Mathematics 9709/11 October/November 2025: question paper and mark scheme

Mark scheme

1 Use of b2 – 4ac *M1 Obtain k 2 − 20k + 64 A1 Attempt solution of quadratic equation or inequality DM1 Solve quadratic using suitable method. E.g., factorisation, quadratic formula, completing the square. Obtain k < 4, k > 16 or clear equivalent B1 B0 for use of ⩽ and/or ⩾. 4

Mark scheme

2(a) 1 B1 Condone poor notation. State or imply that second and fifth terms are a cosθ and a cos 4 θ or r3 = 64 Attempt solution of equation of form cos3θ = k as far as cos θ =… , where 0 < k < 1 M1 Must come from division. Obtain cosθ = 0.25, and hence θ = 1.32 only A1 Allow greater accuracy 1.318116… A0 for 75.5o . 3 2(b) Attempt to find a using their θ and substitute in correct formula for sum to infinity M1 Using exact or approximate value for r. −1 < r < 1 Obtain a = 32 and use exact value of r to obtain 128 3 A1 Or exact equivalent. 2

Mark scheme

3 x4 term from ( px + 3)5 is 5 p 4 x 4 × 3 B1 Or coefficient is 15 p 4 . Not 5C1 × 3 or factorial form or 15(px)4 . Can be in a list. p2 6 p 4 M1 Identify term involving x × 2 as relevant term from x3 + x x Obtain 6 p 2 x 4 or −6 p 2 x 4 A1 Or coefficient is 6 p 2 or −6 p 2 . Not 4C2 or factorial form. Attempt solution for p2 of equation that is quadratic in p 2 M1 Solve quadratic in p2 using suitable method. E.g. factorisation, quadratic formula, completing the square. E.g. 3(5p2 + 18)(p2 – 4) Obtain 15 p 4 − 6 p 2 − 216 = 0 or equivalent and hence p = 2 B1 No other solutions. 5

Mark scheme

4(a) {10} – (x {+}{3}) 2 B1 B1 B1 3 4(b) State reflection in x-axis B1 Obtain m = −3 B1 FT Following their value of b. Obtain n = 10 B1 FT Following their value of a. 3

Mark scheme

5(a) sin 4 θ sin 2θ sin 2θ B1 Candidates can work from LHS to RHS or RHS Use tan 4 θ = or tan 4 θ – 1 = +1 + 2 − 1 to LHS for full marks. If working on both sides cos θ 4 cos θ cos θ 2 simultaneously, maximum of B1 M1 only if a common correct expression is reached. Attempt to express tan 4 θ − 1 as a single fraction in terms of cosθ only M1 (1 − c 2 ) 2 − c 4 (s 2 + c2 )(s 2 − c 2 ) Using or etc. c4 c4 1 − 2cos 2 θ A1 AG – necessary detail needed. Confirm result WWW cos 4 θ Alternative Method for Question 5(a) 1 − 2cos 2 θ cos 4 θ B1 + –1 cos 4 θ cos 4 θ (1 − cos θ ) – 1 M1 2 2 cos 4 θ Confirm result tan 4 θ – 1 A1 AG – necessary detail needed. Alternative Method 2 for Question 5(a) tan 4θ = sec 4θ – 2 sec 2θ + 1 B1 1 2 M1 tan 4θ − 1 = – cos 4θ cos 2θ 1 − 2cos 2 θ A1 AG – necessary detail needed. Confirm result cos 4 θ 3 5(b) Use identity to form equation using cosθ only *M1 1 − 2cos 2 θ Expect = 7 OE cos 2 θ Allow unsimplified. Obtain 9cos 2 θ = 1 or similarly simplified equivalent A1 Condone cos 2θ (1 – 9 cos 2θ ) = 0 Could be implied by one correct solution Solve equation of form cos 2θ = k to obtain at least one solution DM1 Provided 0 < k < 1. Obtain 70.5 and 109.5 A1 Or greater accuracy 70.528…, 109.471…, and no others between 0 and 180. SC B1 for 1.23c and 1.91c only. A0 if cos 2θ (1 – 9 cos 2θ ) = 0 is used. 4

