Cambridge A Level Mathematics 9709 — 2025 Oct/Nov Paper 1 · Variant 5

1 A circle has centre (2, 6) and radius 10. (a) State the equation of the circle. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The circle passes through the point (8, k). Find the two possible values of k. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 2 2 2 B2,1,0 B2 All 3 components correct, = 100 or 10 {( x − 2 ) } + ( y − 6 )    B1 Any 2 components correct. or x 2 − 4 x + y 2 − 12 y −60 = 0 oe    2 1(b) ( 8 − 2 )2 + ( k − 6 ) 2 = 100 or k 2 − 12k − 28 = 0 M1* Substituting the point into their equation to obtain an equation in k. k −=6 8 or ( k − 14 )( k + 2 ) = 0 oe DM1 Simplify to the point where k can be found (allow one sign error). k = 14 , −2 B1 SC B2 only for both answers without working. Allow ‘ y = ’. 3

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2 A geometric progression has first term 3 + 4 2 and second term 5 - 2 . (a) Find the common ratio of the geometric progression. Give your answer in the form 2 + p , where p is an integer to be found. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the sum to infinity of the geometric progression. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 5 − 2 B1 Correct quotient SOI. r = 3 + 4 2 M1 Rationalise their denominator (must see clear 5 − 2 3 − 4 2 ( )( ) = process), or equate their r to 2 + p and solve. 9 − 32 23 − 23 2 B1 = = 2 − 1  p = −1 −23 3 2(b) 3 + 4 2 M1 Uses S  formula. S  = 1 − 2 − 1 FT a = 3 + 4 2 and their r from 2(a), provided ( ) −1 r 1 and r is evaluated. Do not condone missing brackets unless corrected in the calculation.  14 + 11 2   11 2  A1 Allow answers from a correct expression which  =  or  7 +  round to 14.8 to 3sf.  2   2  2

More questions on Series

3 (a) Express 4x 2 + 10x + 6 in the form a ( x + b) 2 + c , where a, b and c are rational constants to be determined. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The curve with equation y = 4x 2 + 10x + 6 and the line y = k have exactly one point of intersection. Using your answer to part (a) or otherwise, state the value of the constant k. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a)   5  2   1  B2,1,0 B2 All 3 components correct, 4  x +   −  B1 Any 2 components correct.  4    4  5 1 Allow a = 4, b = , c = − unless contradicted 4 4 by the algebraic expression. Allow decimals. 2 3(b) 1  1 1  B1 FT Follow through their c from 3(a). k = − or −0.25  allow − or y = −  4  4 4  Don’t allow x = − 1. 4 If solved through restarting answer must be WWW. 1

More questions on Quadratics

4 A B r cm 1 r rad 3 O r cm C The diagram shows the design for a company’s new logo. The sector of the circle, centre O, has radius r cm. The acute angle AOC = 1 r radians. The quadrilateral OABC is a rhombus. 3 (a) Find an expression for the perimeter of the design. Give your answer in terms of r and r. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is now given that the perimeter of the design is 200 cm. Find the area of the design. Give your answer to 3 significant figures. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a)  1   5  M1 Finding the reflex angle AND using arc length 2π − π = πr Arc Length =    r   1   3 3     π formula or 2 π r − r   .  3  5 1 A1 Answer must be exact. Perimeter = πr + 2 r oe e.g. 2πr − πr + 2 r 3 3 2 4(b)  5π  M1 Equating their expression to 200 and attempting to r  + 2  = 200 solve (their expression must be linear).  3  200 A1 CAO r = 27.6 4 allow r = oe 5π + 2 3  1  2  1  M1 Using 2(0.5ab sin C) with a = b = theirr and ( = 661.6 ) Area of Rhombus = 2   ( 27.64 ) sin  π  1  2   3  C = π to get the area of the rhombus (or suitable 3 1 2  π  1 2  1  Or Area of segment = ( 27.64 )   − ( 27.64 ) sin  π  other method), with their value of r which may be 2  3  2  3  substituted later. Expect segment area = 400 – 330.8 = 69.2. Must be an attempt to find and use r.  1  2  5  M1 5 Area of Sector =   ( 27.64 )  π  ( = 2000 ) oe Using sector area formula with their r and π,  2   3  3 Or Total Area = Circle area – Segment area + Triangle area or complete method using the segment. Must be an attempt to find and use r. Total Area = 2000 + 661.6 = 2660 (3sf) A1 CAO WWW AWRT 2660 (3sf). Or Total area = 2400 – 69.2 + 330.8 = 2660 (3sf) 5

