Cambridge A Level Mathematics 9709 — 2025 May/June Paper 1 · Variant 5

dy 2 1 The equation of a curve is such that = 12 ( 2 x - 5 ) + 8 x . It is given that the curve passes through the dx point (2, 4). Find an equation of the curve. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1  12 3  8 2  B1 B1 OE  y =   ( 2 x − 5 )  + x   + c  Terms may be unsimplified.  3  2  2  16 x 3 −116 x 2 +300 x + c  . May see  y =    4 = 2×(2×2 – 5)3 + 4×22 + c [⇒ c = –10] M1 Sub (2, 4) correctly into their integrated expression with a ‘+ c’. y or f or f ( x ) = 2 ( 2 x − 5 )3 + 4 x 2 − 10 A1 May see y = 16 x 3 − 116 x 2 + 300 x − 260. ' y = or 'f ( x ) = or 'f = ' can be implied if seen in working. 4

More questions on Integration

2 In the expansion of ( 3 + ax) 5 + ( 6 - x) 4 , the coefficient of x2 is six times the coefficient of x. Find the possible values of the constant a. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 x2 coeff = 10  33  a 2 and 6  62 −( 1) 2 allow correct terms with x2 B1 Expect 270 a 2 and 216. Terms may be seen separately. Allow if seen in an expansion. Combinations must be evaluated. x coeff = 5  34  a and 4  6 3 −( 1) allow correct terms with x B1 Expect 405a and − 864. Terms may be seen separately. Allow if seen in an expansion. Combinations must be evaluated. 270 a 2 + 216 = 6 ( 405 a − 864 ) M1 OE For forming a correct quadratic equation using their 4 terms only. 45a 2 − 405a + 900 = 0 ⇒ 45 ( a − 4 )( a − 5 ) = 0 DM1 For evidence of a correct method of solving. If quadratic formula used a full substitution must be seen. a = 4 or a = 5 A1 Only dependent on the first M1. 5

More questions on Series

3 (a) Use completing the square to find the exact solutions of the equation 4x 2 - 4 x - 1 = 0 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) Hence solve the equation 4 tan i = 4 + for 0° 1 i 1 180° . [3] tan i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 2 2 M1 OE  1  2  1  1  0 4  x −   −=2 0  or ( 2 x − 1)  −=2 0 or  x −  − =  Must deal with the coefficient of x2 and x correctly to  2   2  2  2 produce an ( ax + b ) term. 1 1 A1 1 1  2 . x =  OE, e.g. x = ( ) 2 2 2 SC B1 only for correct solutions from another method. 2 3(b) 1 M1 Setting tan= their x for at least one value of their x. tan= 1  2 ( ) May restart and solve the quadratic in tan. 2 [ =] 50.4, 168.3 AWRT and no other answers in the range 0   180 A1 A1 SC A1 only for AWRT 0.879 and 2.94 radians. SC M0 B1 B1 for answers only. SC M0 B1(only) for both answers only in radians. 3

More questions on Quadratics

4 y 2 1 y = x - x 2 O 2 x 2 1 The diagram shows part of the curve y = x - . The shaded region is bounded by the curve, the line x 2 x = 2 and the x-axis. Find the volume formed when the shaded region is rotated through 360° about the x-axis, giving your answer correct to 2 decimal places. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 4 B1 WWW Intersects x-axis when x = 1  x = 1 May be seen without working. May be seen as a limit in the integration. Allow ±1 if x = 1 is seen as the lower limit in the integration. 2 *M1 1 1 1     as With some attempt at squaring (accept x 4  x 4 − 2 + dx with attempt at x 2 − 4   4   2 dx = π  V = π  y 2 dx = π  x x x     evidence). integration 1 5 1 A1 x − 2 x − 5 3 x 3  1 5 1   1 1   DM1 Use of limits, x = 2 and their x = 1, min evidence if  2 −2 2 − − − 2 − π   3    283  5 3  2   5 3   correct answer: + 32. If answer incorrect, 120 15 substitution of limits must be clear. DM0 for use of the limit x = 0. π × 4.49… = 14.11 A1 AWRT, WWW 539 Allow π or 14.1, dependent on the first M1 and 120 A1. 5

