Cambridge A Level Mathematics 9709 — 2022 Oct/Nov Paper 3 · Variant 2

3 The equation of a curve is y sin x sin 2x. The curve has a stationary point in the interval 0 x 1 = < < 2π. Find the x-coordinate of this point, giving your answer correct to 3 significant figures. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Use correct product rule on given expression *M1 Obtain correct derivative in any form A1 e.g.cos x sin2 x + 2sin x cos2 x Use correct double angle formulae to express derivative in terms of sin x and *M1 cos x Equate derivative to zero and obtain an equation in one trig variable DM1 dependent on the 2 previous M Marks. Obtain 3sin 2 x = 2 , 3cos 2 x = 1 or tan 2 x = 2 A1 OE Solve and obtain x = 0.955 A1 3 sf only. Final answer in degrees is A0. Ignore any attempt to find the corresponding value of y. Alternative method for the first three marks Use correct double angle formula to obtain y = 2cos x − 2cos 3 x *M1 or y = 2sin 2 x cos x Use chain rule and / or product rule *M1 Obtain derivative y = −2sin x + 6sin x cos 2 x A1 y = −2sin 3 x + 4sin x cos 2 x Alternative method for the second and third M marks Equate derivative to zero and obtain an equation in tan x and tan2x *M1 Use correct double angle formula to obtain an equation in tan x DM1 6

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4 (a) Express 4 cos x x in the form R cos x , where R 0 and State the exact −sin + ! > 0Å < ! < 90Å. value of R and give correct to 2 decimal places. [3] ! ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation 4 cos 2x 2x 3 for x [5] −sin = 0Å < < 180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) State R = 17 B1 Allow if working from an incorrect expansion but not from decimals. Use correct trig formulae to find α (Correct expansion and correct expression for M1 NB: cos= 4 and sin= 1 scores M0A0. trig ratio for ) M0 for incorrect expansion of cos ( x − ) M1 for correct expression for trig ratio for and no errors seen. Obtain = 14.04 A1 2 d.p. required Allow M1A1 for correct answer with no working shown. Correct answer from incorrect working (e.g. 1 ) is awarded M0A0. tan−−1 ( 4 ) 1 is awarded M1 180− tan−1 ( − 4 ) 3 4(b) 1  3  B1 FT FT their R. Accept awrt 43.3 or awrt 316.7 Evaluate cos−   to at least 1 d.p. (43.3138) Can be implied by subsequent working.  17  Use correct method to find a value of x in the interval M1 Must be working with 2x and their . Obtain answer, e.g. 14.6 A1 Accept overspecified answers but they need to be correct. (14.6388…. and 151.3249…). Use a correct method to find a second answer in the interval M1 Must be working with 2x, their and 360−their 43.3 . Obtain second answer in the interval, e.g. 151.3, and no other in the interval A1 Ignore answers outside the given interval. Treat answers in radians (0.255… and 2.64…) as a misread. 5

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5 (a) Solve the equation z2 0, giving the answers in the form x iy, where x and y are real −6iz −12 = + and exact. 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(b) On a sketch of an Argand diagram with origin O, show points A and B representing the roots of the equation in part (a). [1] (c) Find the exact modulus and argument of each root. 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(d) Hence show that the triangle OAB is equilateral. 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Mark scheme: 5(a) 2 M1 Use quadratic formula, or completing the square ( z − 3i ) − 3 = 0 ( ) and use 2i =−1 to find a root Obtain a root, e.g. 3 + 3i A1 Or exact 2 term equivalent e.g. 6i + 12 ISW. 2 2 Obtain the other root, e.g. − 3 + 3i A1 Or exact 2 term equivalent ISW. 3 5(b) Show points representing the roots correctly B1 FT 2 roots consistent with their (a) and with no errors seen on the diagram. B0 if they only have one root or more than 2 roots Must match their scale and 1  3  2 Linear scales seen or implied. Need some indication of scale (numbers or dashes). Scales along an axis must be approximately consistent but scales may be different on the 2 axes. 1 5(c) State modulus of either root is 2 3 , or simplified exact equivalent B1 FT ISW if converted to decimal . Ignore modulus of second root if seen. Follow their root(s) not on either axis (from (a) or (b)). Find the argument of one of their roots – get as far as tan −1 ( ...) M1 SOI but must be correct for their root. 1 2 A1 Must obtain values. Allow degrees. Obtain correct arguments π and π , or simplified exact equivalents 3 3 3 5(d) Give a complete justification that the correct triangle is equilateral B1 Check their diagram in (b). Possible justifications: 3 equal sides, or all angles equal  to 3 , or isosceles and an angle of .3 1 6(a) State or imply AB or AC correctly in component form B1 AB = 2i − 2 j + k , AC = 4i − 3k . ( ) Using the correct process with relevant vectors to evaluate the scalar product M1 or BACA. ( 8 − 3 = 5 ) . AB. AC , M0 for AB.CA . Using the correct process for the moduli, divide their scalar product by the M1  5  product of their moduli to obtain cos or     9 25  Independent of the first M1. 1 A1 ISW. Need to see a value for cos. Obtain answer 5 1 1 3 Accept 15 or 0.333 ( cos− 3 alone is not sufficient) 4 6(b) Use correct method to find an exact value for the sine of angle BAC from their M1 1 ( 1− 9 ) (a) 2 A1 Obtain answer 2 , or equivalent 3 Use correct area formula to find the area of triangle ABC with their versions of M1 1 1 1 1 9 25 their sin or 2 9 25  sin ( cos− 3 ) ( 2 ) relevant vectors Obtain answer 5 2 or 50 A1 Only ISW Alternative method 1 for question 6(b) Use correct method to find the perpendicular distance from A to BC (or B to AC M1  2 + 2  2  or C to AB)    1 −+2 2 . 2 = 0  =       6  1 − 4  −4  7 Obtain 1 i − 5 j + 1 3 75 A1 ( k ) 3 3 3 Use correct area formula to find the area of triangle ABC M1 1 1 ( 2  their 24  their 3 75 ) The length they use for their base must be found correctly. Obtain answer 5 2 or 50 A1