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5 Solve the equation 3 27 x - 28 + = 0 . [3] x 3 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 [ x 6 − 28 x 3 + 27 = 0 ] → u 2 − 28u + 27  = 0 M1* Multiply by 3x and recognise as a quadratic in x 3. quadratic using suitable method. E.g.  x 3 − 1 x 3 − 27 = 0  → ( u − 27 )( u − 1) = 0  DM1 Solve )( )  (  factorisation, quadratic formula, completing the square. x = 1, 3 B1 WWW 3

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6 (a) Show that the equation 1 4 6 sin i + = tan i sin i can be written in the form 6 cos 2i - cos i - 2 = 0 . 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(b) Hence, solve the equation 1 4 6 sin i + = tan i sin i for 0° G i G 360° . 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Mark scheme: 6(a) cos 4 2 M1 Using identity for tan and multiplying through 6sin+ =  6sin + cos= 4 sin sin by sin. 6(1 − cos 2 ) + cos = 4 M1 Uses sin 2 + cos 2 = 1. 6cos 2 − cos− 2 = 0 A1 AG Expands out and simplifies to give the answer. 3 6(b) ( 3cos− 2 )( 2cos+ 1) = 0 M1 Suitable method for solving quadratic. 2 1 B1 Obtaining both values of cos. cos= , cos= − 3 2 = 48.2, 120, 240, 311.8 B1 Any 2 values of  correct. B1 All 4 values of  correct and no other values in the given range. SC B1 All four answers in radians: 0.841, 2π/3, 4π/3, 5.44. 4

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7 r cm h cm A manufacturer wishes to design an open cylindrical tank, as shown in the diagram. The tank will have a base but no top. The outside of the tank will have a fixed surface area of 600r cm2. The radius r cm and height h cm of the tank can vary. (a) Show that the volume, V cm3, of the tank is given by 2 rr ( 600 - r ) V = . 2 [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact value of r which corresponds to the maximum value of V. 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(c) Hence, find the maximum value of V. 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Mark scheme: 7(a) 2  600π − πr 2 600 − r 2  M1* Uses given info and correct formulae to form 2πrh + πr = 600π oe  h = =  equation in h and r.  2πr 2 r  600 − r 2 DM1 Sub their expression for h into V = πr 2 h. ( ) 2 V =  πr oe 2r 2 A1 AG πr 600 − r ( ) = CAO 2 WWW 3 7(b) 1 3 dV 1 2  3πr 2  M1* Differentiate given expression for V. V = π 600r − r = π 600 − 3r or  300π −  1 2 ( ) ( ) 2 dr 2  2  Condone missing π (must be of form a − br ). 2 Alternative: h = r at max V, so 2πr ( r ) + πr 2 = 600π oe 600 − 3r 2 = 0 DM1 Equate their derivative to zero and attempt to solve as far as an equation of the form ‘ r = ’. Alternative: Solving 2πr ( r ) + πr 2 = 600π r = 10 2 A1 CAO (accept 200 ). 3 7(c) π  3  M1 Sub their value of r from 7(b) into the given 10 2 − 10 2 V =  600 ( ) ( )  expression for V, providing their value of r > 0 2   and gives V  0. 8890 ( 3sf ) Accept exact answer 2000 π 2 A1 AWRT 8890 (3sf). 2

