Mathematics 9709/11 May/June 2025: question paper and mark scheme

Mark scheme

1 6sin 2 θ − sin θ − 2[ = 0] ( 2sinθ + 1)( 3sinθ − 2) = 0 M1 For expressing as a 3-term quadratic. Terms need not all be on the same side. 1 2 A1 For both. sin θ = − or sin θ = Allow AWRT 0.667. 2 3 θ = −150° , − 30°, 41.8°, 138.2° A1 A1 AWRT A1 for any correct angle from a correct value of sin θ , A1 for all 4 and no others in the interval −180° < θ < 180° . −5π − π SC B1 for , , 0.730c, 2.41c if use of 6 6 radians is clear. 4

Mark scheme

2(a) 1 M1 dy [Gradient of tangent] = 4 ( 2 × 4 − 5) − 9 × 4 2 [ = 90] Substitute x = 4 into . 3 dx 1 −11 = 4 ( 2 × 4 − 5) − 9 × 4 2 is M0 unless they 3 2 1 reach − . 90 1 A1 AWRT −0.0111. [Gradient of normal] = − 90 2 2(b) 3 B1 B1 Accept unsimplified. [ y ] = ( 2 x − 5)4 1 −6 x 2 [ +c ] 2 11 1 3 M1 11 − = × ( 2 × 4 − 5) − 6 × 4 2 + c 4 Sub x = 4, y = − into an integrated expression 2 2 2 and attempt to find c. 1 3 A1 Condone c = 2 as final answer if ‘y = …’ seen y = ( 2 x − 5) − 6 x 2 + 2 4 previously. 2 Fractions must be simplified. Accept f(x) in place of y. 4

Mark scheme

3(a) ar 2 = 18 , a + ar + ar 2 = 26 or a + ar + 18 = 26 B1 For first and second, or first and third equations. 18 18 M1 For expressing as a 3-term quadratic from two + + 18 = 26 ⇒ 8r 2 − 18r − 18[= 0] ( 4r + 3)( r − 3) = 0 expressions with sign errors only. Terms need r2 r not all be on one side. 3 A1 CAO r=− [or r = 3 ] Allow – 0.75. 4 a = 32 A1 CAO Ignore a = 2. May not find a, but instead use expression for a 18 in terms of r e.g. a = 2 and then find a in r part (b). Tenth term = −2.40 A1 AWRT. Ignore other values. Alternative Method for first three marks in Question 3(a) 2 ar = 18 , 26 = ( a 1 − r3 ) B1 1− r M1 Expressing as a 3-term cubic from two 18 r 2 ( 1 − r3 ) expressions with sign errors only. Terms need 26 = ⇒ 8r 3 − 26r 2 +18 [ = 0 ] not all be on one side 1− r 3 A1 CAO r=− [or r = 3, 1 ] Accept −0.75. 4 5 Note: SC B1 B1 B1 is possible following B1 3 M0 if no method is shown for finding r = − 4 [or r = 3 ]. 3(b) 32 M1 FT on values of a and r, provided −1 < r < 1. S∞ = 3 1− − 4 128 A1 FT FT on values of a and r, provided −1 < r < 0. 7 Condone extra answers. 2

Mark scheme

4 3 3 *M1 Attempt to integrate both terms and subtract Attempt to integrate ( x − 16 ) − 5 x 2 − 20 x = −5 x 2 + 21x − 16 areas. Accept subtraction either way round. B1 B1 B1 for each integral. 5 5 1 2 2 21 2 x − 16 x [ − ] 2 x 2 − 10 x = −2 x 2 + x − 16 x 5 21 2 2 B2,1, 0 for −2 x 2 + x 2 − 16 x. 2 Use of limits 1 and 16 DM1 Limits either way round. Minimum –7.5 – 384. 1 E.g. (( – 16) – (2 – 10)) – ((128 – 256) – 2 (2048 – 2560)) = (15.5 + 8) – (–128 + 512) = –7.5 – 384 = –391.5 Or (–2 + 10.5 – 16) – (–2048 + 2688 – 256) = –7.5 – 384 = –391.5 Or 225 –504 + = –391.5 2 783 A1 CAO 391.5 or SC B1 for correct answer if M1 B1 B1 DM0 2 scored. Alternative Method for Question 4 Height of triangle = 15 B1 1 M1 Area of triangle = × 15 × their height = 112.5 2 3 5 B1 Integrates ± 5 x − 20 x to ± 2 x 2 − 10 x 2 2 Use of limits 1 and 16 on their integral and subtracts area of triangle DM1 Limits either way round. 391.5 A1 CAO SC B1 for correct answer if M1 B1 B1 DM0 scored. 5

