Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 2 · Variant 2
9709/22/F/M/17 · 2 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 scheme7 pages
Answers below. Sit the paper first if you are practising.
Questions as text
Q2 · Given that tan cot 8, show that 34
2 (i) Given that tan cot 8, show that 34. [3] 21 1 = tan21 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence solve the equation tan cot 8 for [2] 21 1 = 0Å < 1 < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 2(i) 1 B1 Use identity cotθ= tanθ Attempt use of identity for tan 2θ M1 Confirm given tan 2 θ = 34 A1 Total: 3 2(ii) Obtain 40.9 B1 Obtain 139.1 B1 Total: 2
Q3 · Solve the inequality 2x x 3
3 (i) Solve the inequality 2x x 3 . [4] −5 < + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence find the largest integer y satisfying the inequality 2 ln y ln y 3 . [2] −5 < + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 3(i) State or imply non-modulus inequality (2 x − 5) 2 < ( x + 3) 2 or B1 corresponding equation or pair of linear equations Attempt solution of 3-term quadratic inequality or equation M1 or of 2 linear equations Obtain critical values 23 and 8 A1 State answer 23 < x < 8 A1 Total: 4 3(ii) Attempt to find y from ln y = upper limit of answer to part (i) M1 Obtain 2980 A1 Total: 2
What was in this paper
The subtopics covered by these 2 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 2017 Feb/March, Paper 2 · Variant 2. A higher threshold means an easier paper — the bar moves with how the cohort did.