More questions on Integration

5 D A 5 cm C 5 cm B The diagram shows a sector ABD of a circle with centre A and radius 10 cm. The perpendicular bisector of AB passes through D. (a) Find the perimeter of the shaded region BCD, giving your answer correct to 1 decimal place. 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(b) Find the area of the shaded region BCD, giving your answer correct to 1 decimal place. 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Mark scheme: 5(a) π B1 OE Angle CAD = or 60o, CD = 75 or 5 3 ˆ or CD. Allow CADˆ = 1.05, CD = 8.7 3 For either CAD or 8.66.  π   60  B1 B1 OE + 75 + 5 + 75 + 5 [Perimeter =] 10     or   2π (10 )    B1 for the two sides and B1 for arc length (allow 10.5).  3   360  = 24.1 cm B1 4 5(b) 1 2 π 1 60 2 1 M1 OE Area =  10  − 5 75 or  π  10 − 5 75 Use of sector area formula minus triangle area 2 3 2 360 2 formulae, with their angle CAD and their side CD. = 30.7 cm2 A1 2

More questions on Circular measure

6 Each year, on her birthday, Ananya receives some money from each of her parents. On Ananya’s first birthday, her father gives her $10. Every subsequent year, her father gives her $5 more than he gave her the previous year. On Ananya’s first birthday, her mother also gives her $10. Every subsequent year, her mother gives her 20% more than she gave her the previous year. (a) Show that on Ananya’s eleventh birthday she receives more from her mother than from her father. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the total amount of money Ananya receives up to and including her eighteenth birthday. 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Mark scheme: 6(a) Eleventh birthday: Father 10+10×5 [= 60] M1 For correct use of AP formula OE. Mother 10×1.210 [= 61.9(2)] M1 For correct use of GP formula OE. 60 and 61.9 A1 Both answers. Accept 2 sf answers. 3 6(b) 18 M1A1 A1 may be implied by a correct final answer. Father ( 2  10 + 17  5 ) = 945 2 18 M1A1 A1 may be implied by a correct final answer. 10 1 − 1.2 ( ) Mother = 1281.17 (1 − 1.2 ) Total = 2226.17 A1 Accept 3 or more sf answers. Ignore S17 ( = 1909.3 ) as an extra answer. 5

More questions on Series

7 In the parallelogram ABCD, the coordinates of A are (3, 7), the coordinates of B are (6, p) and the coordinates of D are (1, p). It is given that the gradient of AB is - 2 . 3 (a) Find the value of p. 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(b) Find the coordinates of C. 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(c) Find the area of the triangle formed by the perpendicular bisector of AB and the x- and y-axes. 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Mark scheme: 7(a) 7 − p 2 2 M1 OE = − or use of straight line equations with m = − , ( x , y ) = ( 3,7 ) 3 − 6 3 3  2  x − 3 ) with (6, p) substituted or Expect y − 7 =  −  ( with (6, p) substituted.  3  −2 y = x + 9 with (6, p) substituted. 3 p = 5 A1 2 7(b) ({4}, {their 2p-7}) B1 WWW B1FT Extra solutions lose both marks. 2 7(c) Mid-point AB is (4.5, 6) B1   B1  − 1  3 Gradient of perp bisector  =  = 2 2  −   3  3 M1 OE. Must be using their mid-point and their Equation y − 6 = ( x − 4.5 ) perpendicular gradient. 2 Crosses axes at (0, − 0.75), (0.5, 0) DM1 Correct use of their perpendicular bisector equation to find the x- and y-intercept. Area = 0.1875 (accept 3 sf accuracy) A1 3 OE, e.g. . Last 2 marks can be gained by integrating 16 the line equation between zero and 0.5. 5

