Cambridge A Level Mathematics 9709 — 2025 Feb/March Paper 3 · Variant 2

1 Solve the equation ln `1 - e -2 xj + 3 = 0 . Give your final answer correct to 4 decimal places. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 State that 1 – e–2x = e–3 B1 OE, with ln removed. Use correct method to solve an equation of the form e±2x = a, where a > 0, and a M1 E.g. [e–2x = 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–2x = e–3 B1 OE, without ln. Rearrange to obtain an expression for ex and solve an equation of the form e±x = a, M1  3  x 1 x e where a > 0, and a reasonable attempt at the B1, for x E.g.  e = −3 , e = 3  ,  1 − e e − 1    1 x = ln 3 1 − e− 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

More questions on Logarithmic and exponential functions

2 The equation of a curve is xy 2 + ln `x + 2yj = 1. Find the gradient of the curve at the point where x = 0 . 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Mark scheme: 2 dy B1 1 + 2 dx State or imply as the derivative of ln (x + 2y) x + 2 y d y B1 2 dy 2 State derivative of xy2 is x2y + y2 x y + y d x dx dy B1 OE 1 + 2 2 dy dx May be implied by correct final answer. Obtain y + 2 xy + = 0 dx x + 2 y 1 B1 OE Obtain y = e when x = 0 1 Allow 2  2.718 or 1.36 or better e.g. 1.359… 2 May be implied by correct final answer. dy 1 3 B1 OE Obtain = − e + 4 ( ) Accept AWRT –3.01. ISW. dx 8 5

More questions on Differentiation

3 Im 4i 3i 2i i – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 Re – i – 2i – 3i The shaded region on the Argand diagram shows points representing complex numbers z defined by two inequalities. The shaded region is bounded by a circle and a line parallel to the real axis. The boundaries of the region are included in the shaded region. (a) Find two inequalities in terms of z that define the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the greatest value of z for points in this region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

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

More questions on Complex numbers

4 By first expressing the equation tan ( x - 60°) = 2 cot x as a quadratic equation in tanx, solve the equation for 0° G x G 180° . [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

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 B1 May be implied by further work. Reduce the equation to tan² x – 3 3 tan x – 2 = 0, or three-term equivalent 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

More questions on Trigonometry

5 The square roots of - 4 + 6 5i can be expressed in the Cartesian form x + yi , where x and y are real and exact. By first forming a quartic equation in x or y, find the square roots of - 4 + 6 5i in exact Cartesian form. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 M1 Square x + iy and equate real and imaginary parts to – 4 and 6 5 respectively A1 Or x2 + y2 = 14. Obtain equations x2 – y2 = – 4 and 2xy = 6 5 2 2 2 2 6 5 or x + y = ( −4 ) + ( ) 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

More questions on Complex numbers

6 The variables x and i satisfy the differential equation d x 1 2 = b x + 1l sin 2 i , d i 5 and x = 5 when i = 0 . Solve the differential equation and obtain an expression for x in terms of i. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 6 Use correct double angle formula to express sin2 2θ in terms of cos 4θ B1 1 sin2 2θ = (1 – cos 4θ) 2 1 Separate variables correctly and reasonable attempt at integration of at least one side M1 Position of (x + 5) or ( 5 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 (− sin4) Allow   sin 4 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  ,  and sin4  5  Obtain correct answer in any form A1 1 1 E.g. 5ln ( x + 5 ) = (− sin4) + 5 ln 10 2 4  1  1   A1 FT  1  1   Obtain final answer x = 10exp − sin4 − 5 or equivalent x = 10exp  sin4 − 5         10  4    2  4   with ln removed Must remove ln from 1 1 ln ( x + 5 ) = (  sin4 ) +  ln 10. 2 4 7

More questions on Differential equations

8 Two lines have equations r = f 3p + m f 3p and r = f- 3p + n f- 2p. - 4 - 1 - 1 1 (a) Show that the lines are skew. 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(b) Find the obtuse angle between the directions of the two lines. 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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     or –9. Carry out correct process for evaluating the scalar product of 3 and −2          − 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

More questions on Vectors

9 The polynomial 6 x 3 + ax 2 + bx + 9 is denoted by p ( )x , where a and b are constants. It is given that ( x - 3 ) is a factor of p ( )x , and when the first derivative pl ( )x is divided by ( x - 3 ) the remainder is 72. (a) Find the values of a and b. 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(b) When a and b have the values found in part (a), factorise p ( )x completely. 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(c) Hence solve the inequality p ( )x 1 0 . 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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 3 1 Special Case: If only( x –3)( x + 2 )( x – 3 ) 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 1 B1 FT 3 1 Obtain both regions x − 3,  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

More questions on Quadratics

- 7x 2 + 2 x - 6 10 Let f ( x) = . `1 + x`j4 + x 2j (a) Express f ( )x in partial fractions. 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(b) Hence find the exact value of f ( )x d x . Give your answer in the form ar - ln b , where a and b 2y 0 are constants. 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Mark scheme: 10(a) A Bx + C B1 State or imply the form + 4 + x 2 (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, 4 + x 2 (1 + x ) ( ) M1 A1. Max 2/5. A Bx Special Case 2: + + Find A, 4 + x 2 (1 + x ) ( ) 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−dx1 with d ≠ 1 following separation into two M1 d = 12 or 2 only. expressions −x1 A1 FT C −1 x Obtain 3tan FT tan 2 2 2 1 where Substitute correct limits correctly in an expression (obtained correctly) of the form M1 a ln (3) + b ln (8) − b ln (4) + c( 4 π ) , 1 where a, b, c ≠ 0. a ln (1 + x), b ln (4 + x2), and c tan −1 ( 2 x ) , a, b, c ≠ 0. 1 Do not allow slips, and must get to c ( 4 π ) . 3 A1 Must be in the form aπ – ln b. Obtain answer π − ln108 4 6

More questions on Integration

r 11 Find the exact value of x 2 cos 1 x dx . 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Mark scheme: 11 2 1 1 *M1 OE Commence integration by parts and reach Ax sin x  Bx sin x dx BOD on ± otherwise scores 0/6. 3 3 2 1 1 A1 OE Obtain 3 x sin x − 6 x sin x dx Allow 3×2 for 6. 3 3 2 1 1 1 *DM1 OE Complete integration by parts and reach Ax sin x + Bx cos x + C sin x 3 3 3 2 1 1 1 A1 Allow 6 × 3 for 18, 9 × 6 for 54 OE. Obtain 3 x 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. 2 1 1 1 π 3 π 1 Ax sin x + Bx cos x + C sin x , where ABC ≠ 0 Need to use sin = , and cos = to obtain 3 3 3 3 2 3 2 2 3 Bπ C 3 Aπ + + . 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: 2 1 1 *M1 Commence integration by parts and reach Ax sin x + Bx cos x 3 3 2 1 1 A1 OE Obtain 3 x sin x + 18 x cos x Allow 6 × 3 for 18. 3 3 2 1 1 1 *DM1 Complete integration by parts and reach Ax sin x + Bx cos x + C sin x 3 3 3 11 2 1 1 1 A1 OE Obtain 3 x sin x + 18 x cos x − 54sin x Allow 6 × 3 for 18, 9 × 6 for 54, OE. 3 3 3 6

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