Mark scheme

6(a) State f ( x) − 3 or y − 3 or clear equivalent B1 Must be ⩾ rather than >. Allow ⩾ –3 but not x ⩾ –3. 1 6(b) Attempt to arrange to x = ... in terms of y or equivalent M1 Or in terms of x already. Sign errors only. Obtain −3 + x + 12 A1 Now in terms of x. A0 for ± in final answer. Condone y = … 2 6(c) Attempt expression for gf ( x) *M1 Expect 2((x + 3)2 – 12) – 5. Sign errors only. Obtain 2( x + 3) 2 − 29 = 69 A1 OE E.g. 2x2 + 12x – 80 [= 0]. Attempt solution of quadratic equation to find at least one value of x DM1 No method needed. Obtain x = 4 only A1 Alternative Method for Question 6(c) −1 Attempt solution of g( x) = 69 or evaluation of g (69) *M1 Incorrect order of f −1 and then g −1 scores *M0. Obtain 37 A1 −1 Attempt solution of f ( x) = their 37 or evaluation of f (their 37) DM1 Obtain x = 4 only A1 4

Mark scheme

7(a) Obtain correct 12 r 2 × 23 π − 12 r 2 sin 23 π M1 0.614r 2 A1 AG Greater accuracy is 0.61418... r 2 . 2 7(b)(i) dA B1 Or greater accuracy Obtain = 1.228r dr dA dA dr M1 dA Use = × , or equivalent, with r = 20 ‘their’ × 0.4 with r = 20 dt dr dt dr Obtain 9.8 A1 AWRT 3 7(b)(ii) State or imply 23 πr for length of arc AB B1 dl 2π Could be implied by = . dθ 3 Differentiate and apply correct use of chain rule M1 Obtain 0.84 A1 AWRT 3

Mark scheme

8(a) 3 Integrate to obtain expression 12 x 2 ÷ 23 B1 OE 3 M1 Evaluate integral of the form k x 2 between limits 0 and 9 and subtract result from 272 3 Obtain 13 x 2 giving area under curve is 9 and shaded area is 92 A1 Alternative Method for Question 8(a) Obtain x = 4 y 2 B1 Integrate to obtain expression of form k y 3 and evaluate between limits 0 and 32 M1 Obtain 43 y 3 and hence shaded area is 92 A1 3 8(b) Attempt to express x 2 in terms of y M1 Condone ky4 . Attempting y 2 dx scores 0/4. Obtain or imply volume is [π] 16 y [dy] 4 A1 5 Integrate to obtain ky and evaluate between limits 0 and 32 M1 Obtain 165 y5 and hence 243 π or exact equivalent A1 Condone omission of π except for last mark. 10 4

Mark scheme

9(a) Attempt to express each of S2 − 1, S4 , S9 in terms of d (with at least one correct) *M1 Allow unsimplified. Obtain S2 − 1 = 3 + d , S4 = 8 + 6d , S9 = 18 + 36d A1 Must be simplified but can be implied by later work. Attempt equation in d by linking the three values DM1 Obtain correct (3 + d ) + (18 + 36d ) = 2(8 + 6d ) or equivalent, and hence d = − 15 A1 4 9(b) 2 + 14 × their d M1 Second progression: obtain either first term = 145 or common difference = 4 B1 Must be from correct d in part (a). Second progression: attempt to find 15th term M1 Using a = 3 + their d from part (a). Obtain 58.8, and hence difference = 59.6 A1 4

Mark scheme

10(a) Attempt to substitute for x or y to produce quadratic equation in one variable *M1 Allow unsimplified. Obtain 5 x 2 − 40 x + 75 = 0 or 5 y 2 − 20 = 0 A1 Or similarly simplified equivalent. Find two values of one variable B1 Obtain coordinates (3, 2) and (5, − 2) B1 4 10(b) Obtain centre (0, − 2) [and radius 5] B1 Attempt area of ABC using mid-point of AB or 12 × base × height M1 OE Use of correct formula for area of a triangle. Obtain 12 20 × 20 or 12 × 5 × 4 (or equivalent) and hence 10 A1 3

Mark scheme

11(a) Substitute 4 to obtain gradient of curve is 101 B1 Attempt equation of normal using (4, 3) and − m1 for their gradient M1 y = −10 x + 43 A1 3 11(b) −16x +40(2x − 3) −3 −3 B1 B1 OE 7 Substitute 4 to obtain 100 B1 3 11(c) [ y =]−8x−1+5(2 x − 3)−1[+c] B1 B1 Substitute x = 4, y = 3 in an integrated expression to find value of c M1 Obtain 3 = −2 + 1 + c and hence y = −8 x −1 + 5(2 x − 3) −1 + 4 A1 OE For finding c = 4. Substitute x = −1 to obtain q = 11 A1 Not y = 11. 5