Mathematics 9709/35 October/November 2025: question paper and mark scheme

Mark scheme

1(a) y B1 Straight lines. Symmetrical, and extending into the second quadrant. 2 and 6 marked correctly on the axes. 6 Ignore y = 3x − 6 below the axis if intention is clear, e.g. dotted line or the required lines are clearly bolder. O 2 x Ignore any attempt to sketch y = 5 x − 3. 1 1(b) Solve the linear equation 6 − 3x = 5x − 3, or solve the quadratic equation M1 Or corresponding inequality. (5x − 3) = (3x − 6) 2 2 Obtain critical value x = 98 A1 Ignore x = − 32 if seen. Obtain final answer x < 98 A1 No other answer. 3

Mark scheme

2 Show a circle with centre 1 + 3i B1 Im(z) If shaded region Some indication of scale seen or implied on both axes 3i shown with no circles drawn, full Show a circle with correct radius 2, and centre not at the origin, soi. B1 marks available if -4 1 4 Re(z) correct with correct Show a circle or semi-circle in the first 2 quadrants, with radius 4 and centre at the B1 descriptions of the origin circles which -4i formed it. Shade the correct region on a correct diagram B1 4

Mark scheme

3(a) Take logarithms of both sides to obtain ln A + ln y = x ln b M1 Or three-term equivalent with no indices. Compare with Y = mX + c , pY + qX + r = 0 A1 AG Condone use of y = mx + c in comparison. 2 3(b) Form equations in A and b and solve for A or b M1 ln A + 0.86 = 3.4ln b and ln A + 2.56 = 5.7 ln b. Or Ae0.86 = b3.4 and Ae2.56 = b5.7 . Obtain b = 2.1 or A = 5.2 A1 Allow for 2 sf or more. Obtain b = 2.1 and A = 5.2 A1 Accept A = 5.3. 3

Mark scheme

4 d 1 dy B1 SOI x 2 ln 2 y = 2 x ln 2 y + x 2 × dx y dx d B1 dx ( ) ddyx × ln ( 2 + x ) + 22+xyx y ln 2 + x 2 = 2 2 Equate derivative of right-hand side to zero and substitute x = 2, y = 3 M1 Must be using implicit differentiation and have attempted differentiation of products where appropriate. Need to see evidence of substitution. 4 dy dy A1 OE Obtain 4 ln 6 + . − ln 6. − 2 = 0 3 dx dx 6 − 12 ln 6 10 A1 OE Simplify to obtain or 4 − Do not ISW. 4 − 3ln 6 4 − 3ln 6 No fractions within fractions. 5

Mark scheme

5 Substitute z = x + iy and z* and expand M1* use i 2 = −1 twice DM1 x 2 + y 2 + 5ix − 5 y + 2 − 10i = 0 Compare real and imaginary parts and equate to zero M1 Must obtain at least one equation. Obtain x + y − 5 y + 2 = 0 and 5(i) x − 10(i) = 0 2 2 A1 OE Obtain z = 2 + 3i and z = 2 + 2i A1 5