Mark scheme

5(a)(i) 32 − 80 px + 80 p 2 x 2 B2,1,0 B2 for all correct. B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(a)(ii) 3 2 B2,1,0 OE 1 − 2x + x B2 for all correct. 2 B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(b) 3 *M1 3 terms FT their values. Coefficient of x 2 = 32 × + ( −2 × − 80 p ) + 80 p 2 2 48 + 160 p + 80 p 2 = 93 ⇒ 5 ( 4 p + 9 )( 4 p − 1) [ = 0] DM1 Set their 3-term coefficient to 93 and attempt to solve by factorising or other accepted method for solving their 3-term quadratic. 9 1 A1 SC B1 following M0 if method for solving p=− and p = quadratic is not shown. 4 4 3

Mark scheme

6(a) 2 x 2 − 2 x (11x + 3) + 2 [ = 0] *M1 Substitutes k = 2, p = 11 and eliminates y or x. 2 x2 + 2 Note: = 11x + 3. 2x −20 x 2 − 6 x + 2 [ = 0 ] ⇒ [( 5 x − 1)( 2 x + 1) = 0 ] DM1 Simplifies to a 3-term quadratic. Terms need not all be on one side. 1 5 1 26 A1 A1 A1 for either both x-values correct or for both Coordinates − , − , , coordinates of one point correct. 2 2 5 5 Need not be written as coordinates. Fractions must be simplified. 4 6(b) 2 x 2 − kx ( 4 x + 3) + 2 = 0 ⇒ ( 2 − 4k ) x 2 − 3kx + 2 [= 0] *M1 Substitute and reduce to 3-term quadratic. Terms need not all be on one side. Allow 2x2 – 4kx2 – 3kx + 2 [= 0]. ( 2 − 4k ) x 2 − 3kx + 2 [= 0] A1 Correct quadratic. All terms to one side. Allow 2x2 – 4kx2 – 3kx + 2 [= 0]. b 2 − 4ac = 9k 2 − 4 × ( 2 − 4k ) × 2 DM1 Use of b2 − 4ac. Must be correct for their a, b, c. a term must have two components. 9k 2 + 32k − 16 [< 0] ⇒ ( k + 4 )( 9k − 4 ) [< 0] M1 Attempt to solve a 3-term quadratic in k by factorising or other accepted method for solving their 3-term quadratic. 4 A1 SC B1 following M0 if no method shown for −4 < k < solving quadratic. 9 A0 for correct answer following incorrect quadratic. Must be k. Allow other correct notation. 5

Mark scheme

7(a) dy 18 B1 OE dx = 8 x − x3 Accept unsimplified. dy 55 M1 OE At x = 2, = dx 4 dy dx For evaluating their or . dx dy dy dy dt 55 dt M1 For correct use of chain rule with ±5 and their = × ⇒ = −5 × dx dt dx 4 dx dy (may be algebraic). dx Condone missing brackets. dx 4 A1 4 dt = − 11 Or decreasing at a rate of . 11 AWRT – 0.364. 4 7(b) 18 9 M1 dy 8x − = 0 x 4 = Equating their 2-term to zero. x 3 4 dx 3 6 A1 AWRT ± 1.22. x=± or ± 2 2 y = 4 (for both) A1 A0 A1 if one point correct. AWRT 4.00. d2 y 54 M1 For differentiation. 2 =8+ 4 At least one correct term needed. dx x So both are minima A1 No need for reason. WWW on x-values. 5

