Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 1 · Variant 1

1 Find the set of values of the constant k for which the quadratic equation 3kx 2 + ( k + 8) x + 3 = 0 has two distinct real roots. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Use of b2 – 4ac *M1 Obtain k 2 − 20 k + 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

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2 A geometric progression has first term a and common ratio cosi, where 0 1 i 1 1 r . It is given that 2 the second term is 8 and the fifth term is 1. 8 (a) Find the value of i. Give your answer correct to 3 significant figures. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact value of the sum to infinity. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 4 1 B1 Condone poor notation. State or imply that second and fifth terms are a cos and a cos  or r3 = 64 Attempt solution of equation of form cos 3 = k as far as cos θ =… , where 0  k  1 M1 Must come from division. Obtain cos= 0.25, and hence = 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 1283 A1 Or exact equivalent. 2

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3 In the expansion of 5 3 p 4 ( px + 3) - b x + l , x the coefficient of x4 is 216. Find the value of the positive constant p. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

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. 2 4 M1 6  3 p  Identify term involving x p 2 as relevant term from  x +  x  x  Obtain 6p 2 x 4 or −p6 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

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4 (a) Express 1 - 6x - x 2 in the form a - ( x + b ) 2 , where a and b are constants. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The graph of y = x2 is transformed to the graph of y = 1 - 6x - x 2 by a reflection followed by a m translation of e o. Give details of the reflection and determine the values of m and n. [3] n ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

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

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2 4 1 - 2 cos i 5 (a) Show that tan i - 1 / . [3] cos 4 i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence solve the equation cos 2 i (tan 4 i- 1) = 7 for 0° 1 i 1 180° . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 4 sin 4  4  sin 2   sin 2  B1 Candidates can work from LHS to RHS or RHS Use tan  = 4 or tan θ – 1 =  2 +1  +  2 − 1  to LHS for full marks. If working on both sides cos   cos    cos   simultaneously, maximum of B1 M1 only if a common correct expression is reached. Attempt to express tan 4 −as1 a single fraction in terms of cos only M1 (1 − c 2 ) 2 − c 4 (s 2 + c 2 )(s 2 − c 2 ) Using or etc. c 4 c 4 1 − 2cos 2  A1 AG – necessary detail needed. Confirm result 4 WWW cos  Alternative Method for Question 5(a) 1 − 2cos 2  cos 4  B1 + – 1 cos 4  cos 4  2 2 M1 1 − cos  ( ) – 1 cos 4  4 A1 AG – necessary detail needed. Confirm result tan θ – 1 Alternative Method 2 for Question 5(a) tan 4 = sec 4 – 2 sec 2 + 1 B1 4 1 2 M1 tan −=1 – cos 4 cos 2 2 A1 AG – necessary detail needed. 1 − 2cos  Confirm result 4 cos  3 5(b) Use identity to form equation using cos only *M1 1 − 2cos 2  Expect = 7 OE cos 2  Allow unsimplified. Obtain 9cos 2 = 1 or similarly simplified equivalent A1 Condone cos 2 (1 – 9 cos 2) = 0 Could be implied by one correct solution Solve equation of form cos 2 θ = 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 2(1 – 9 cos 2) = 0 is used. 4

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6 Functions f and g are defined by f ( x) = ( x + 3 ) 2 - 12 for x H 0 , g ( )x = 2x - 5 for x ! R . (a) State the range of f. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find an expression for f -1 ( )x . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Solve the equation gf ( )x = 69 . [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) Attempt solution of g( x ) = 69 or evaluation of g − 1 (69) *M1 Incorrect order of f −1 and then g− 1 scores *M0. Obtain 37 A1 Attempt solution of f( x ) = their 37 or evaluation of f −1 ( their 37) DM1 Obtain x = 4 only A1 4

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7 A B r cm 2 r rad r cm 3 O The diagram shows a sector of a circle with centre O and radius r cm. The shaded region is bounded by the chord AB and the arc AB. The size of angle AOB is 2 r radians. 3 (a) Show that the area of the shaded region is approximately 0.614r 2 cm2. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ It is given that the radius of the circle is increasing at a rate of .04 cm s -1 . (b) (i) Find the rate of increase of the area of the shaded region at the instant when r = 20 . Give your answer correct to 2 significant figures. [3] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Find the rate of increase of the length of the arc AB. Give your answer correct to 2 significant figures. 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Mark scheme: 1 7(a) Obtain correct 2 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)  d A  B1 Or greater accuracy Obtain = 1.228r    d r  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 = . d 3 Differentiate and apply correct use of chain rule M1 Obtain 0.84 A1 AWRT 3

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8 y 3 P 2 O x 9 The diagram shows the curve with equation y = 1 x and the point P with coordinates b,9 3 l. The 2 2 shaded region is bounded by the curve and the lines x = 0 and y = 3 . 2 (a) Find the area of the shaded region. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The shaded region is rotated through 360° about the y-axis. Find the exact volume of the solid produced. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) 1 32 3 B1 OE Integrate to obtain expression 2 x  2 3 M1 Evaluate integral of the form k x 2 between limits 0 and 9 and subtract result from 272 1 32 9 A1 Obtain 3 x giving area under curve is 9 and shaded area is 2 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 4 3 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 4 [dy] A1 Integrate to obtain ky 5 and evaluate between limits 0 and 32 M1 Obtain 16 5 y 5 and hence 24310 π or exact equivalent A1 Condone omission of π except for last mark. 4

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9 An arithmetic progression has first term 2 and common difference d. The sum of the first n terms is denoted by Sn. (a) It is given that ( S - 1 ) , are the first three terms of a second arithmetic progression. 2 S4, S9 Find the value of d. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the difference between the values of the 15th terms of the two arithmetic progressions. 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Mark scheme: 9(a) Attempt to express each of S 2 − 1, S 4 , S9 in terms of d (with at least one correct) *M1 Allow unsimplified. Obtain S 2 −=1 3 + d , S 4 = 8 + 6 d , 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

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10 A circle has equation x 2 + y 2 + 4y - 21 = 0 and a straight line has equation 2x + y - 8 = 0 . The line intersects the circle at two points. (a) Find the coordinates of these two points of intersection. 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(b) The circle has centre C and the two points of intersection are denoted by A and B. Find the area of the triangle ABC. 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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. 1 1 A1 Obtain 2 20  20 or 2 5 4 (or equivalent) and hence 10 3

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11 A curve passes through the point P (4, 3) and is such that dy 8 10 = - . dx x 2 ( 2 x - 3 ) 2 (a) Find the equation of the normal to the curve at P. Give your answer in the form y = mx + c . 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(b) Find the rate of change of the gradient of the curve when x = 4 . 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(c) Given that the curve also passes through the point (-1, q), find the value of q. 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Mark scheme: 11(a) Substitute 4 to obtain gradient of curve is 1 B1 10 Attempt equation of normal using (4, 3) and −m1 for their gradient M1 y = −10 x + 43 A1 3 11(b) −16 x −3 +40(2 x − 3) −3 B1 B1 OE    Substitute 4 to obtain 1007 B1 3 −8 x +5(2 x − 3) + c 11(c)  y =  −1 −1 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

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