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8 y O a 2a x The diagram shows the curve with equation y = 6x + 5 . The shaded region is bounded by the curve, the x-axis and the lines x = a and x = 2a , where a is a positive constant. The shaded region is rotated through 360° about the x-axis to form a solid. The solid has volume, V, such that V H 46 r . (a) Show that 9a 2 + 5a - 46 H 0 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the range of possible values of a. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) 2 a B1 CAO SOI V = π  a ( 6 x + 5 ) d x  Must include π and the correct limits. 2 a 2 M1* Correct integral of 6 x + 5 (can be awarded if π is = π 3 x + 5 x  a missing and the limits are missing or incorrect).  2 a  6 x + 5 ) 2  or = (π    12  a = π 12a 2 + 10a − 3a 2 − 5a 46π  DM1 Correct substitution of correct limits. ( )  9 a 2 + 5 a − 46  0 A1 AG Can only be awarded if the argument leading to the statement is complete and clear. 4 8(b) Solve inequality (or equation) M1 Solve quadratic by suitable method. B1 Obtain − 23, 2 Condone absence of − 23. 9 9 Condone use of x. Final answer a  2 B1 WWW Don’t allow x 2. 3

More questions on Integration

9 In the expansion of ( p + qx) 4 , the coefficient of x is equal to the coefficient of x 2. The constants p and q are both positive. (a) Find the ratio p : q. Give your answer in its simplest form. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is given that the coefficient of x 3 is 486. Find the values of p and q. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 9(a) 4  p 3  q or 6  p 2  q 2 M1 Attempt to find either term in x or 2x – must use binomial coefficients. May see x terms. p 3 DM1 Equate coefficients and simplify to an equation = oe equivalent to 2 p = 3q (no x’s should be present). q 2 p : q = 3:2 A1 Must be in this form only. 3 9(b) 4 p  q 3 = 486 B1 Set coefficient of the term in 3x equal to 486. May include 3x on both sides of the equation. 3q 3 M1 Substitute their 9(a) equation, providing this is of 4   q = 486 the form p = aq where a is positive and rational. 2 4 9 A1 Solve for q (or p). q = 81  q = 3, p = 2 A1 Solve for p and q with no negative answer. SC B1 for answers without working. 4

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10 The function f is defined by 7 f ( x) = 3 + x - 2 for x 2 2 . (a) It is given that f ( a) = 4 . Find the value of a. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find an expression for f – 1 ( x) and state the domain of f – 1 . 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(c) The function g is defined by 1 + 4x g ( x) = 2x - 3 for x 2 3 . 2 Show that fg ( x) / kx , where k is a constant to be determined. 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Mark scheme: 10(a) 7 M1 Forms an equation in a. 4 = 3 + a − 2 a = 9 A1 Condone x = 9. 2 10(b) Swap x and y and change the subject. This can be viewed as four stages: M1 Any two stages. 7 7 7 7 e.g. x = 3 + , x − 3 = , y − 2 = , y = 2 + M1 The other two stages. y − 2 y − 2 x − 3 x − 3 Condone one sign error present in the complete 7 2 x + 1 or x = 3 + , x ( y − 2 ) = 3 y − 6 + 7, xy − 3 y = 2 x + 1, y = method. y − 2 x − 3 One algebraic slip would result in only one M1. −1 7 A1  2 x + 1  f ( x ) = 2 + CAO  Could be written as  x − 3  x − 3  Need to see 'f −1 ( x ) = ' or 'f −1 : x → ', not just' y = '. Domain: x  3 B1 Allow equivalent statement but must involve x. 4 10(c) 7 B1 Substitute g ( x ) into f ( x ) . fg ( x )  3 +   1 + 4 x − 2  2 x − 3  −1 1 + 4 x 7 Alternative 1: g ( x ) = f ( kx ) → = 2 + 2 x − 3 kx − 3 14 x − 21 7 ( 2 x − 3 ) M1 Simplify to expression with no embedded 3 + or 3 + oe fractions (working may be minimal). 1 + 4 x − 4 x + 6 1 + 4 x − 2 ( 2 x − 3 ) Alternative 1: Clearly amending fractions to a linear equation to solve 7 Alternative 2: Solve 3 + = kx to a linear equation   1 + 4 x − 2  2 x − 3  14 x − 21 A1 Cannot just state k = 2. Must have supporting 3 +  2 x or k = 2 algebra. 7 Trying several values to establish k = 2 scores M0A0. 3

