Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 2 · Variant 1
9709/21/O/N/19 · 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 scheme10 pages
Answers below. Sit the paper first if you are practising.










Questions as text
Q1 · Solve the inequality 2x 2x
1 (i) Solve the inequality 2x 2x . [3] −7 < −9 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence find the largest integer n satisfying the inequality 2 ln n 2 ln n . [2] −7 < −9 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Mark scheme: 1(i) State or imply non-modular inequality 2 2 (2 7) (2 9) x x − < − or corresponding equation or linear equation (with signs of 2x different) M1 Obtain critical value 4 A1 State 4 x < only A1 3 1(ii) Attempt to find n from lnn = their critical value from part (i) M1 Obtain or imply 4 e n < and hence 54 A1 2
Q3 · 2 ln x3 A curve has equation y +
3 2 ln x3 A curve has equation y + . Find the exact gradient of the curve at the point for which y 4. = 1 ln x = + [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Mark scheme: 3 Use quotient rule (or product rule) to find first derivative *M1 Must have correct u and v Obtain 2 1 (1 ln ) x x − + or (unsimplified) equivalent A1 Use 4 y = to obtain 1 2 ln x = − or exact equivalent for x B1 Substitute for x in their first derivative DM1 Obtain 1 2 4e − or exact equivalent A1 Must be simplified to contain a single exponential term 5
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 2019 Oct/Nov, Paper 2 · Variant 1. A higher threshold means an easier paper — the bar moves with how the cohort did.