More questions on Coordinate geometry

8 The equation of a curve is y = x 3 + ax 2 + bx + 5 . The curve has a stationary point at (1, 9). (a) Find the values of the constants a and b. 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(b) Find the coordinates of the other stationary point. 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(c) A point P is moving along part of the curve in such a way that the y-coordinate of P is increasing at a constant rate of 6 units per second. Find the rate at which the x-coordinate of P is increasing when x = 5 . 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Mark scheme: 8(a) 9 = 1 + a + b + 5 B1 dy 2 B1 = 3 x + 2 ax + b dx Gradient = 0 at (1, 9) so 0 = 3 + 2a + b M1 d y Setting their to zero and substituting x = 1. d x Attempt to solve their linear equations simultaneously DM1 Can be implied by their answers. a = −6, b = 9 A1 WWW 5 8(b) dy 2 M1 d y = 3 x − 12 x + 9 = 0 Setting their to zero. dx d x Solution [leading to x = 1 or x = 3] DM1 Solving their 3-term quadratic (3, 5) or x = 3, y = 5 A1 WWW Ignore (1, 9 ) if given as a second answer. Only dependent on the first M1. 3 8(c) dy 2 M1 dy At x = 5, = 3  5 − 12 +5 9 Substituting x = 5 into their . May be implied. dx dx dx dt M1 OE 6 = their 24  or their 24 = 6  dt dx  dy dy dx  Use of chain rule SOI  =   .  dt dx dt  d x d t Linking correctly (or ), their 24 and 6. d t d x d x 1 A1 OE = d t 4 3

More questions on Differentiation

9 Functions f and g are defined as follows. f ( x) = cos x for 0 G x G r g ( x) = 3 cos ( x - r) + 2 for r G x G 2r (a) Describe fully the transformations that have been combined to transform the graph of y = f ( x) to the graph of y = g ( x) . 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(b) On the given axes, sketch the graphs of y = f ( x) and y = g ( x) . [4] y x O r 2r rj. [4](c) Find g -1 f `13 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Explain why the composite function fg cannot be formed. 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Mark scheme: 9(a) {Stretch}{factor 3} {in y-direction} B2,1,0 If 2 or more stretches or extra transformations, give B0 for stretches.  π  B2,1,0 π {Translation or shift}   Translation may be split to  before the stretch and   2  0 0  after the stretch. 2 If both vectors correct but ‘translation’/’shift’ not stated, then B1 only for the translations. Alternative Method for Question 9(a)  π   0  B2,1,0 If two or more stretches or any extra incorrect   π   transformation is given, then B0 for the stretches. and {Translation or shift} 2 or    2      If order incorrect, maximum of 3/4. 0     3   3  followed by {stretch} {factor 3} {in y direction} B2,1,0 4 7 y9(b) B1† Graph of f(x) with correct domain. 6 5 4 B1† For g(x) being a decreasing function in the domain π to 3 2π. 2 1 x B1 Domain π to 2π for g(x). π/2 π 3π/2 2π −1 −2 B1 Range for g approximately correct (should be from 5 to −3 −1). 4 †Sketches must be curves and have zero gradient at the ends of the given domains. 9(c)  π   1  B1 May be implied by correct substitution later. [g-1] f   = [g-1 ]    3   2  −1 −1  x − 2  −1  y − 2  M1 Finding inverse of g. [g ( x ) or y = ] cos   + π or  x =  cos   + π Allow one sign error.  3   3  1 Alt: 3cos ( x − π ) + 2 = 2  1  DM1 1 − 2 Substituting their x = into their expression for −1  1 −1  2  2 [g  = ] cos   + π −1  2   3  g ( x ) .   1 2π Alt: Use of arccos for cos ( x − π ) = − ⇒ x − π = 2 3 −1 2π 5π A1 [g ( x ) = ] + π = or 5.24 AWRT 3 3 4 9(d) The domain of f does not include the whole of the range of g B1 OE, but must mention range of g and domain of f. Alternative Method for Question 9(d) Show clearly that a particular value or set of values in the domain of g gives a B1 Value of x for substitution into g ( x ) must be in the value of g ( x ) which is outside the domain of f ranges: π x  4.322 or 5.447  x 2π. 1

