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Cambridge AS & A Level Mathematics (9709), Paper 2, October/November 2019. Read both free below, or download them. Done the paper? Get it marked by AI against this mark scheme.
9709_w19_qp_239709_w19_ms_239709/23/O/N/19
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
(i) Solve the inequality 2x − 7 < 2x − 9. [3] (ii) Hence find the largest integer n satisfying the inequality 2 ln n − 7 < 2 ln n − 9. [2] 2
Mark scheme
1(i) State or imply non-modular inequality (2 x − 7) 2 < (2 x − 9) 2 or corresponding M1 equation or linear equation (with signs of 2x different) Obtain critical value 4 A1 State x < 4 only A1 3 1(ii) Attempt to find n from ln n = their critical value from part (i) M1 Obtain or imply n < e4 and hence 54 A1 2
Question
Find the exact value of Ó 2e2x − 12 dx. Show all necessary working. [5] 1 3 + 2 ln x
Mark scheme
2 Expand integrand to obtain 4e4 x − 4e2 x + 1 B1 Integrate to obtain at least two terms of form k1e 4 x + k2 e 2 x + k3 x *M1 Obtain correct e4 x − 2e2 x + x A1 Apply both limits correctly to their integral DM1 Obtain e8 − 3e4 + 2e2 + 1 A1 5
Question
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 1 A1 Obtain − 2 or (unsimplified) equivalent x(1 + ln x) Use y = 4 to obtain ln x = − 12 or exact equivalent for x B1 Substitute for x in their first derivative DM1 1 A1 Must be simplified to contain a single exponential Obtain −4e 2 or exact equivalent term 5
Question
The polynomial p x is defined by p x = ax3 + ax2 − 15x − 18, where a is a constant. It is given that x − 2 is a factor of p x. (i) Find the value of a. [2] (ii) Using this value of a, factorise p x completely. [3] (iii) Hence solve the equation p e y = 0, giving the answer correct to 2 significant figures. [2] a
Mark scheme
4(i) Substitute x = 2 , equate to zero and attempt solution M1 Obtain a = 4 A1 2 4(ii) Divide by x − 2 at least as far as the x term M1 By inspection or use of identity Obtain 4 x 2 + 12 x + 9 A1 Conclude ( x − 2)(2 x + 3) 2 A1 Each factor must be simplified to integer form 3 4(iii) Attempt correct process to solve e y = k where k > 0 M1 For y = ( ln k )2 Obtain 0.48 and no others A1 AWRT 2
Question
It is given that Ó 3x2 + 4 cos 2x − sin x dx = 2, where a is a constant. 0 (i) Show that a = 3 3 − 2 sin 2a − cos a. [4] (ii) Using the equation in part (i), show by calculation that 0.5 < a < 0.75. [2] (iii) Use an iterative formula, based on the equation in part (i), to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3]
Mark scheme
5(i) Integrate to obtain form x 3 + k1 sin 2 x + k2 cos x *M1 Obtain correct x3 + 2sin 2 x + cos x A1 Apply limits correctly and equate to 2 DM1 Confirm given result A1 AG; necessary detail needed 4 5(ii) Consider sign of a − 3 3 − 2sin 2a − cos a or equivalent for 0.5 and 0.75 M1 Obtain –0.26 and 0.10 or equivalents and justify conclusion A1 AG; necessary detail needed 2 5(iii) Use iterative process correctly at least once M1 Need to see a correct x3 , may be implied by x1 = 0.5 so x3 = 0.65256 or x1 = 0.75 so x3 = 0.64897 OE Must be working with radians Obtain final answer 0.651 A1 Show sufficient iterations to 5sf to justify answer or show a sign change in the A1 interval [0.6505, 0.6515] 3
Question
(a) Showing all necessary working, solve the equation sec ! cosec ! = 7 for 0Å < ! < 90Å. [5] (b) Showing all necessary working, solve the equation sin " + 20Å + sin " − 20Å = 6 cos " for 0Å < " < 90Å. [4]
Mark scheme
6(a) 1 B1 OE; May be implied by subsequent work Express equation as =7 cos α sin α Attempt use of identity for sin 2α or attempt to obtain a quadratic equation in M1 From equation of form sin 2α = k where 0 < k < 1 or terms of any one of the following: from use of correct identities sin 2 α , cos 2 α , cot 2 α or tan 2 α Obtain sin 2α = 72 or a correct 3 term quadratic equation, equated to zero in any A1 one of the following: sin 2 α , cos 2 α , cot 2 α or tan 2 α Attempt correct process to find at least one correct value of α M1 Obtain 8.3 and 81.7 and no others between 0 and 90 A1 5 6(b) Simplify left-hand side to obtain 2sin β cos 20° B1 Attempt to form equation where tan β is only variable, tan β ≠ 3 M1 3 A1 OE Obtain tan β = cos 20° Obtain β = 72.6 and no others between 0 and 90 A1 5
Question
The equation of a curve is x2 − 4xy − 2y2 = 1. dy (i) Find an expression for and show that the gradient of the curve at the point −1, 2 is − 52 . [5] dx (ii) Show that the curve has no stationary points. [3] (iii) Find the x-coordinate of each of the points on the curve at which the tangent is parallel to the y-axis. [2] 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(i) dy Obtain −4 y − 4 x dx from use of the product rule B1 dy Differentiate −2 y 2 to obtain −4 y dx B1 Obtain 2 x, = 0 with no extra terms B1 dy Rearrange to obtain expression for dx and substitute x = −1 , y = 2 M1 dy 2 x − 4 y A1 Obtain = OE and hence − 52 dx 4 x + 4 y 5 7(ii) Equate numerator of derivative to zero to produce equation in x and y M1 Substitute into equation of curve to produce equation in x or y M1 Obtain −6 y 2 = 1 or − 32 x 2 = 1 OE and conclude A1 3 7(iii) Use denominator of derivative equated to zero with equation of curve to produce M1 equation in x 1 A1 OE Obtain 3x2 = 1 and hence x = ± 3 2
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Mathematics 9709/23 for October/November 2019 is out of 50 marks, with 1 hour 15 minutes to answer it.
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