Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 3 · Variant 3

1 2x2 Expand + in ascending powers of x, up to and including the term in x2, simplifying the 1 −2x coefficients. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 1 B1 2 State a correct unsimplified term in x or x 2 of the expansion of either (1 + 2 x ) 1 − 2 or (1 −x2 ) 1 B1 2 up to the term in x 2 State correct unsimplified expansion of (1 + 2 x ) 1 B1 − 2 up to the term in x 2 State correct unsimplified expansion of (1 −x2 ) Obtain sufficient terms of the product of the expansions M1 Obtain final answer 1 + 2 x + 2 x 2 A1 Alternative method for question 2 1 B1 − 2 and state a term of the 1 − 4 x 2 State that the expression equals (1 + 2 x )( ) expansion 1 B1 + B1 − State correct unsimplified expansion of 1 −x4 2 2 up to the term in x 2 ( ) Obtain sufficient terms of the product of (1 + 2x) and the expansion M1 Obtain final answer 1 + 2 x + 2 x 2 A1 5

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14π 3 Find the exact value of x sec2x dx. [5] Ó 0 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Commence integration by parts and reach x tan x  tan x.1 dx *M1 Use a correct method to integrate tan x M1 Obtain integral x tan x − lnsec x , or equivalent A1 Use limits correctly, having integrated twice DM1 1 1 A1 Obtain answer π − ln 2 , or exact equivalent 4 2 5

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5 (a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z 2 and Im z [4] + ≤2 ≥1. (b) Find the greatest value of arg z for points in the shaded region. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) Show a circle with centre – 2 B1 Show a circle with radius 2 and centre not the origin B1 Show the line y = 1 B1 Shade the correct region B1 4 5(b) Identify the correct point and carry out a correct method to find the argument M1 11 A1 2.88 radians or 165°. Obtain answer π 12 2

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7 (a) Show that the equation 5 sec x tan x 4 can be expressed as R cos x 5, where R 0 + = + ! = > and Give the exact value of R and the value of correct to 2 decimal places. [4] 0Å < ! < 90Å. ! ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 7(a) B1 Rearrange and obtain 4cos x − sin x = 5 State R = 17 B1 Use trig formulae to find α M1 Obtain α =14.04 A1 4 7(b)  5  B1 FT FT their R. Evaluate cos− 1      17  Carry out a correct method to find a value of x in the given interval M1 Obtain answer, e.g. 21.6 A1 Obtain a second answer, e.g. 144.4 and no other in the interval A1 Treat answers in radians as a misread. Ignore answers outside the given interval. 4

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x3 8 The curve with equation y has a stationary point at x p, where p 0. = ex = > −1 (a) Show that p 3 1 . [3] = −e−p ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Verify by calculation that p lies between 2.5 and 3. 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(c) Use an iterative formula based on the equation in part (a) to determine p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative at x = p to zero and obtain the given equation A1 3 8(b) Evaluate a relevant expression or pair of relevant pair of expressions at p = 2.5 M1 and p = 3 Complete the argument with correct calculated values A1 2 8(c) pn M1 correctly at least once 1 − e− Use the iterative formula np +1 = 3 ( ) Obtain final answer p = 2.82 A1 Show sufficient iterations to 4 d.p.to justify 2.82 to 2 d.p., or show there is a sign A1 change in the interval (2.815, 2.825) 3

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9 With respect to the origin O, the position vectors of the points A, B and C are given by ` a ` a ` a −−¿OA 05 , −−¿OB 10 and −−¿OC 4 . = = = −3 2 1 −2 The midpoint of AC is M and the point N lies on BC, between B and C, and is such that BN 2NC. = (a) Find the position vectors of M and N. 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(b) Find a vector equation for the line through M and N. 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(c) Find the position vector of the point Q where the line through M and N intersects the line through A and B. 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Mark scheme: 9(a) 2 B1  State OM = 1   0 Use a correct method to find ON M1  3  A1   Obtain answer − 2     − 1   3 9(b) Carry out a correct method to form a vector equation for MN M1 2  1  A1 OE    Obtain a correct equation in any form, e.g. r = 1 + λ − 3       0 − 1    2 9(c) 1  − 1  B1    State a correct vector equation for AB in any form, e.g. r = 0 + µ 5       1  1  Equate components of AB and MN and solve for λ or for µ M1 Obtain λ = – 3 or µ = 2 A1  − 1  A1   Obtain position vector 10 , or equivalent, for Q      3  4

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10 A gardener is filling an ornamental pool with water, using a hose that delivers 30 litres of water per minute. Initially the pool is empty. At time t minutes after filling begins the volume of water in the pool is V litres. The pool has a small leak and loses water at a rate of 0.01V litres per minute. dV The differential equation satisfied by V and t is of the form a dt = −bV. (a) Write down the values of the constants a and b. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the differential equation and find the value of t when V 1000. 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(c) Obtain an expression for V in terms of t and hence state what happens to V as t becomes large. 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Mark scheme: 10(a) a = 30 and b = 0.01 B1 1 10(b) Separate variables and integrate one side M1 Obtain terms −100ln ( 30 − 0.01V ) and t, or equivalent A1 FT FT their a and b. + A1 FT Evaluate a constant, or use t = 0, V = 0 as limits, in a solution containing terms c M1 ln ( 30 − 0.01V ) and dt where cd ≠ 0 Obtain solution 100ln30 − 100ln ( 30 − 0.01V ) = t , or equivalent A1 Substitute V = 1000 and obtain answer t = 40.5 A1 6 10(c) Obtain V = 3000 1 − e−0.01t B1 OE ( ) State that V approaches 3000 B1 2

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