Mathematics 9709/32 February/March 2025: question paper and mark scheme

Mark scheme

1 State that 1 – e–2 x = e–3 B1 OE, with ln removed. Use correct method to solve an equation of the form e±2 x = a, where a > 0, and a M1 E.g. [e–2 x = 1 − e–3 ] reasonable attempt at the B1, for ± 2x ln e or ± x ln e −2x = ln(1 − e–3 ) … OE. Can be numerical ln (1 − 0.049787) = ln 0.9502. Evidence of method must be seen. Obtain answer 0.0255 A1 CAO Must be 4 decimal places. No working seen scores 0. Alternative Method for Question 1 State that 1 – e–2 x = e–3 B1 OE, without ln. Rearrange to obtain an expression for ex and solve an equation of the form e±x = a, M1 1 e3 where a > 0, and a reasonable attempt at the B1, for x E.g. e x = , e x = , 1 − e −3 e3 − 1 1 x = ln 1 − e −3 Can be numerical. Evidence of method must be seen. Obtain answer 0.0255 A1 CAO Must be 4 decimal places. No working seen scores 0. 3

Mark scheme

2 dy B1 1+ 2 State or imply dx as the derivative of ln (x + 2y) x + 2y dy B1 dy State derivative of xy2 is x2y + y2 x2 y + y2 dx dx dy B1 OE 1+ 2 May be implied by correct final answer. dy dx = 0 Obtain y 2 + 2 xy + dx x + 2 y 1 B1 OE Obtain y = e when x = 0 Allow 12 × 2.718 or 1.36 or better e.g. 1.359… 2 May be implied by correct final answer. ( ) dy 1 B1 OE Obtain = − e3 + 4 Accept AWRT –3.01. ISW. dx 8 5

Mark scheme

3(a) Obtain Im (z) ⩽ –1 B1 Condone strict inequalities throughout (a). Obtain answer of the form z − a b M1 Accept equation or any inequality sign. a = ± 2 ± i and b = 3, e.g. z + 2 − i = 3 or z + 2 − i > 3. Obtain answer z + 2 − i ≤ 3 A1 Accept z − ( −2 + i ) 3 as final answer. Do not ISW. 3 3(b) Identify the coordinates of correct point M1* ( −2 − 5, −1) , if correct. From solving (x ± 2)² + (y ± 1)² = 3² (or = 3) with y = –1, or attempt to get 2 + 5 using a right- angled triangle. Carry out a correct method for finding the greatest value of | z | DM1 A1 AWRT 4.35, e.g. 4.3525… Obtain answer 4.35 or 10 + 4 5 3

Mark scheme

4 tan x − 3 B1 OE State Allow decimals throughout. 1 + 3 tan x 2 B1 SOI 2 cot x replaced by tan x Reduce the equation to tan² x – 3 3 tan x – 2 = 0, or three-term equivalent B1 May be implied by further work. Solve a three-term quadratic in tan x, for x M1 FT their 3-term quadratic. Allow tan–1 (..). Obtain answer, e.g. 79.8° A1 AWRT 79.8. Obtain the second answer, e.g. 160.2° and no other in the interval A1 Allow 160, or AWRT 160.2. Treat answers in radians as a misread. Ignore answers outside the given interval. 6

Mark scheme

5 Square x + iy and equate real and imaginary parts to – 4 and 6 5 respectively M1 Obtain equations x2 – y2 = – 4 and 2xy = 6 5 A1 Or x2 + y2 = 14. ( −4 ) + ( 6 5 ) 2 2 or x2 + y 2 = Eliminate one variable and find a horizontal equation in the other M1 Allow slips in e.g. signs, powers etc. Obtain x4 + 4x2 – 45 = 0 or y4 – 4y2 – 45 = 0 or three-term equivalents, A1 May be implied by further work. or 2x2 = 10 or 2y2 = 18 Obtain answers ± ( 5 + 3i ) A1 Accept e.g. x = 5, y = 3 and x = − 5 , y = –3 or ± ( 5,3) , but must be clearly paired. Can be implied by (e.g.) column working. 5

