Mathematics 9709/32 October/November 2023: question paper and mark scheme

Mark scheme

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. 2 Some indication of scale on both axes – accept dashes. Must go beyond (0, 2) and (1, 2). Ignore any attempt to sketch y = 1 + 3x . 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 − 4x 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) > (1 + 3x ) , or corresponding M1 e.g. 7 x 2 − 22 x + 3 = 0 . 2 2 quadratic equation Available if they start with the correct equation / inequality, have a correct method for squaring (i.e. not ( a + b ) = a 2 + b2 ) and a correct method for 2 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

Mark scheme

2 dx 2 B1 Any equivalent form. Obtain = ln t dt t dy B1 Any equivalent form. = −2te2−t 2 Obtain dt M1 dy −2e 2 e 2 − e 2 dy dy dx = ÷ and substitute t = e Correct use of chain rule for . dx dt dt dx 2ln e Condone an error between correct combination of the derivatives and attempt to substitute e. Obtain −e 4−e 2 A1 ISW Accept −0.0337(405..) . −e 4 Accept −e 4 e − e , and −e2 e2−e . 2 2 e2 e Allow M1A1 for a correct decimal answer following B1B1 seen. 4

Mark scheme

3 Substitute x = 12 and equate the result to zero ( ) M1 Or divide as far as ( 2 x − 1) x 2 + px + q and equate constant remainder to zero. Obtain a correct equation with powers evaluated A1 e.g. 14 + a4 − 112 + b = 0 or a + 4b = 21 . Substitute x = −1 and equate result to 12 ( ) M1 Or divide as far as ( x + 1) 2 x 2 + rx + s and equate constant remainder to 12. Obtain a correct equation with powers evaluated A1 e.g. −2 + a + 11 + b = 12 or a + b = 3 . Obtain a = −3, b = 6 A1 5 x2 ( a2+1 ) x 1 2 ( − 212 + a2 ) 2x 2 ( a − 2) x ( −9 − a ) 2x − 1 2x3 ax 2 −11x b x +1 2x3 ax 2 −11x b 2x3 − x2 2x3 2x2 −11x ( a − 2) x2 −11x ( a + 1) x2 ( a + 1) x2 − ( a2+1 ) x ( a − 2) x2 ( a − 2) x ( − 212 + a2 ) x b ( −9 − a ) x b ( −9 − a ) x ( −9 − a ) ( − 212 + a2 ) x − 12 ( − 21 2 + a2 ) b+9+a b − 21 4 + a4

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 e.g tan−1 34 + sin−1 52 ( 0.6435 + 0.4115 ) 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 i 5 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). The equation ( x − 4) + ( y − 3) = 4 will have one root. 2 2 ( ) Hence 1 + m2 x 2 − (8 + 6m ) x + 21 = 0 , discriminant = 0 = 48m2 − 96m + 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

Mark scheme

5 x−4 B1 Split fraction to obtain 1 + x2 + 4 x ( ) Allow for p ln x 2 + 4 from M1 ( ) x Attempt integration and obtain p ln x 2 + 4 or q tan −1 from correct dx but only if a 2 x +4 2 working correct method for splitting has been used. ( Obtain 12 ln x 2 + 4 ) A1 FT Follow through is on their coefficients in the partial fraction. x2 x Allow from 2 + 2 even if the split of the x +4 x +4 4 x fraction is not complete. If 1 − 2 + 2 later seen x +4 x +4 or implied, award the B1. Only available from a correct split, not from an approach using parts that is incomplete. x A1 FT Only available from a correct split, not from an approach Obtain −2 tan −1 using parts that is incomplete. 2 ( Correct use of correct limits 0 and 6 in an expression involving p ln x 2 + 4 , ) M1 p and q should be constants. The x term is not required at this stage. x q tan −1 and no incorrect terms. 2 Obtain 6 + 12 ln10 − 2tan −1 3 A1 ISW Or three term equivalent. (Must combine the ln terms.) Accept with 12 ln 10 . 5 Alternative method for question 5 Use the substitution x = 2tanθ to obtain 2 tan θ + tan θ dθ B1 2 Attempt integration and obtain p tan θ or r ln ( secθ ) from correct working M1 Obtain 2 tan θ ( −2θ ) and A1 FT Follow through on their coefficients after the substitution. Obtain lnsecθ A1 FT Follow through on their coefficients after the substitution. Use correct limits 0 and tan–1 3in an expression involving u tan θ, v ln sec θ M1 u and v should be constants. The θ term is not required at and no incorrect terms this stage. ( ) Obtain 6 + ln sec tan −1 3 − 2 tan −1 3 A1 ISW Or three term equivalent. ( ) Not required to simplify ln sec tan −1 3 . 6

