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Cambridge AS & A Level Mathematics (9709), Paper 2 Pure Mathematics 2, May/June 2023. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_s23_qp_219709_s23_ms_219709/21/M/J/23
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
Use logarithms to solve the equation 12x = 32x+1 . Give your answer correct to 3 significant figures. [4] 2 + 3 ln x A curve has equation y = 1 + 2x
Mark scheme
1 Apply logarithms to both sides and apply power law correctly at least once *M1 OE with x not in a power. Obtain x ln12 = (2 x + 1)ln 3 A1 Attempt solution of linear equation DM1 Obtain 3.82 A1 Do not condone incorrect use of logarithms or greater accuracy. 4
Question
. Find the equation of the tangent to the curve at the point 1, 32 . Give your answer in the form ax + by + c = 0, where a, b and c are integers. [5] It is given that Ó 3e2x − 1 dx = 12, where a is a positive constant. a
Mark scheme
2 Use quotient rule (or equivalent) to find first derivative M1 (1 + 2 x) 3x − (2 + 3ln x)2 A1 OE Obtain 2 (1 + 2 x) Substitute x = 1 and obtain 59 A1 Attempt equation of tangent through (1, 23 ) with their numerical gradient M1 Must have made an attempt at differentiation. Obtain 5 x − 9 y + 1 = 0 or equivalent of required form A1 5
Question
0 (a) Show that a = 12 ln 9 + 23 a . [4] (b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3]
Mark scheme
3(a) Integrate to obtain the form k1e2 x + k2 x *M1 Where k1k 2 ≠ 0 . Obtain correct 3 2x e −x A1 2 Use limits correctly and attempt rearrangement at least as far as e 2 a = ... DM1 For DM1, must be equated to 12 and simplified using a correct method. Do not condone verification. Confirm given result a = 12 ln(9 + 23 a ) with sufficient detail A1 AG 4 3(b) Use iteration process correctly at least once M1 Need to see 1.13434 and 1.13895 . Obtain final answer 1.139 A1 Final answer needed to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show a sign change in A1 interval [1.1385, 1.1395] 3
Question
The polynomial p x is defined by p x = 2x3 + 3x2 + kx − 30, where k is a constant. It is given that x − 3 is a factor of p x. (a) Find the value of k. [2] (b) Hence find the quotient when p x is divided by x − 3 and factorise p x completely. [3] (c) It is given that a is one of the roots of the equation p x = 0. Given also that the equation 4y − 5 = a is satisfied by two real values of y, find these two values of y. [3]
Mark scheme
4(a) Substitute x = 3 , equate to zero and attempt solution M1 51 Condone − . 3 Obtain k = −17 A1 2 4(b) Divide by x − 3 at least as far as 2 x 2 + mx M1 Obtain 2 x 2 + 9 x + 10 A1 Obtain ( x − 3)(2 x + 5)( x + 2) A1 SC – no attempt at division (or equivalent) and only correct fully factorised form shown: award B1 only. 3 4(c) Solve linear equation 4 y − 5 = 3 to obtain y = 2 B1 FT following any positive root from their factorised p( x) . Attempt solution of linear equation 4 y − 5 = −3 or equivalent M1 with RHS of equation being – (their positive root). Obtain y = 12 A1 Alternative method for Question 4(c) State or imply (4 y − 5) 2 = 32 B1 FT following any positive root from their factorised p( x) . Attempt solution of 3-term quadratic equation (4 y − 5) 2 = 32 M1 with RHS of equation involving their positive root. Obtain 12 and 2 and no other solutions A1 3
Question
y M x O The diagram shows the curve with parametric equations x = 4e2t , y = 5e−t cos 2t, for − 14 π ≤ t ≤ 14 π. The curve has a maximum point M . dy (a) Find an expression for in terms of t. [3] dx (b) Find the coordinates of M , giving each coordinate correct to 3 significant figures. [5] 1π @ A Show that Ô 4 cos2 2x + dx = 34 3 + 16 π − 1. 3 1
Mark scheme
5(a) dy M1 Use product rule to find dt dy A1 or (unsimplified) equivalent (do not condone poor use of Obtain = −5e−t cos 2t − 10e−t sin 2t brackets. dt dy −5e −t cos 2t − 10e −t sin 2t A1 dy Obtain = OE following their expression for . dx 8e 2t dt 3 5(b) dy M1* dx Equate to zero and simplify at least as far as tan 2t = ... now condoning any error with . dx dt Obtain tan 2t = − 12 A1 Obtain t = −0.231... A1 allow t = −0.232 . Substitute negative value of t in expressions for x and y DM1 Obtain x = 2.52 and y = 5.64 A1 or greater accuracy. 5
Question
2 [7] 1π cos x 4
Mark scheme
6 Express 4cos 2 2x in the form k1 cos 4 x + k2 M1 where k1k 2 ≠ 0 . Obtain correct 2 cos 4 x + 2 A1 Allow unsimplified. 1 B1 Maybe implied by integration. State or imply 2 = sec 2 x cos x Integrate to obtain k3 sin 4 x + k 4 x + tan x *M1 where k3 k4 ≠ 0 . Obtain correctly 12 sin 4 x + 2 x + tan x A1 Use limits correctly with correct values of sin 43 π and tan 13 π indicated DM1 Confirm given result 34 3 + 16 π − 1 with sufficient detail A1 AG 7
Question
(a) Express 7 cos 1 + 24 sin 1 in the form R cos 1 − !, where R > 0 and 0Å < ! < 90Å. Give the value of ! correct to 2 decimal places. [3] (b) Solve the equation 7 cos 1 + 24 sin 1 = 18 for 0Å < 1 < 360Å. [4] (c) As " varies, the greatest possible value of 150 7 cos 12 " + 24 sin 21 " + 50 is denoted by V . Find the value of V and determine the smallest positive value of " (in degrees) for which the value of V occurs. [4] 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) State R = 25 B1 Use appropriate trigonometry to find α M1 Allow if found in radians . Obtain α = 73.74 A1 or greater accuracy. 3 7(b) Use correct method to find one value of θ M1 Obtain 29.8 (or 117.7) A1 or greater accuracy. Use correct method to find second value of θ between 0 and 360 M1 Obtain 117.7 (or 29.8) A1 or greater accuracy; and no others between 0 and 360. 4 7(c) 150 B1 FT following their R and α . State or imply expression is 25cos( 12 β − 73.74) + 50 Obtain V = 6 B1 Attempt complete method to find positive value from cos( 12 β − 73.74) = −1 M1 for their α . Obtain 507.5 A1 or greater accuracy. 4
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Mathematics 9709/21 for May/June 2023 is out of 50 marks, with 1 hour 15 minutes to answer it.
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