Cambridge A Level Mathematics 9709 — 2023 Oct/Nov Paper 3 · Variant 2

1 (a) Sketch the graph of y 4x . [1] = −2 (b) Solve the inequality 1 3x 4x . [4] + < −2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) B1 Show a recognizable sketch graph of y = 4 x − 2 . y Roughly symmetrical. Should extend into the second quadrant. Ignore y = 4 x − 2 below the axis if intention is clear e.g. dashed or the required lines are clearly bolder. Some indication of scale on both axes – accept dashes. 2 Must go beyond (0, 2) and (1, 2). Ignore any attempt to sketch y = 1 + 3 x . x 1 2 1 1(b) Obtain critical value x = 3 B1 Allow incorrect inequality. Allow if later rejected. Allow 217 . Solve the linear equation 1 + 3x = 2 − 4 x M1 Or corresponding linear inequality. Obtain critical value 17 A1 Allow 0.143 or better. Allow incorrect inequality. Allow if later rejected. Obtain final answer x  17 [or] x  3 A1 Or equivalent. Allow with a comma, or nothing between. Strict inequalities only. Exact values. A0 for 17  x  3 A0 for x  17 and x .3 Alternative method for question 1(b) Solve the quadratic inequality ( 4 x − 2 ) 2  (1 + 3 x ) 2 , or corresponding M1 e.g. 7 x 2 − 22 x + 3 = 0 . Available if they start with the correct equation / quadratic equation inequality, have a correct method for squaring 2 2 2 (i.e. not ( a + b ) = a + b ) and a correct method for solving. Need to obtain at least one critical value. Obtain critical value x = 3 A1 Allow incorrect inequality. Allow if later rejected. Allow 217 . Obtain critical value 17 A1 Allow 0.143 or better. Allow incorrect inequality. Allow if later rejected. Obtain final answer x  17 [or] x  3 A1 Or equivalent. Strict inequalities only. Allow with a comma, or nothing between. Exact values. A0 for 17  x  3 A0 for x  17 and x .3 4

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4 (a) On a sketch of an Argand diagram, shade the region whose points represent complex numbers z satisfying the inequalities z and Re z [4] −4 −3i ≤2 ≤3. (b) Find the greatest value of arg z for points in this region. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Show a circle with centre 4 + 3i . Accept a curved shape with correct point B1 Im roughly in the middle. 5i Show a circle with radius 2 and centre not at the origin. B1 The shape should be consistent with their scales 3i Show correct vertical line. Enough to meet correct circle twice or complete B1 line for any other circle. Shade the correct region on a correct diagram B1 i Any other shading must be accompanied by words to explain which region is required 3 4 Re Need some indication of scale e.g. label the centre, mark key points on the axes or dashes on the axes. Condone dotted lines in place of solid lines Condone correct shaded shape but not an entire circle 4 4(b) Carry out a complete method for finding the greatest value of arg( )z M1 Im 0.6435 + 0.4115 ) e.g tan −1 34 + sin −1 52 ( 5i Obtain answer 1.06 (accept 1.055 or 1.056) radians A1 2 or 60.45 (accept 60.4or 60.5) 3i 21 5 i 3 4 Re Alternative method for question 4(b) Tangent to circle passing through origin has equation y = mx . M1 Complete method for finding the greatest value of arg(z). 2 2 The equation ( x − 4 ) + ( y − 3) = 4 will have one root. Hence 1 + m 2 x 2 − ( 8 + 6 m ) x + 21 = 0 , discriminant = 0 = 48m 2 − 96 m + 20 ( ) 6  21 and m = with the larger value needed to give greatest arg(z). 6 Required angle is tan −1 m . Obtain answer 1.06 radians or 60.45 A1 Accept 1.055 or 1.056 radians. Accept 60.4or 60.5. 2