Mark scheme

6(a) Sketch a relevant graph. B1 e.g. y = cot x: x intercept should be correct. y Not touching the y-axis. No incorrect curvature. Ignore anything outside 0 < x ⩽ 12 π . y = cotx 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 2 π 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 > 1 from tan x ( 2 − cos x ) = 1 . Need all relevant values but only one (pair) needs to be correct to award M1. 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 1 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

Mark scheme

7(a) Use correct expansion for cos ( 2θ + θ ) *M1 Use correct double angle formulae to express cos3θ in terms of cosθ and DM1 sinθ Show sufficient working to confirm cos3θ ≡ 4cos3 θ − 3cosθ A1 AG 3 7(b) Use the identity and correct double angle formula to obtain an equation in *M1 e.g. 4 cos3 θ – 3 cos θ + cos θ (2 cos2 θ – 1) = cos2 θ cosθ only. Must come from using all three terms in the given equation. 6 cos3 θ – cos2 θ – 4 cos θ = 0 or 6 cos2 θ – cos θ – 4 = 0. Obtain θ = 90° B1 Allow if cosθ 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 cosθ to obtain a value of θ 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

Mark scheme

8(a) Multiply both sides by a + 2i and attempt expansion of right-hand side *M1 Use of i2 = −1 seen at least once (or implied) DM1 e.g. 2 + 3ai = λ ( 2a + 2 ) + λi ( −a + 4 ) Compare real and imaginary parts to obtain an equation in a only M1 3a −a + 4 e.g. = . Any equivalent form. 2 = λ (2a + 2), 3a = λ ( −a + 4) 2 2a + 2 Obtain 3a 2 + 4a − 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 i2 = −1 seen at least once or implied DM1 e.g. λ ( 2 − i ) = ( 8a + i 3a 2 − 4 ). a +4 2 Compare real and imaginary parts to obtain an equation in a only ( ) M1 e.g. 8a = −2 3a 2 − 4 . Any equivalent form. Obtain 3a 2 + 4a − 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 λ only (based on an equation seen in their working in (a) or (b)) value for λ seen. Obtain a = −2, λ = −1 or a = 23 , λ = 53 A1 Allow 106 and 0.6. Obtain second correct pair of values A1 3

Mark scheme

9(a) Use correct product rule *M1 As far as p cos x cos 2 x + q sin x sin 2 x or full working (u, v, du/dx, dv/dx) shown. dy A1 OE Obtain = cos x cos 2 x − 2sin x sin 2 x dx Equate derivative to zero and use correct double angle formulae DM1 Allow if only have one double angle in their derivative. ( ) Obtain cos x 1 − 6sin 2 x = 0 or equivalent ( ) A1 e.g. cos x 6cos2 x − 5 = 0 , 5 tan 2 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 ( ) Obtain cos x 1 − 6sin 2 x = 0 A1 OE Obtain a = 0.42 A1 Only. Accept x = 0.42 . 5 9(b) Use double angle formula and obtain p cos3 x + q cos x correctly *M1 e.g. from 2cos 2 x sin x − sin x dx . 2 A1 Obtain ± − cos3 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 2 −1 3 1 3 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 cos2x + s sin x sin 2x *M1 Seen, not just implied. 1 2 A1 Obtain cos x cos 2 x + sin x sin 2 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. Alternative method 2 for question 9(b) Use factor formula and integrate to obtain g cos3x + h cos x *M1 (sin 3x − sin x ) dx . 1 2 Obtain ± ( − 16 cos3x + 12 cos x ) A1 Correct for their integral. Correct use of limits 14 π and 34 π (or use double the integral from 14 π to 12 π ) DM1 OE 1 2 ( 16 + 12 + 16 + 12 ) . 2 2 A1 Obtain Or exact equivalent. Final answer must be positive. 3 4