Mark scheme

6 Use correct double angle formula to express sin 2 2θ in terms of cos 4θ B1 1 sin2 2θ = (1 – cos 4θ) 2 Separate variables correctly and reasonable attempt at integration of at least one side M1 Position of (x + 5) or ( 15 x + 1) and sin2 2θ sufficient for correct separation. 1 B1 OE Obtain term 5ln x + 1 May see 5ln ( x + 5 ) . 5 1 1 B1 FT 1 1 1 Obtain term (θ − sin 4θ ) Allow ±θ ± sin 4θ from = (± 1 ± cos 4θ). 2 4 2 4 2 Use x = 5 when θ = 0 to evaluate a constant or as limits in a solution containing M1 OE 1 terms of the form ln x + 1 , θ and sin 4θ 5 Obtain correct answer in any form A1 1 1 E.g. 5ln ( x + 5 ) = (θ − sin 4θ ) + 5 ln 10 2 4 1 A1 FT 1 Obtain final answer x = 10exp θ − sin 4θ − 5 or equivalent 1 1 x = 10exp ±θ ± sin 4θ − 5 10 4 2 4 with ln removed Must remove ln from 1 1 ln ( x + 5 ) = (±θ ± sin 4θ ) + α ln 10. 2 4 7

Mark scheme

7(a) Use correct product rule M1 d 3x2 cos (2x) + x3 (cos 2x) OE dx Obtain derivative 3x2 cos (2x) – 2x3 sin (2x) A1 OE 1 3 A1 AG Equate correct derivative to zero and obtain p = tan −1 after full and correct Must reach tan 2p = … or tan 2x = … before given 2 2p answer. working 3 7(b) Calculate the values of a relevant expression or pair of expressions at x = 0.5 and M1 1 3 x = 0.7 Must have at least 1 of 2 values correct or 3 of 4 values correct f(p) = p − tan −1 2 2p f(0.5) = –0.124… < 0, f(0.7) = 0.132.. > 0 1 3 or comparing p and tan −1 . 2 2p 1 −1 3 0.5 tan = 0.624 0.5 < 0.624… 2 2p 1 −1 3 0.7 tan = 0.567 0.7 > 0.567 2 2p Complete the argument correctly with correct calculated values A1 2 7(c) Use the iterative formula correctly at least twice M1 Obtain final answer 0.596 A1 Show sufficient iterations to at least 5 decimal places to justify 0.596 to 3 decimal A1 E.g. 0.5, 0.62452, 0.58814, 0.59856, 0.59556, places or show there is a sign change in the interval (0.5955, 0.5965) 0.59642 0.6, 0.59514, 0.59654, 0.59614, 0.59626 0.7, 0.56708, 0.60467, 0.59380, 0.59693, 0 59603, 0.59629 Can recover from wrong values. 3

Mark scheme

8(a) Express general point of a line in component form, B1 e.g. (–1 + 2λ, 3 + 3λ, – 4 – λ) or (2 – μ, –3 – 2μ, –1 + μ) Equate at least two pairs of components and solve for λ or for μ M1 Obtain correct answer for λ or for μ A1 Possible answers are 6, 12, 0 for λ and –9, –21, –3 for μ. Verify that one component equation is not satisfied A1 E.g. show 21 ≠ 15 for (11, 21, –10) and Can show by correctly obtaining 2 values of λ or 2 values of μ (11, 15, –10), or show –16 ≠ –22 for (23, 39, –16) and (23, 39, –22), or show –1 ≠ 5 for (–1, 3, – 4) and (5, 3, – 4). Show that the lines are not parallel B1 2 −1 E.g. 3 ≠ k −2 at least 2 components −1 1 required. Just a statement that direction vectors are not scalar multiple of each other insufficient, if direction vectors have not been clearly identified. Also, told answer is skew. 5 8(b) 2 −1 M1 E.g. (2 × –1) + (3 × –2) + (–1 × 1) or –2 – 6 – 1 Carry out correct process for evaluating the scalar product of 3 and −2 or –9. −1 1 Using the correct process for the moduli, divide the scalar product by the product of M1 Allow for any pair of vectors here but must be the moduli and evaluate the inverse cosine of the result consistent between scalar product and magnitudes. Obtain answer AWRT 169.1° or 2.95c A1 Allow 169°. 3