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11 y P O x The diagram shows the curve with equation y = 4x 2 - x 3 and the tangent to the curve at the point P. The point P has x-coordinate 3. (a) Find the equation of the tangent to the curve at the point P. Give your answer in the form y = mx + c . 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(b) The shaded region is bounded by the curve, the x-axis and the tangent to the curve at P. Find the exact area of the shaded region. 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The graph of y = 4x 2 - x 3 is transformed by a stretch of scale factor 1 in the x-direction. The point Q is 3 the image of P under this transformation. The transformed shaded region is bounded by the transformed curve, the x-axis and the tangent to the transformed curve at Q. (c) (i) Find the equation of the transformed curve in the form y = mx 2 + nx 3 , where m and n are integers to be found. [1] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (c) (ii) State the coordinates of Q and the area of the transformed shaded region. [2] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................

Mark scheme: 11(a) dy 2 B1 CAO = 8 x − 3 x dx dy B1 = 24 − 27 = −3  when x = 3 dx y = 9 [when x = 3] B1 SOI y − 9 = −3 ( x − 3 ) or y = −3 x + c → 9 = −+9 c →=c 18 oe M1 d y Uses their y and their numerical to find d x equation of the tangent; condone one sign error. y = −3 x + 18 A1 5 11(b) 4 M1* Must obtain ax 3 + bx 4 and indicate the limits 3 2 3 Area between curve and x-axis = 4 x − x dx  and attempt to integrate )   ( and 4. 3 4 A1 SC B1 for use of wrong or no limits (only for 3 4  4 x x  =  −  correct integral).  3 4  3  256   81  DM1 Correct sub of correct limits (allow one slip).  − 64  −  36 −  64  3   4  Minimum acceptable: − 63. 3 4 67 A1 SOI = May be implied by a correct final answer if the 12 two areas are combined. SC B1 if substitution of the limits is not seen. 9 67 DM1 27 Shaded region = ( 6 − 3)  −their Expect −‘their integral’, but must be ‘area 2 12 2 6 3 under their line’ minus ‘their area under the  −3 x  67 or  + 18 x  – their curve’, where ‘their integral’ is an attempt at the  2  3 12 area under the curve between x = 3 and x = 4. May use the lengths from their tangent equation. 95 A1 Calculating area of triangle – (correct) area under = any equivalent exact answer the curve. 12 11(b) Alternative Method for Question 11(b): Finds area between curve and tangent between x = 3 and x = 4 M1* Integrate at least two of the four terms correctly. 4 Area under the line could be found from the 4 x 2 − x 3 3 x + 18 ) − ( ) dx ( 4 − 3  ( − 3 trapezium area )( 9 + 6 ) . 2 4 2 3 4 A1 Integrating all four terms correctly.  3 x 4 x x  =  − + 18 x − +   4 x 3 x 4   2 3 4  3 SC B1 for the correct integral − +  .  3 4   256   27 81  DM1 Correct sub of limits (allow one slip). =  −24 + 72 − + 64  −− + 54 − 36 +   3   2 4  23 A1 SOI = SC B1 if substitution of the limits is not seen. 12 1 23 DM1 Calculating area of triangle between x = 4 and Shaded region = ( 6 − 4 ) 6 + 2 12 x = 6 + their area, providing limits of 3 and 4 are used to find the area between the curve and the tangent. 95 A1 Must be exact. = 12 6 11(c)(i) y = 36 x 2 − 27 x 3 or state m = 36, n = −27 B1 CAO (must be expanded) 1 11(c)(ii) Q(1, 9) B1 CAO coordinates of Q. 95 B1 FT 1 Area = of their area from 11(b). 36 3 Allow 2.64. 2

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