12 correctly A1 Attempt integration of expression M1 Obtain tan x + 14 sin 2 x − 12 x A1 Use limits correctly for integral involving at least tan x and sin 2x M1 Obtain 54 − 18 π or exact equivalent A1 [7] (ii) State or imply volume is ∫ π (tan x + cos x 2) dx B1 Attempt expansion and simplification M1 Integrate to obtain one term of form k cos x M1 Obtain π ( 54 − 18 π ) + π ( 2 − 2 ) or 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 2. A higher threshold means an easier paper — the bar moves with how the cohort did.