Mark scheme

8 Centre of circle is ( 3, −5) B1 Seen or implied. ( y − 1)( y + 11) = 0 x = −2 ⇒ y 2 + 10 y − 11[ = 0 ] M1 3-term quadratic. Terms need not all be on one side. or (y + 5)2 = 36. ( 3, −5) must be correct if it is used. y = 1 and y = −11 A1 No method needed for solving quadratic. −5 − 1 −6 −5 + 11 6 M1 At least one correct. Gradient of line PC is = or gradient of line QC is = 3 − ( −2 ) 5 3 − ( −2 ) 5 5 5 A1 At least one correct. Gradient of tangent at P is or gradient of tangent at Q is − 6 6 5 46 M1 Distance from point of intersection to line Equation of tangent is y − 1 = ( x + 2 ) . Crosses y = −5 at x = − , so 6 36 6 5 x = −2 is 6 × = . 36 5 5 distance = 5 5 8 −5 38 Note y = x + and y = x– . 5 5 6 3 6 3 or equations of two tangents are y − 1 = ( x + 2 ) and y + 11 = − ( x + 2 ) and these 6 6 46 36 meet when x = − so distance = 5 5 1 36 M1 46 Area = × 12 × their Condone use of . 2 5 5 8 432 216 A1 or or 43.2 10 5 8

Mark scheme

9(a) 1 2 B1 r α = 8α ⇒ r = 4 2 1 2 π B1 r sin α = 4 ⇒ α = 2 6 1 2 π M1 Using their r and their α . Area of segment = ×4 × −4 2 6 1 π Condone 4 – × 42 × . 2 6 Allow use of 8 × their α – 4 4 A1 Fraction must be simplified. = π−4 3 4π − 12 Allow . 3 4 9(b) 1 3 *M1 2 1 r 2 + r 2 − 2r 2 cos α = r 2 2cos α = Or α = 2 sin -1 or 2 sin -1 2 2 4 2 2 π Using α = or r = 4 implies 0/4. 6 1 2 DM1 1 2 α = 0.723[0.72273] r sin ( their 0.723) [ = 4] Or α = 41.4 r sin ( their 41.4 ) [ = 4] 2 2 r = 3.48 [3.4777 ] A1 Accept r 2 = 12.1 AWRT. 1 A1 1 π Area of segment = × 3.482 × 0.723 − 4 = 0.371 [0.37068…] Or × 3.482 × × 41.4 − 4 = 0.371 AWRT. 2 2 180 4

Mark scheme

10(a) {Stretch}{factor 3} {parallel to y-axis/in y-direction/vertically} B2,1,0 −2 B2,1,0 Translation −5 B1 Transformations correct and in the correct order. Alternative solution for Question 10(a) −2 B2,1,0 Translation 5 − 3 {Stretch} {factor 3} {parallel to y-axis/in y-direction/vertically} B2,1,0 B1 Transformations correct and in the correct order. 5 The translation parallel to the x-axis can be made anywhere in the sequence. Note: If 3 or more transformations are given then maximum 2/5 for any correct one. 10(b) y B1 For line or curve in correct quadrants only. B1 Must not pass through (0, 0). Fully correct including the line y = x . No label needed. x Approximate reflection of y = f(x). Curve must not come back on itself. 2 10(c) y+5 M1 Allow x/y swap. y =3 x+ 2 −5 = x+2 3 x+5 2 A1 Must be in terms of x. [g −1 ( x )] = −2 Not ‘x = …’. 3 2 10(d) [Range of g −1 is g −1 ( x )] −2 B1 FT x+a 2 Following their g ( x ) = −1 − c where a, b b, c are non-zero. Not x − 2. Not > − 2. Accept other notations, e.g. [−2, ∞]. 1 10(e) 31 B1 AWRT 3.44. [g −1h ( 4 ) = g −1 ( 2 )] = 9 1 10(f) hg −1 is impossible since the range of g −1 is x − 2 is not within the domain of h, B1 Minimum acceptable: ‘The range of g −1 is not which is x 0 within the domain of h’. 1