More questions on Functions

10 The equation of a circle is x 2 + y 2 + 4x - 8y - 12 = 0 . (a) Find an equation of the tangent to the circle at the point (2, 8), giving your answer in the form ax + by + c = 0 . 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(b) Given that the line x + 3y = k does not intersect the circle, show that k 2 - 20 k - 220 2 0 . 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Mark scheme: 10(a) Centre is (–2, 4) B1 8 − 4 M1 OE, e.g. using equation of the radius. Gradient of perpendicular = [ = 1] 2 −−( 2 ) Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from d y an attempt to find , or the negative reciprocal of d x their perpendicular gradient. x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required. Answer without working, allow maximum of 2/4. Alternative Method for Question 10(a) 1 1 B1 OE; a correct expression for y. y =  28 − x 2 − 4 x 2 + 4 32 − ( x + 2 ) 2 ( ) 2 + 4 or y =  ( ) −1 M1 OE dy 1 2 2 =  28 − x − 4 x −2 x − 4 ) [= −1 at x = 2] Differentiating y with no more than one sign error. ( ) ( dx 2 Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from d y an attempt to find , or the negative reciprocal of d x their perpendicular gradient. x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required. Answer without working, allow maximum of 2/4. Alternative Method 2 for Question 10(a) d y d y B1 Differentiating implicitly. 2 x + 2 y + 4 − 8 = 0 d x d x d y ( 2 x + 4 ) M1 d y = =  −1 at ( 2,8 )  Rearranging to make d x the subject, d x ( 8 − 2 y ) dy May substitute (2, 8) and then rearrange to find . dx Equation of tangent is y – 8 = –1(x – 2) M1 Using a correct point and their tangent gradient from d y an attempt to find , or the negative reciprocal of d x their perpendicular gradient. x + y − 10 = 0 or –x – y + 10 = 0 A1 CAO – this form is required. Answer without working, allow maximum of 2/4. 4 10(b) ( k − 3 y ) 2 + y 2 + 4 ( k − 3 y ) − 8 y − 12 = 0 M1* k − x Sub x = k − 3 y or y = into circle equation. 2 2 3 or ( k − 3 y + 2 ) + ( y − 4 ) = 32 2 k − x ( k − x ) k − x 2 y = gives: − 8 + x + 4 x − 12 = 0 3 9 3 2 2  k − x  or ( x + 2 ) +  − 4  = 32.  3  9 y 2 − 6 ky − 20 y + y 2 + k 2 + 4 k − 12 = 0 DM1 OE All squared brackets expanded to give a quadratic equation in y (or x, which gives: 9 x 2 + 60 x − 2kx + k 2 + x 2 − 24k − 108 = 0 or x 2 + k 2 − 2 kx 8k − 8 x 2 − + 16 + x + 4 x + 4 = 32. ) 9 3 2 2 2 M1** OE k + 4 k − 12 b − 4 ac = ( 6 k + 20 ) −4 10  ( ) Factorising out y (or x) from their quadratic (this is the quadratic where the previous DM1 or possibly DM0 was awarded) to identify ‘b’ and then finding the discriminant. From equation in x, this is: k 2 − 24 k − 108 . ( 60 − 2 k ) 2 − ( 4 )(10 )( ) −4k 2 + 80k + 880  0 DM1 OE Simplify the discriminant and setting it to less than zero. From eliminating y this is: −36k 2 + 720k + 7920  0 Only dependent on the previous M1. k 2 − 20k − 220  0 A1 AG WWW 10(b) Alternative Method for Question 10(b) Centre of circle (–2, 4) and radius of circle is 32 B1 1( −2 ) + 3 ( 4 ) − k M1* Correct use of ‘distance of a line from a point formula’ Distance of x + 3 y − k from ( −2,4 ) = with their centre coordinates. (12 + 32 1( −2 ) + 3 ( 4 ) − k DM1 Setting distance to be greater than their radius. Setting  32 (12 + 32 ) Squaring both sides DM1 Removing the square roots. 2 A1 AG Rearranging to k − 20k − 220  0 5

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