Cambridge A Level Mathematics 9709 — 2008 Oct/Nov Paper 2 · Variant 1
9709/21/O/N/08 · 8 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 |x −3| > |2x|
1 Solve the inequality |x −3| > |2x|. [4]
Mark scheme: 1 EITHER: State or imply non-modular inequality (x – 3)2 > (2x)2 or corresponding quadratic equation or pair of linear equations (x – 3) = ± 2x M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 Obtain critical values x = 1 and x = –3 A1 State answer –3 < x < 1 A1 OR: Obtain critical value x = –3 from a graphical method, or by inspection, or by solving a linear inequality or linear equation B1 Obtain the critical value x = 1 similarly B2 State answer –3 < x < 1 B1 [4]
Q2 · The polynomial 2x3 −x2 + ax −6, where a is a constant, is denoted by p(x)
2 The polynomial 2x3 −x2 + ax −6, where a is a constant, is denoted by p(x). It is given that (x + 2) is a factor of p(x). (i) Find the value of a. [2] (ii) When a has this value, factorise p(x) completely. [3]
Mark scheme: 2 (i) Substitute x = –2 and equate result to zero, or divide by x + 2 and equate constant remainder to zero M1 Obtain answer a = –13 A1 [2] (ii) Obtain quadratic factor 2x2 – 5x – 3 B1 Obtain linear factor 2x + 1 B1 Obtain linear factor x –3 B1 [3] [Condone omission of repetition that x + 2 is a factor.] [If linear factors 2x + 1, x – 3 obtained by remainder theorem or inspection, award B2 + B1.]
Q3 · Ln y (0, 1.3) (1.6, 0.9) x O The variables x and y satisfy the equation y = A(b−x), where…
3 ln y (0, 1.3) (1.6, 0.9) x O The variables x and y satisfy the equation y = A(b−x), where A and b are constants. The graph of ln y against x is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. Find the values of A and b, correct to 2 decimal places. [5]
Mark scheme: 3 State or imply ln y = ln A – xln b B1 State ln A = 1.3 B1 Obtain A = 3.67 B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.28 A1 [5]
Q4 · Show that the equation sin(x + 30◦) = 2 cos(x + 60◦) can be written in the form (3 √3)…
4 (i) Show that the equation sin(x + 30◦) = 2 cos(x + 60◦) can be written in the form (3 √3) sin x = cos x. [3] (ii) Hence solve the equation sin(x + 30◦) = 2 cos(x + 60◦), for −180◦≤x ≤180◦. [3] 2
Mark scheme: 4 (i) Use correct sin(A + B) and cos(A + B) formulae M1 Substitute exact values for sin 30° etc. M1 Obtain given answer correctly A1 [3] (ii) Solve for x M1 Obtain answer x = 10.9° A1 Obtain second answer x = –169.1° and no others in the range A1 [3] [Ignore answers outside the given range.]
Q5 · 4 18 −5 Show that dx = ln 25
1 4 18 −5 Show that dx = ln 25. [6] x 2x + 1 1
Mark scheme: 5 Integrate and state term ln x B1 Obtain term of the form kln (2x + 1) M1 State correct term –2ln (2x + 1) A1 Substitute limits correctly M1 Use law for the logarithm of a product, quotient or power M1 Obtain given answer correctly A1 [6] − 1 x 1 x
Q6 · D2y 2x6 Find the exact coordinates of the point on the curve y = xe−1 at which = 0
d2y 2x6 Find the exact coordinates of the point on the curve y = xe−1 at which = 0. [7] dx2
Mark scheme: x x 6 At any stage, state the correct derivative of e 2 or 2e B1 Use product or quotient rule M1 Obtain correct first derivative in any form A1 Obtain correct second derivative in any form B1√ Equate second derivative to zero and solve for x M1 Obtain x = 4 A1 Obtain y = 4e– 2, or equivalent A1 [7] GCE A/AS LEVEL – October/November 2008 9709 02
Q7 · By sketching a suitable pair of graphs, show that the equation cos x = 2 −2x, where x is…
7 (i) By sketching a suitable pair of graphs, show that the equation cos x = 2 −2x, where x is in radians, has only one root for 0 ≤x ≤12π. [2] (ii) Verify by calculation that this root lies between 0.5 and 1. [2] (iii) Show that, if a sequence of values given by the iterative formula xn+1 = 1 −12 cos xn converges, then it converges to the root of the equation in part (i). [1] (iv) Use this iterative formula, with initial value x1 = 0.6, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
Mark scheme: 7 (i) Make a recognizable sketch of a relevant graph, e.g. y = cos x or y = 2 – 2x B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of cos x – (2 – 2x) at x = 0.5 and x = 1, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to x = 1 – 1 cos x, or vice versa B1 [1] 2 (iv) Use the iterative formula correctly at least once Ml Obtain final answer 0.58 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.575, 0.585) B1 [3]
Q8 · Prove the identity + sin x sec2x + sec x tan x ≡1
8 (i) (a) Prove the identity + sin x sec2x + sec x tan x ≡1 . cos2x (b) Hence prove that 1 sec2x + sec x tan x ≡ 1 −sin x. [3] 1 dy (ii) By differentiating cos x, show that if y = sec x then dx = secx tan x. [3] (iii) Using the results of parts (i) and (ii), find the exact value of 14π 1 dx. [3] 1 −sin x 0
Mark scheme: 8 (i) (a) Use trig formulae and justify given result B1 (b) Use 1 – sin2 x = cos2 x M1 Obtain given result correctly A1 [3] (ii) Use quotient or chain rule M1 Obtain correct derivative in any form A1 Obtain given result correctly A1 [3] (iii) Obtain integral tan x + sec x B1 Substitute limits correctly M1 Obtain exact answer 2 , or equivalent A1 [3]
What was in this paper
The subtopics covered by these 8 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 2008 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.