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6 (a) By sketching a suitable pair of graphs, show that the equation cotx 2 = −cosx has one root in the interval 0 x [2] < ≤12π. (b) Show by calculation that this root lies between 0.6 and 0.8. 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P Q 1 (c) Use the iterative formula to determine the root correct to 2 decimal tan−1 2 xn xn+1 = −cos places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 6(a) Sketch a relevant graph. B1 y e.g. y = cot x: x intercept should be correct. Not touching the y-axis. No incorrect curvature. y = cotx Ignore anything outside 0 < x ⩽12 π . Sketch a second relevant graph and justify the given statement B1 e.g. y = 2 − cos x : Condone if looks almost straight, but not if drawn with a 2 ruler and not incorrect curvature. Correct y intercept. y = 2 - cosx Needs to be drawn for 0 < x ⩽12 π . Ignore outside this. 2 1 1 x 2π 2nd B1 requires a mark at the point of intersection or a suitable comment for the justification. 6(b) Calculate the value of a relevant expression or values of a pair of expressions M1 e.g. 1.17  1.46, 1.30  0.971 , −0.29  0, 0.33  0 . at x = 0.6 and x = 0.8 Must be working in radians. Values correct to at least 2 −0.20  0, 0.342  0 from tan x ( 2 − cos x ) −=1 0 . significant figures. 0.80  1, 1.34 from1 tan x ( 2 − cos x ) = 1 . Need all relevant values but only one (pair) needs to be correct to award M1. 1 . Complete set of values for their expression. If not comparing with 0 or 1 then 0.146  0, − 0.105  0 from x − tan −1 ( 2 − cos x ) the pairing must be clear, not just embedded values. Complete the argument correctly with correct calculated values (awrt 2 s.f.). A1 Accept truncated values. If comparing with 0 can either Clear comparison for their expression. Allow work on a smaller interval. indicate different signs or a negative product. 2 6(c) Use the iterative process correctly at least once. Must be working in radians M1 Obtain final answer 0.68 A1 Must be a clear conclusion. Show sufficient iterations to at least 4 d.p. to justify 0.68 to 2 d.p. or show A1 e.g. 0.7, 0.6806, 0.6855, 0.6843, 0.6846 there is a sign change in the interval (0.675, 0.685). 0.6, 0.7053, 0.6792, 0.6858, 0.6842, 0.6846 Allow recovery. 0.8, 0.6545, 0.6920, 0.6826, 0.6850, 0.6844, 0.6845. Allow truncation. Allow small differences in the 4th s.f. 3 Question Answer Marks Guidance

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7 (a) By expressing as prove the identity cos 4 cos [3] 31 21 + 1, 31  cos31 −3 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation cos cos cos 31 + 1 21 = cos21 for [5] 0Å ≤1 ≤180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Use correct expansion for cos ( 2+ ) *M1 Use correct double angle formulae to express cos3in terms of cosand DM1 sin Show sufficient working to confirm cos3 4cos 3 − 3cos A1 AG 3 7(b) Use the identity and correct double angle formula to obtain an equation in *M1 e.g. 4 cos3 – 3 cos + cos  (2 cos2 – 1) = cos2  cos only. Must come from using all three terms in the given equation. 6 cos3 – cos2 – 4 cos  = 0 or 6 cos2 – cos – 4 = 0. Obtain = 90 B1 Allow if cosobtained correctly as a factor of their expression (even if there is an error in the quadratic factor). Can follow M0. Solve a 3-term quadratic in cos to obtain a value of  DM1 Obtain one value e.g. 25.3 A1 Accept awrt 25.3. Obtain a second value e.g. 137.5and no extras in range A1 Accept awrt 137.5. Ignore values outside the range. Mark solutions in radians as a misread ( 0.442,1.57, 2.40 ) . 5

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2 3ai 8 It is given that 2 , where a and are real constants. + a 2i = , −i , + (a) Show that 3a2 4a 0. 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(b) Hence find the possible values of a and the corresponding values of [3] ,. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Multiply both sides by a + 2i and attempt expansion of right-hand side *M1 Use of 2i = −1 seen at least once (or implied) DM1 e.g. 2 + 3ai = ( 2 a + 2 ) + i ( −+a 4 ) Compare real and imaginary parts to obtain an equation in a only M1 3a −+a 4 e.g. = . Any equivalent form. = 2 2 a + 2  2 (2a + 2), 3a = ( −+ a 4 )  Obtain 3a 2 + 4 a − 4 = 0 from correct working A1 AG Alternative method for question 8(a) Multiply top and bottom of the left-hand side by a − 2i and attempt both *M1 Do not need the right-hand side at this stage. expansions Use of 2i = −1 seen at least once or implied DM1 8a + i 3a 2 − 4 ( ) . e.g.   ( 2 − i ) = 2 a + 4 Compare real and imaginary parts to obtain an equation in a only M1 2 e.g. 8a = −2 3a − 4 . Any equivalent form. ( ) Obtain 3a 2 + 4 a − 4 = 0 from correct working A1 AG 4 8(b) Solve given quadratic to obtain a value of a and use this to form an equation M1 Can be implied by relevant working seen or a correct in  only (based on an equation seen in their working in (a) or (b)) value for  seen. Obtain a = −2, = −1 or a = 23 , = 53 A1 Allow 106 and 0.6. Obtain second correct pair of values A1 3