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7 The variables x and satisfy the differential equation 1 dx x cot sin21 = tan21 −2 1, d1 for 0 1 and x 0. It is given that x 2 when 1 < 1 < 2π > = 1 = 4π. d cot (a) Show that 1 . cot21 = −2 d1 sin21 (You may assume without proof that the derivative of cot with respect to is [1] 1 1 −cosec21.) ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Solve the differential equation and find the value of x when 1 [7] 1 = 6π. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 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Mark scheme: 7(a) Show sufficient working to justify the given statement B1 e.g. see 2cot − cosec 2 in the working or express in terms of sinand cos and use quotient rule to obtain the given result. Solution must have θ present throughout and must reach the given answer. 1 7(b) Separate variables correctly B1 tan 2  2cot  xdx = 2 − 2 d  sin  sin  Check for relevant working in (a) Condone incorrect notation e.g. missing dx. Need either the integral sign or the dx, dθ. 1 2 B1 Obtain term x 2 Obtain terms tan+ cot 2 B1 + B1  2cot   2cos 1 d = − C ) Alternative:  2 d =  3 2 ( +  sin   sin  sin  1 M1 Need to have 3 terms. Constant of correct form. Form an equation for the constant of integration, or use limits x = 2, = π , in 4 a solution with at least two correctly obtained terms of the form ax 2 , btan and ccot 2, where abc ≠ 0 A1 1 or 12 x 2 = tan+ cosec 2− 1 x 2 = tan+ cot 2 State correct solution in any form, e.g. 2 If everything else is correct, allow a correct final answer to imply this A1. 1 A1 18 + 2 3 Substitute = π and obtain answer x = 2.67 2.6748… 3 If see a correctly rounded value ISW. 6 7

More questions on Integration

8 y x O a The diagram shows part of the curve y sin x. This part of the curve intersects the x-axis at the point = where x a. = (a) State the exact value of a. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Using the substitution u x, find the exact area of the shaded region in the first quadrant = bounded by this part of the curve and the x-axis. 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Mark scheme: 2 .8(a) State (a =) π 2 B1 Allow 32400, 180 2 . Accept ( x = ) 1 8(b) State or imply dx =2u du or equivalent B1 e.g. ddux = 2 1 x Incorrect statements e.g. du = 1 is B0. 2 x Substitute for x and dx throughout the integral M1 Obtain u2 sin u du A1 Allow with missing du. Commence integration of ku sin u du by parts and reach *M1 ku cos u  k cos u du Obtain integral −ku cos u + k sin u A1 Substitute limits u = 0 and u = their a , a ≠0 , a in radians DM1 −2πcosπ + 2sin π ( +0 − 2sin0 ) or x = 0 and their a Need limits stated but condone if zeros not shown in in the complete integral substitution. Obtain answer 2π A1 7

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9 C r 1 rad A O B The diagram shows a semicircle with diameter AB, centre O and radius r. The shaded region is the minor segment on the chord AC and its area is one third of the area of the semicircle. The angle CAB is radians. 1 (a) Show that 1 sin . 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(b) Verify by calculation that 0.5 0.7. 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(c) Use an iterative formula based on the equation in part (a) to determine correct to 3 decimal 1 places. Give the result of each iteration to 5 decimal places. 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Mark scheme: 9(a) State or imply angle AOC = π − 2 B1 Might be seen on the printed diagram. Use correct formulae for the area of a sector and triangle, or of a segment, and M1 1 2 1 2 r sin ( π − 2) or 2 r ( π − 2) − 2 find the area of the shaded region 1 1 r 2 ( 2) + 2 πr 2 − 12 2 r 2 sin ( π − 2)  M0 if subtraction the wrong way round. 1 A1 1 e.g. 1 π − 2) − r 2 sin ( π − 2) . 6 πr 2 = 12 r 2 ( 2 Equate to π r 2 and obtain a correct equation in any form 6 Obtain  = 13 ( − 1.5sin 2) correctly A1 AG Condone if state / imply sin ( π − 2) = sin 2 . 4 9(b) Evaluate a relevant expression or pair of expressions at = 0.5 and = 0.7 M1 Allow work on a smaller interval. Need to evaluate for both limits, with at least one correct. When using x = f ( x ) embedded values are not sufficient e.g. f(0.5) …. is accepted but 1 3 (− 1.5sin 2  0.5 ) = ... is not. Complete the argument correctly with correct calculated values A1 e.g. 0.5  0.626, 0.7  0.554 or 0.126  0, − 0.146  0 If using pairs then the pairing must be clear. Need to see the inequalities or an appropriate comment. Need to see values calculated to at least 2 sf. 2 9(c) M1 i.e obtain one value and use that value to obtain a second Use the iterative process n +1 = 1(π − 1.5sin2n ) correctly at least once value. Must be working in radians. 3 Obtain final answer 0.586 A1 Show sufficient iterations to 5 d.p. to justify 0.586 to 3 d.p., or show there is a A1 0.5,0.62646,0.57225,0.59195,0.58416,0.58715, sign change in the interval (0.5855, 0.5865). e.g. 0.58599,0.58644 0.6,0.58118,0.58833,0.58553,0.58661,0.58619,0.58636 0.7,0.55447,0.59958,0.58133,0.58827,0.58556, 0.58661,0.58620,0.58636 Allow working to more than 5 dp, but not less. 3

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