Mark scheme

9(a) Substitute x = 3 or –3 into p (x) and equate to 0 or into p' (x) and equate to 72 M1* Obtain 162 + 9a + 3b + 9 = 0 A1 OE Obtain 162 + 6a + b = 72 A1 OE Solve simultaneous equations to obtain either a or b after using p (±3) = 0 and DM1 pꞌ (±3) = 72 Obtain a = –11 and b = –24 A1 5 9(b) Equate (x – 3)(6x2 + Ax + B) to 6x3 – 11x2 – 24x + 9 and obtain equations to solve for M1 Using their a and their b. A and B A – 18 = a = − 11, B – 3A = − b = − 24, − 3B = 9. A = 7 and B = –3. or divide 6x3 – 11x2 – 24x + 9 by x – 3 and reach 6x2 ± 7x Or reach 6x2 ± (their a + 18) x. (x – 3)(6x2 + 7x – 3) A1 SOI Obtain (x – 3)(2x + 3)(3x – 1) A1 Special Case: If only ( x – 3) ( x + 32 )( x – 13 ) or (x – 3)(2x + 3)(3x – 1) seen, SC B1 only (but can gain two marks in (c)). 3 9(c) 3 1 B1 FT Must be final answer not in working. Obtain one correct region x < − or < x < 3 FT is on the last two brackets (not ( x – 3)). 2 3 3 1 B1 FT 3 1 Obtain both regions x < − , < x < 3 Allow x < − and < x < 3. 2 3 2 3 3 1 SC B1 for x − , x 3. 2 3 FT is on the last two brackets (not ( x – 3)). If incorrect factor or factors in (b) but correct regions here, allow SC B1 only. 2

Mark scheme

10(a) A Bx + C B1 State or imply the form + ( ( ) 4 + x2 1 + x ) Use a correct method for finding a constant Even with incorrect PF denominators M1 A( 4 + x 2 ) + (Bx + C)(1 + x) + = −7 x 2 +2x − 6 Obtain one of A = –3, B = – 4 and C = 6 A1 Obtain a second value A1 Obtain a third value A1 A C Special Case 1: + + Find A, ( ) (1 + x ) 4 + x 2 M1 A1. Max 2/5. A Bx Special Case 2: + + Find A, ( ) (1 + x ) 4 + x 2 M1 A1. Max 2/5. 5 10(b) Obtain term –3ln (1 + x) B1 FT OE FT A ln (1 + x) Obtain term –2ln (4 + x2 ) B1 FT OE B FT ln (4 + x2 ) 2 Obtain integral of the form c tan −1 dx with d ≠ 1 following separation into two M1 d = 12 or 2 only. expressions x A1 FT C −1 x Obtain 3tan −1 FT tan 2 2 2 Substitute correct limits correctly in an expression (obtained correctly) of the form M1 a ln (3) + b ln (8) − b ln (4) + c ( 1 π ) , where a ln (1 + x), b ln (4 + x2 ), and c tan −1 ( 12 x ) , where a, b, c ≠ 0. 4 a, b, c ≠ 0. Do not allow slips, and must get to c ( 14 π ). 3 A1 Must be in the form aπ – ln b. Obtain answer π − ln108 4 6

Mark scheme

11 1 1 *M1 OE Commence integration by parts and reach Ax 2 sin x ± Bx sin xdx BOD on ± otherwise scores 0/6. 3 3 1 1 A1 OE Obtain 3x 2 sin x − 6 x sin x dx Allow 3×2 for 6. 3 3 1 1 1 *DM1 OE Complete integration by parts and reach Ax 2 sin x + Bx cos x + C sin x 3 3 3 1 1 1 A1 Allow 6 × 3 for 18, 9 × 6 for 54 OE. Obtain 3x2 sin x + 18 x cos x − 54sin x oe 3 3 3 Substitute limits correctly in an expression of the form DM1 Dependent on both previous M1 marks. 1 1 1 Ax 2 sin x + Bx cos x + C sin x , where ABC ≠ 0 π 3 π 1 Need to use sin = , and cos = to obtain 3 3 3 3 2 3 2 3 Bπ C 3 Aπ 2 + + . 2 2 2 3 3 2 A1 27 3 3 18 54 Obtain answer π + 9π − 27 3 or exact equivalent ISW Allow for , for 9, for 27, 2 2 2 2 2 2187 for 27 3 etc. Alternative Method for first 4 marks: 1 1 *M1 Commence integration by parts and reach Ax 2 sin x + Bx cos x 3 3 1 1 A1 OE Obtain 3x 2 sin x + 18x cos x Allow 6 × 3 for 18. 3 3 1 1 1 *DM1 Complete integration by parts and reach Ax 2 sin x + Bx cos x + C sin x 3 3 3 11 1 1 1 A1 OE Obtain 3x2 sin x + 18 x cos x − 54sin x Allow 6 × 3 for 18, 9 × 6 for 54, OE. 3 3 3 6