Mark scheme

10(a) Use the correct process to calculate the scalar product of the direction vectors M1 ( −2 + 4 + 2c ) . Divide the scalar product by the product of the moduli and equate the result M1 Or equivalent e.g. 2 + 2c = 6 20 + c2 cos60° . to cos60° Allow for the correct process using 60° but the wrong vectors. Obtain correct equation in c A1 2 + 2c 1 e.g. = or 10c 2 + 32c − 104 = 0 . 6 20 + c 2 2 Obtain c = 2 A1 Only. 4 10(b) Calling ( 6, − 3, 6 ) A, find AP for a general point P on l B1 −3 + λ e.g. 1 + λ . −5 + 2λ Equate the scalar product of their AP and a direction vector for l to zero and *M1 e.g. ( −3 + λ ) + (1 + λ ) + ( −10 + 4λ ) = 0 . obtain an equation in λ Solve and obtain λ = 2 A1 Carry out a method to calculate AP DM1 e.g. ( −1)2 + 32 + ( −1)2 or 12 + 32 + 12 . Obtain 11 from correct working A1 AG Alternative method 1 for question 10(b) Calling ( 6, − 3, 6 ) A, find AP for a general point P on l B1 −3 + λ e.g. 1 + λ −5 + 2λ Differentiate the modulus of AP or the square of the modulus and equate the *M1 e.g. 2 ( −3 + λ ) + 2 (1 + λ ) + 4 ( −5 + 2λ ) = 0 derivative to zero Solve and obtain λ = 2 A1 Carry out a method to calculate AP DM1 e.g. ( −1)2 + 32 + ( −1)2 or 12 + 32 + 12 Obtain 11 from correct working A1 AG Alternative method 2 for question 10(b) Vector from ( 6, − 3, 6 ) to ( 3, − 2, 1) is −3i + j − 5k B1 The method works for vector from ( 6, − 3, 6 ) to any point on l. Use scalar product to find the angle between their vector and the direction of M1 l 3 − 1 + 10 24 11 A1 Obtain cosθ = = or sin θ = 35 6 35 35 Correct use of trig to find the projection of their vector on the normal to l M1 11 35 sin θ = 35 × . 35 Obtain 11 from correct working A1 AG Alternative method 3 for question 10(c) Vector from ( 6, − 3, 6 ) to ( 3, − 2, 1) is −3i + j − 5k B1 Find the vector product of their vector and the direction of l M1 Obtain i ( 2 + 5 ) − j ( −6 + 5 ) + k ( −3 − 1) ( = 7i + j − 4k ) A1 Correct use of trig to find the perpendicular distance M1 their vector product . direction vector 66 A1 Distance = = 11 6 5

Mark scheme

11(a) Correct separation of variables. B1 1 dy = 1 − 2 dx . 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 +y 2 ( y ± y) 2 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 y y +1 ( ) ( ) but followed by an error. 2 2 y + 12 − 12 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 xp , q ln y and r ln (1 + y ) If they rearrange the equation before finding the constant 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 −1 ex A1 1 1 Obtain y = Or equivalent e.g. y = , y = 1− 1 + ln 2 1− 1x . 2 − ex 1 −1 2e − 1 e x −1 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 1 FT on an expression in e x . 1