Mark scheme

6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. dy ( 5 + 2 sin x ) sec 2 x − tan x × 2 cos x A1 Obtain = ( 5 + 2 sin x ) 2 dx Express derivative in terms of sin x and cos x, and simplify to an expression in DM1 1 sin x sin x only dy ( 5 + 2sin x ) 2 − × 2 cos x = cos x cos x ( 5 + 2sin x ) 2 dx dy 5 + 2 sin 3 x A1 OE without fractions within a fraction. = Denominator must be in factorised form. dx (1 − sin x ) ( 5 + 2 sin x ) 2 2 ISW once correct form is seen. Alternative Method for Question 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. dy cos x ( 5 + 2 sin x ) cos x − sin x ( cos x ( 2 cos x ) + ( 5 + 2 sin x )( − sin x ) ) A1 Obtain = cos 2 x ( 5 + 2 sin x ) 2 dx Simplify to an expression in sin x only DM1 dy 5 + 2 sin 3 x A1 OE without fractions within a fraction. = Denominator must be in factorised form. dx (1 − sin x ) ( 5 + 2 sin x ) 2 2 ISW once correct form is seen. 6(a) Alternative Method 2 for Question 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. dy ( 5 cos x + sin 2 x ) cos x − sin x ( −5sin x + 2 cos 2 x ) A1 = cos 2 x ( 5 + 2 sin x ) 2 dx Simplify to an expression in sin x only DM1 dy 5 + 2 sin 3 x A1 OE without fractions within a fraction. = Denominator must be in factorised form. dx (1 − sin x ) ( 5 + 2 sin x ) 2 2 ISW once correct form is seen. 4 6(b) Set numerator of their derivative equal to zero and attempt to solve for sin x M1 Must be a polynomial in sin x. Show that there are no solutions and hence no stationary points. A1 AG From correct working only using a correct numerator from part (a). Reference to −1 ≤ sin x ≤ 1 OE must be made. Alternative Method for Question 6(b) Use of −1 sin x 1 leading to 3 5 + 2 sin 3 x 7 M1 5 + 2 sin 3 x ≠ 0 A1 OE 2

Mark scheme

7(a) du B1 State or imply = 2x dx Obtain x = 7 u = 4 ( = a ) and x = 12 u = 9 ( = b ) B1 Allow if just written down. du M1 12 Use = 2 x and substitute u for x in the integral 2x2 × 2x dx E.g., using dx. 7 x2 − 3 Condone missing or incorrect limits. A1 AG 2 ( u + 3) du 12 9 4 x3 Allow from working in reverse if fully correct. Obtain dx = from full and correct working with 7 x −3 2 4 u sufficient detail 4 7(b) Integrate to obtain cu + du 3 2 1 2 *M1 May integrate by parts. Must be in terms of u as ‘Hence’. 3 Obtain 43 u + 12u 2 1 2 A1 Use their limits correctly DM1 For example: 4 104 152 ( 27 − 8 ) + 12 ( 3 − 2 ) or 72 − or 88 − . 9 3 3 Obtain 112 3 or 37 13 A1 Exact answer only. Answer only is 0/4. 4

Mark scheme

8(a) Expand sin (θ + 16 π ) and substitute trigonometric values *M1 3 sin θ + 12 cos θ 2 Obtain 52 sin θ + 3 2 3 cos θ A1 Obtain R = 13 B1 From 1 25 + 27. 2 Correct use of trig to obtain α DM1 5 3 3 E.g. cos α = , sin α = 2 13 2 13 3 3 or tan α = . 5 3 3 5 M0 if it comes from sin α = , cos α = 2 2 9 3 3 or sin θ + cos θ . 2 2 α = 0.805 A1 5 8(b) 6 6 B1FT Follow their R (and α ). 2 x + α = sin −1 = sin −1 = 0.747 Accept AWRT 0.747 (0.746898….). R 13 Solve for x to get an answer in the range M1 2 x + 0.805 = 0.747, 2.395, 7.030 Obtain 0.795 or 3.11 A1 Accept AWRT 0.795 or 3.11 from correct work. Obtain 0.795 and 3.11 and no others in range A1 Accept AWRT 0.795 or 3.11 from correct work. 4

