Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 3 · Variant 5

1 (a) Sketch the graph of y = 3x - 6 . [1] (b) Solve the inequality 5x - 3 1 3x - 6 . [3] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) y B1 Straight lines. Symmetrical, and extending into the second quadrant. 2 and 6 marked correctly on the axes. Ignore y = 3 x − 6 below the axis if intention is 6 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 = 5 x − 3, or solve the quadratic equation M1 Or corresponding inequality. ( 5 x − 3) 2 = ( 3x − 6 ) 2 Obtain critical value x = 98 A1 Ignore x = − 32 if seen. Obtain final answer x  98 A1 No other answer. 3

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2 On a sketch of an Argand diagram, shade the region which represents complex numbers z satisfying both the inequalities z - 1 - 3 i G 2 and z H 4. [4]

Mark scheme: Im(z) If shaded region2 Show a circle with centre 1 + 3i B1 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

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3 The variables x and y satisfy the equation Ay = bx , where A and b are constants. (a) Show that the graph of lny against x is a straight line. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... (b) When x = 3 .4 , ln y = 0.86 and when x = 5. 7, ln y = 2.56 . Find the value of A and the value of b. Give your answers correct to 2 significant figures. [3] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

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 = 5.7ln b. Or Ae 0.86 = b 3.4 and Ae 2.56 = b 5.7 . 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

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4 The equation of a curve is x 2 ln 2y - y ln `2 + x 2j = ln 6. Find the exact value of the gradient of the curve at the point (2, 3). Give your answer in simplified form. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 d 2 2 1 d y B1 SOI x ln 2 y = 2 x ln 2 y + x  d x y dx d 2 dy 2 2 xy B1 y ln ( 2 + x ) =  ln ( 2 + x ) + 2 dx dx 2 + x 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 4ln 6 + . − ln 6. − 2 = 0 3 dx dx 6 − 12ln 6 10 A1 OE Simplify to obtain or 4 − Do not ISW. 4 − 3ln 6 4 − 3ln 6 No fractions within fractions. 5

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5 Find the complex numbers, z, which satisfy the equation zz * + 5 iz + 2 - 10 i = 0. Give your answers in the form x + yi , where x and y are real. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 Substitute z = x + iy and z* and expand M1* use 2i = −twice1 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 2 + y 2 − 5 y + 2 = 0 and 5(i) x − 10(i) = 0 A1 OE Obtain z = 2 + 3i and z = 2 + 2i A1 5

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tan x 6 A curve has equation y = . 5 + 2 sin x d y (a) Express as a simplified fraction in terms of sinx. [4] d x ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... (b) Show that the curve has no stationary points. [2] ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. 2 x − tan x  2cos x A1 dy ( 5 + 2sin x ) sec Obtain = 2 dx ( 5 + 2sin x ) Express derivative in terms of sin x and cos ,x and simplify to an expression in DM1 1 sin x ( 5 + 2sin x ) 2 −  2cos x sin x only dy cos x cos x = 2 dx ( 5 + 2sin x ) 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) 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 + 2sin x ) cos x − sin x ( cos x ( 2cos x ) + ( 5 + 2sin x )( − sin x ) ) A1 Obtain = dx cos 2 x ( 5 + 2sin x ) 2 Simplify to an expression in sin x only DM1 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) 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 ( 5cos x + sin 2 x ) cos x − sin x ( −5sin x + 2cos2 x ) A1 = dx cos 2 x ( 5 + 2sin x ) 2 Simplify to an expression in sin x only DM1 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) 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 + 2sin 3 x 7 M1  5 + 2sin 3 x  0 A1 OE 2

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7 (a) Use the substitution u = x 2 - 3 to show that 12 4x 3 b 2 ( u + 3) dx = du , y 2 y 7 x - 3 a u where a and b are values to be found. 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Mark scheme: 7(a) du B1 State or imply = 2 x dx B1 Allow if just written down. Obtain x = 7  u = 4 ( = a ) and x = 12  u = 9 ( = b ) 12 2 du M1 Use = 2 x and substitute u for x in the integral  2 x  2 x E.g., using d x. dx 2  7 x − 3 Condone missing or incorrect limits. 12 3 9 A1 AG  4 x 2 ( u + 3 ) du from full and correct working with Allow from working in reverse if fully correct. Obtain  2 dx =   7 x − 3  4 u sufficient detail 4 7(b) 32 12 *M1 May integrate by parts. Integrate to obtain cu + du Must be in terms of u as ‘Hence’. 4 32 12 A1 Obtain 3 u + 12u Use their limits correctly DM1 For example: 4 104 152 ( 27 − 8 ) + 12 ( 3 − 2 ) or 72 − or 88 − . 9 3 3 Obtain 1123 or 37 13 A1 Exact answer only. Answer only is 0/4. 4

