Cambridge A Level Mathematics 9709 — 2012 Oct/Nov Paper 2 · Variant 2
9709/22/O/N/12 · 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 [3] |2x + 1| < |2x −5|
1 Solve the inequality [3] |2x + 1| < |2x −5|.
Mark scheme: 1 EITHER State or imply non-modular inequality (2 x + 1)2 < (2 x − 5 )2 , or M1 corresponding equation or pair of linear equations Obtain critical value 1 A1 State correct answer x < 1 A1 OR State the critical value x = 1, by solving a linear equation (or inequality) or from a graphical method or by inspection B2 State correct answer x < 1 B1 [3]
Q2 · Sin 2x 2 The curve with equation y has one stationary point in the interval 0 2π
sin 2x 2 The curve with equation y has one stationary point in the interval 0 2π. Find the exact = e2x ≤x ≤1 x-coordinate of this point. [4]
Mark scheme: 2 Use quotient rule or product rule, correctly M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 π Obtain x = A1 [4] 8 2 2
Q3 · The polynomial x4 3x2 4x is denoted by −4x3 + + −4 p(x)
3 The polynomial x4 3x2 4x is denoted by −4x3 + + −4 p(x). (i) Find the quotient when is divided by x2 2. [3] p(x) −3x + (ii) Hence solve the equation 0. [3] p(x) =
Mark scheme: 3 (i) Attempt division by x2 – 3x + 2 or equivalent, and reach a partial quotient of x 2 + kx M1 Obtain partial quotient x 2 − x A1 Obtain x 2 −x − 2 with no errors seen A1 [3] (ii) Correct solution method for either quadratic e.g. factorisation M1 One correct solution from solving quadratic or inspection B1 All solutions x = 2, x = 1 and x = –1 given and no others A1 [3]
Q4 · Y x O 1p 2 The diagram shows the part of the curve y for 0 2π
4 y x O 1p 2 The diagram shows the part of the curve y for 0 2π. = √(2 −sin x) ≤x ≤1 (i) Use the trapezium rule with 2 intervals to estimate the value of 12π dx, ã 0 √(2 −sin x) giving your answer correct to 2 decimal places. [3] (ii) The line y x intersects the curve y at the point P. Use the iterative formula = = √(2 −sin x) xn+1 = √(2 −sin xn) to determine the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]
Mark scheme: 4 (i) State or imply correct ordinates 1.4142…, 1.1370…, 1 B1 π Use correct formula, or equivalent, correctly with h = and three ordinates M1 4 Obtain answer 1.84 with no errors seen A1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 1.06 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.055, 1.065) B1 [3]
Q5 · Ln y (1, 2.9) (3.5, 1.4) x O The variables x and y satisfy the equation y where A and b…
5 ln y (1, 2.9) (3.5, 1.4) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y against x is a straight line passing through the= A(b−x),points and as shown in the diagram. Find the values of A and b, correct to 2 decimal places.(1, 2.9) (3.5, 1.4), [6]
Mark scheme: 5 State or imply ln y = ln A − x ln b B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.82 A1 Use gradient and one point correctly to find ln A M1 Obtain ln A = 3.5 A1 Obtain A = 33.12 A1 [6] GCE AS LEVEL – October/November 2012 9709 22 1 − x
Q6 · X6 (a) Find 4e−1 dx
2x6 (a) Find 4e−1 dx. [2] ã 3 6 (b) Show that dx ln 16. [5] 3x = ä1 −1
Mark scheme: 6 (a) Obtain integral ke 2 with any non-zero k M1 Correct integral A1 [2] (b) State indefinite integral of the form k ln (3x – 1), where k = 2, 6 or 3 M1 State correct integral 2 ln (3x – 1) A1 Substitute limits correctly (must be a function involving a logarithm) M1 Use law for the logarithm of a power or a quotient M1 Obtain given answer correctly A1 [5] dy 2
Q7 · The equation of a curve is 3x2 2y2 0
7 The equation of a curve is 3x2 2y2 0. −4xy + −6 = dy 3x (i) Show that [4] −2y dx 2x = −2y. (ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the x-axis. [5]
Mark scheme: dy 27 (i) State 4 y as derivative of 2y , or equivalent B1 dx dy State 4 y + 4 x as derivative of 4xy, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [4] (ii) State or imply that the coordinates satisfy 3x – 2y = 0 B1 Obtain an equation in x2 (or y2) M1 Solve and obtain x2 = 4 (or y2 = 9) A1 State answer (2 , 3) A1 State answer (−2, −3) A1 [5]
Q8 · Given that tan A t and 4, find tan B in terms of t
8 (a) Given that tan A t and 4, find tan B in terms of t. [3] = tan(A + B) = (b) Solve the equation 2 3 tan x, tan(45◦−x) = giving all solutions in the interval [6] 0◦≤x ≤360◦.
Mark scheme: 8 (a) Use tan (A + B) formula to obtain an equation in tan B M1 t + tan B State equation = 4 , or equivalent A1 1 − t tan B 4 − t Solve to obtain tan B = A1 [3] 1 + 4t tan 45 − tan x (b) State equation 2 = 3 tan x , or equivalent B1 1 + tan 45 tan x Transform to a quadratic equation M1 Obtain 3tan2 x + 5tan x – 2 = 0 (or equivalent) A1 Solve the quadratic and calculate one angle, or establish that tan x = ⅓, –2 M1 Obtain one answer, e.g. x = 18.4o A1 Obtain other 3 answers 116.6o, 198.4o, 296.6o and no others in range A1 [6]
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 2012 Oct/Nov, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.