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9 y M x O a π R The diagram shows the curve y sin x cos 2x, for 0 and a maximum point M, where x a. The shaded region between the curve= and the x-axis is≤xdenoted≤π, by R. = (a) Find the value of a correct to 2 decimal places. 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(b) Find the exact area of the region R, giving your answer in simplified form. 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Mark scheme: 9(a) Use correct product rule *M1 As far as p cos x cos2 x + q sin x sin2 x or full working (u, v, du/dx, dv/dx) shown. dy A1 OE Obtain = cos x cos2 x − 2sin x sin2 x dx Equate derivative to zero and use correct double angle formulae DM1 Allow if only have one double angle in their derivative. 2 A1 2 2 Obtain cos x 1 − 6sin x = 0 or equivalent e.g. cos x 6cos x − 5 = 0 , 5tan x = 1 . ( ) ( ) Simplified but not necessarily factorised - like terms must be collected. Obtain a = 0.42 A1 Only. Accept x = 0.42 . Alternative method for question 9(a) Use correct double angle formula *M1 Obtain sin x − 2sin 3 x or equivalent A1 Use correct chain rule or product rule to differentiate and equate the DM1 derivative to zero 2 A1 OE Obtain cos x 1 − 6sin x = 0 ( ) Obtain a = 0.42 A1 Only. Accept x = 0.42 . 5 2cos 2 x sin x − sin x dx .9(b) Use double angle formula and obtain p cos 3 x + q cos x correctly *M1 e.g. from   2 3  A1 Obtain − cos x + cos x  Correct for their integral.  3  Correct use of limits 14 π and 34 π (or use double the integral from 14 π to 12 π ) DM1 OE   3 3   2  −1   1  1 1 −    −    − −  .   3   2   2   2 2  2 2 A1 Or simplified exact equivalent. Obtain 3 Final answer must be positive. Alternative method 1 for question 9(b) Use integration by parts twice and obtain r cos x cos2 x + s sin x sin2 x *M1 Seen, not just implied. 1 2 A1 Obtain cos x cos2 x + sin x sin2 x Accept  (correct for their integral). 3 3 Correct use of limits 14 π and 34 π (or use double the integral from 14 π to 12 π ) DM1 OE 1  1 1   0 + 2  −−1 0 − 2   1 .   3  2 2  2 2 A1 Or simplified exact equivalent. Obtain 3 Final answer must be positive.

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11 The variables x and y satisfy the differential equation dy x2 y2 y 0. dx + + = It is given that x 1 when y 1. = = (a) Solve the differential equation to obtain an expression for y in terms of x. 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(b) State what happens to the value of y when x tends to infinity. Give your answer in an exact form. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 11(a) Correct separation of variables. B1  1 1 − dx .  2 dy =  2  y + y  x Condone missing integral signs or missing dx, dy, but not both. 1 B1 Obtain x 1 *M1  1 Express in partial fractions Allow for the correct split of . y 2 + y y 2  y ( ) or express the denominator of the fraction as a difference of two squares 1 1 1 A1 Allow if coefficients for the partial fractions are correct Obtain − or 1 1 y y + 1 ( y + 2 ) 2 − ( 2 ) 2 but followed by an error. Obtain ln y − ln ( y + 1) A1 Or equivalent, dependent on where they left the minus sign. Use x = 1, y = 1 to find constant of integration or as limits in a definite DM1 ln 12 = 1 + C integral in an expression containing terms of the form p , q ln y and r ln (1 + y ) If they rearrange the equation before finding the constant x of integration then the constant must be of the correct form. Correct equation in x and y A1 y 1 1 ln = −+1 ln . 1 + y x 2 1 x −1 A1 1 1 e . Or equivalent e.g. y = , y = Obtain y = 1 x x −1 2e1− 1 − 1 e1−+1 x ln2 − 1 2 − e Accept with decimal value for e-1. 8 11(b) 1 B1 FT 1 State that y approaches Or exact equivalent. Condone y = 2e − 1 2e −.1 FT on an expression in 1ex . 1

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