Cambridge A Level Mathematics 9709 — 2019 May/June Paper 2 · Variant 3
9709/23/M/J/19 · 4 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 paper12 pages
Mark scheme11 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q2 · Solve the equation 4 2x 3
2 (i) Solve the equation 4 2x 3 . [3] + = −5x ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence solve the equation 4 2e3y 3 , giving the answer correct to 3 significant figures. + = −5e3y [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 2(i) State or imply non-modular equation 2 2 (4 2 ) (3 5 ) x x + = − or pair of linear equations Attempt solution of 3-term quadratic eqn or pair of linear equations M1 Obtain 7 1 7 3 , − A1 SC B1 for 1 7 x = − from one linear equation 3 2(ii) Attempt correct process to solve 3e y k = where 0 k > from (i) M1 Obtain 0.282 and no others A1 2
Q3 · X 3 Find the exact coordinates of the stationary point of the curve with equation y [5]…
3x 3 Find the exact coordinates of the stationary point of the curve with equation y [5] ln x. = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Mark scheme: 3 Use quotient rule to find first derivative or equivalent *M1 Obtain 1 2 3ln 3 d d (ln ) x x x y x x − × = or equivalent A1 Condone lack of brackets in denominator unless specifically simplified to 2ln x Equate first derivative to zero and attempt value of x from ln x k = oe DM1 Must get as far as x = Obtain e x = A1 Allow 1e Obtain 3e y = A1 Allow 3 1e SC1: If 1 3ln 3 0 x x x − × = seen with no reference to d d y x , then allow M1 A1 then following marks SC2: If denominator incorrect and numerator correct/reversed/added then max marks M0A0M1A1A1 SC3: If numerator reversed then max marks M1A0M1A1A1 5
Q6 · Y P x O The diagram shows the curve with parametric equations x 3t y = −6e−2t, = 4t2e−t…
6 y P x O The diagram shows the curve with parametric equations x 3t y = −6e−2t, = 4t2e−t, for 0 At the point P on the curve, the y-coordinate is 1. ≤t ≤2. 1 1 (i) Show that the value of t at the point P satisfies the equation t 2t. [2] 2e = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 1 2tn (ii) Use the iterative formula tn+1 = 2e with t1 = 0.7 to find the value of t at P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the gradient of the curve at P, giving the answer correct to 2 significant figures. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 6(i) Equate 2 4 e t t − to 1, rearrange to 2 ... t = and hence ... t = M1 Allow M1 for 1 4 t t e− = Confirm 1 2 1 2 e t t = with necessary detail needed as answer is given A1 2 6(ii) Use iterative process correctly at least once M1 Obtain final answer 0.715 t = A1 Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval [0.7145, 0.7155] A1 SC: M1A1 from iterations to 4sf resulting in 0.71 3 Question Answer Marks Guidance 6(iii) Obtain 2 d d 3 12e t x t − = + B1 Use product rule to find d d y t M1 Obtain 2 8 e 4 e t t t t − − − A1 Divide correctly to obtain d d y x M1 Substitute value from part (ii) to obtain 0.31 A1 Allow greater accuracy 5
Q7 · Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + !
7 (a) (i) Express 4 sin 4 cos in the form R sin , where R 0 and [3] 1 + 1 1 + ! > 0Å < ! < 90Å. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) Hence find the smallest positive value of satisfying the equation 4 sin 4 cos 5. [2] 1 1 + 1 = ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Solve the equation 4 cot 2x 5 tan x = + for 0 x showing all necessary working and giving the answers correct to 2 decimal places. < < 0, [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 7(a)(i) State 32 R = or equivalent or 5.657… Use appropriate trigonometry to find α M1 Obtain 45 α = A1 3 7(a)(ii) Carry out correct process to find one value of θ M1 Obtain 17.1 A1 Ignore other positive values greater than 17.1 2 Question Answer Marks Guidance 7(b) Use or imply 1 cot 2 tan2 x x = B1 Use identity of form 2 2tan tan 2 1 tan x x x ± = ± to obtain equation in tan x M1 Obtain 2 6tan 10tan 4 0 x x + − = or equivalent A1 Attempt solution of 3-term quadratic equation for tan x M1 Obtain 1 3 tan x = and hence 0.32 A1 Allow greater accuracy Obtain tan 2 x = − and hence 2.03 and no others between 0 and π A1 Allow greater accuracy 6
What was in this paper
The subtopics covered by these 4 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 2019 May/June, Paper 2 · Variant 3. A higher threshold means an easier paper — the bar moves with how the cohort did.