Mark scheme

9(a) Calculate the scalar product of the direction vectors to form an equation in a. M1 2 + 6 − 3a = 0 8 A1 Exact answer only, but ISW if go on to give a a= decimal. 3 2 9(b) Express general point of a line in component form B1 2+λ 3 + 2μ 1 + 2λ or −1 + 3μ 4 − 3λ 5 + aμ Equate i and j components of l2 and l1 and solve for λ or μ M1 Obtain λ = −7, μ = −4 A1 SOI Substitute in the equation for the k components and obtain a = −5 A1 4 9(c) Use the scalar product to obtain an expression for the cosine of the angle *M1 2 + 6 − 3a cos θ = 14 13 + a 2 Form a quadratic equation in a DM1 E.g. 14 ( 8 − 3a ) = 5 14 13 + a 2 . Obtain 101a 2 − 672a + 571 = 0 A1 Or three-term equivalent, e.g. 1414a 2 − 9408a + 7994 = 0. 571 A1 Accept 1 and 5.65. Obtain a = 1, a = ISW once correct values seen. 101 SC B1 if a = 1 spotted from 14 ( 8 − 3a ) = 5 14 13 + a 2 OE, 3/4 max. 4

Mark scheme

10(a) *M1 1 1 x x Use integration by parts to obtain pxe + q e dx2 2 A1 1 1 x x Obtain 2 xe − 2 e dx 2 2 Complete the integration to obtain 2 xe − 4e 1 2 x 1 2 x A1 Use limits 0 and a correctly and equate the answer to 6 DM1 ( −0) + 4 = 6 1 1 a a 2ae − 4e 2 2 Obtain a = 2 + e − 12 a from full and correct working A1 AG 5 10(b) Calculate the value of a relevant expression or values of a relevant pair of M1 Example 1 2.2 < 2.33 and 2.4 > 2.30 expressions at 2.2 and 2.4. Values to at least 2 sf. 1 − a Need all relevant values but only one (pair) needs to be correct to score M1. Example 2 f ( a ) = a − 2 − e 2 OE Complete set of values for their expression. If not comparing with 0, then the f ( 2.2) = −0.132... comparison must be clear. f ( 2.4) = 0.0988... 1 1 a a Example 3 f ( a ) = 2ae 2 − 4e 2 + 4 f ( 2.2) = 5.20... f ( 2.4) = 6.65... 1 1 a a Example 4 f ( a ) = 2ae 2 − 4e 2 − 2 f ( 2.2) = −0.80... f ( 2.4) = 0.65... Complete the argument correctly with correct calculated values. Accept truncated A1 Example 1 A clear explanation is needed. values. Allow work on a smaller interval. Example 2 f ( 2.2) = −0.132... < 0 f ( 2.4) = 0.0988... > 0 OE Example 3 f ( 2.2) = 5.20... < 6 f ( 2.4) = 6.65... > 6 Example 4 f ( 2.2) = −0.80... < 0 f ( 2.4) = 0.65... > 0 2 10(c) Use iterative process correctly at least once M1 Obtain final answer 2.31 A1 Show sufficient iterations to at least 4 dp to justify 2.31 to 2 dp A1 Allow recovery. Allow truncation. E.g. 2.3, 2.3166, 2.3140, 2.3144. 3

Mark scheme

11(a) dx B1 AG Obtain = kx (1 − x ) , stating where the factors come from Factor 1 – x explained. dt 1 11(b) Separate variables correctly B1 1 dx = k dt x (1 − x ) Condone missing integral signs or dx and dt but not both. 1 M1 Condone a sign error between the fractions. Correct method to express in partial fractions x (1 − x ) 1 1 A1 Obtain + x 1− x Obtain general solution ln x − ln (1 − x ) = kt + A A1 OE Use t = 0, x = 14 in an expression containing pt , q ln x and q ln (1 − x ) to obtain the *M1 1 A = ln constant of integration 3 Use t = 2, x = 13 in an expression containing pt , q ln x and q ln (1 − x ) to calculate DM1 1 3 k = ln the value of k 2 2 x 1 3 1 A1 Must be seen, as part of question demand. Obtain ln = ln t + ln x 1− x 2 2 3 OE, e.g. ln = (0.2027...)t − 1.0986... 1− x Substitute x = 0.75 and obtain t = 11 years (to the nearest year). A1 10.838… may be seen. 8 SC M1DM1A0A1 for candidates who use t = 1 and t = 3 rather than t = 0 and t = 2 to obtain t = 12 (11.838…) years (to the nearest year).