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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, February/March 2022. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_m22_qp_229709_m22_ms_229709/22/F/M/22
Each question as it's printed, and the mark scheme's points for it, as text. Diagrams, tables and graphs are in the PDFs above.
Question
Solve the equation 5x − 2 = 4x + 9. [3]
Mark scheme
1 Solve 5 x − 2 = 4 x + 9 to obtain x = 11 B1 Attempt solution of linear equation where signs of 5x and 4x are M1 different Obtain final value x = − 79 A1 Alternative method for question 1 State or imply non-modulus equation (5 x − 2) 2 = (4 x + 9) 2 B1 Attempt solution of 3-term quadratic equation M1 Obtain x = − 79 and x =11 A1 3
Question
A curve has equation y = 7 + 4 ln 2x + 5. Find the equation of the tangent to the curve at the point −2, 7, giving your answer in the form y = mx + c. [5]
Mark scheme
2 k M1 Any non-zero constant k Differentiate to obtain form 2x + 5 8 A1 Obtain correct 2x + 5 Substitute x = −2 to obtain gradient 8 A1 Attempt equation of tangent through (−2, 7) with numerical gradient M1 Obtain y = 8 x + 23 A1 5
Question
The variables x and y satisfy the equation y = 32a ax , where a is a constant. The graph of ln y against x is a straight line with gradient 0.239. (a) Find the value of a correct to 3 significant figures. [3] (b) Hence find the value of x when y = 36. Give your answer correct to 3 significant figures. [2]
Mark scheme
3(a) State or imply equation is ln y = ln 32 a + x ln a B1 Equate gradient of line involving a to 0.239 M1 Obtain ln a = 0.239 and hence a = 1.27 A1 3 3(b) Substitute y = 36 in ln y = ... equation and solve for x M1 Or substitute in original equation with necessary manipulation Obtain 3.32 A1 2
Question
(a) Show that sin 21 cot 1 − cos 21 1. [3] (b) Hence find the exact value of sin 16 π cot 12 1 π. [2] (c) Find the smallest positive value of 1 (in radians) satisfying the equation sin 21 cot 1 − 3 cos 21 = 1. [2] (a) Given that y = tan2 x, show that = 2 tan x + 2 tan3 x. dy
Mark scheme
4(a) Use at least two of sin 2θ = 2sin θ cos θ , cos 2θ = cos 2 θ − sin 2 θ , B1 OE cos θ cot θ = sin θ Express LHS in terms of sin θ and cos θ only and attempt valid M1 simplification Obtain cos 2 θ + sin 2 θ or equivalent and hence 1 A1 AG – necessary detail needed 3 4(b) Substitute θ = 121 π and show or imply sin 16 π cot 121 π = 1 + cos 16 π M1 Obtain 1 + 12 3 or exact equivalent A1 2 4(c) Use the identity from part (a) to obtain −2cos 2θ = 0 or equivalent M1 Or alternative starting again, using valid simplification and reaching single trigonometric function. Obtain θ = 14 π A1 2
Question
[2] dx 1π (b) Find the exact value of Ó tan x + tan2 x + tan3 x dx. 3 [6] 1π 4
Mark scheme
5(a) Differentiate to obtain 2 tan x sec 2 x B1 Use sec 2 x = 1 + tan 2 x to confirm 2 tan x + 2 tan 3 x B1 AG – necessary detail needed 2 5(b) Attempt to use part (a) result to integrate tan x + tan 3 x M1 Obtain 12 tan 2 x A1 Use relevant identity to integrate tan 2 x M1 Obtain sec 2 x − 1 and hence tan x − x A1 Use limits correctly for integrand of form k1 tan 2 x + k2 tan x + k3 x M1 Obtain 3 − 121 π A1 Or exact equivalent 6
Question
The polynomial p x is defined by p x = 4x3 + 16x2 + 9x − 15. (a) Find the quotient when p x is divided by 2x + 3, and show that the remainder is −6. [3] (b) Find Ô p x 2x + 3 dx. [2] (c) Factorise p x + 6 completely and hence solve the equation p cosec 21 + 6 = 0 for 0Å < 1 < 135Å. [5]
Mark scheme
6(a) Carry out division at least as far as 2 x 2 + kx M1 Obtain quotient 2 x 2 + 5 x − 3 A1 Confirm remainder is –6 A1 AG – necessary detail needed 3 6(b) Integrate to obtain at least k1 x 3 and k2 ln(2 x + 3) terms M1 Obtain 23 x 3 + 52 x 2 − 3 x − 3ln(2 x + 3) A1 condone absence of ... + c and modulus signs 2 6(c) State or imply p( x ) + 6 = (2 x + 3)(2 x 2 + 5 x − 3) B1 FT FT their quotient Conclude (2 x + 3)(2 x − 1)( x + 3) B1 State or imply sin 2θ = − 23 or sin 2θ = − 13 or both B1 FT FT their relevant factors Carry out correct process to find θ in at least one case M1 Obtain 99.7 and 110.9 A1 Or greater accuracy and no others between 0° and 135° 5
Question
A curve has equation e2x y − ey = 100. = y dy 2e2x y dx e − e2x (a) Show that . [3] (b) Show that the curve has no stationary points. [2] It is required to find the x-coordinate of P, the point on the curve at which the tangent is parallel to the y-axis. (c) Show that the x-coordinate of P satisfies the equation x = ln 10 − 12 ln 2x − 1. [4] (d) Use an iterative formula, based on the equation in part (c), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures. [3] Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
Mark scheme
7(a) Use product rule to differentiate e 2 x y M1 dy A1 Obtain 2e 2 x y + e 2 x dx dy dy A1 AG – necessary detail needed Obtain 2e 2 x y + e 2 x − ey = 0 and rearrange to confirm given result dx dx 3 7(b) Consider e 2 x y = 0 and either state e 2 x ≠ 0 or substitute y = 0 in equation M1 of curve Complete argument with e 2 x ≠ 0 or e 2 x > 0 and substitution to show y A1 AG – necessary detail needed cannot be zero 2 7(c) State or imply e y − e 2 x = 0 and hence y = 2 x B1 Substitute for y in equation of curve and attempt rearrangement as far as M1 e 2 x = ... Use relevant logarithm properties M1 Confirm equation x = ln10 − 12 ln(2 x − 1) A1 AG – necessary detail needed 4 7(d) Use iteration process correctly at least once M1 Obtain final answer 1.82 A1 Answer required to exactly 3 sf Show sufficient iterations to 5 sf to justify answer or show sign change in A1 interval [1.815, 1.825] 3
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Mathematics 9709/22 for February/March 2022 is out of 50 marks, with 1 hour 15 minutes to answer it.
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