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8 (a) Express 3 3 sin bi + 1 rl - 2 sin i in the form R sin ( i + a) , where R 2 0 and 0 1 a 1 1 r . Give 6 2 the exact value of R and state the value of a correct to 3 decimal places. 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(b) Hence, solve the equation 3 3 sin b2x + 1 rl - 2 sin 2x = 6 for 0 1 x 1 r. 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Mark scheme: 18(a) Expand sin (+ 6 π ) and substitute trigonometric values *M1 2 3 sin+ 12 cos 5 3 3 A1 Obtain 2 sin+ 2 cos B1 1 Obtain R = 13 From 2 25 + 27. Correct use of trig to obtain  DM1 5 3 3 E.g. cos = , sin  = 2 13 2 13 3 3 or tan = . 5 3 3 5 M0 if it comes from sin = , cos= 2 2 9 3 3 or sin + cos. 2 2 = 0.805 A1 5 8(b) −1 6  −1 6  B1FT Follow their R (and ). 2 x +  = sin  = sin = 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

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9 The equations of two lines are given by l : r = ( 2i + j + 4k ) + m ( i + 2j - 3k ), 1 l : r = ( 3i - j + 5k ) + n ( 2i + 3j + ak ). 2 (a) Find the value of a for which l1 is perpendicular to l2. 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(b) Find the value of a for which l1 and l2 intersect. 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(c) Find the values of a for which the acute angle between l1 and l2 is equal to cos -1 b 145 l . 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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 +    3 + 2      1 + 2 or −+1 3          4 − 3  5 + a  Equate i and j components of 2l and 1l and solve for or  M1 Obtain = −7, = −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 = 14 13 + a 2 Form a quadratic equation in a DM1 2 E.g. 14 ( 8 − 3a ) = 5 14 13 + a . Obtain 101a 2 − 672 a + 571 = 0 A1 Or three-term equivalent, e.g. 1414 a 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 2 OE, 3/4 max. 14 ( 8 − 3a ) = 5 14 13 + a 4

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a 1 10 The constant a is such that xe 2 x dx = 6. y 0 - 21 a (a) Show that a = 2 + e . 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(b) Verify by calculation that a lies between 2.2 and 2.4. 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(c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 10(a) 12 x 12 x *M1 e dx Use integration by parts to obtain pxe + q  1 2 x 12 x A1 Obtain 2 xe 2 e dx − 1 2 x 12 x A1 Complete the integration to obtain 2 xe − 4 e 1 Use limits 0 and a correctly and equate the answer to 6 DM1 2ae 2 a − 4e 12 a ( −0 ) + 4 = 6 − 12 a A1 AG Obtain a = 2 + e from full and correct working 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 2 OE Need all relevant values but only one (pair) needs to be correct to score M1. Example 2 f ( a ) = a −−2 e 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 2 − 4e 2 + 4 Example 3 f ( a ) = 2ae f ( 2.2 ) = 5.20... f ( 2.4 ) = 6.65... 1 1 a a 2 − 4e 2 − 2 Example 4 f ( a ) = 2ae 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

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11 A fungal disease is affecting some of the trees in a forest. The fraction of the trees affected after t years is denoted by x. The rate of increase of x is proportional to the product of the fraction of the trees affected and the fraction of the trees not affected. d x (a) Explain why, after t years, = kx ( 1 - x) , where k is a constant. [1] d t ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... (b) When the disease is first detected, one quarter of the trees are affected. Two years later, one third of the trees are affected. Solve the differential equation to find the number of years from the time when the disease is first detected until the time when three quarters of the trees are affected. Give your answer correct to the nearest year. 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Mark scheme: 11(a) d x B1 AG Obtain = kx (1 − x ) , stating where the factors come from Factor 1 – x explained. d t 1 11(b) Separate variables correctly B1 1 k dt dx =   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 1 3 Use t = 2, x = 13 in an expression containing pt , q ln x and q ln (1 − x ) to calculate DM1 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).

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