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

2 1 Solve the equation 6 sin i = 1 + for - 180° 1 i 1 180° . [4] sin i .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 6sin 2 − sin− 2  = 0  ( 2sin+ 1)( 3sin− 2 ) = 0 M1 For expressing as a 3-term quadratic. Terms need not all be on the same side. 1 2 A1 For both. sin = − or sin= Allow AWRT 0.667. 2 3 = −150−, 30, 41.8, 138.2 A1 A1 AWRT A1 for any correct angle from a correct value of sin, A1 for all 4 and no others in the interval −180  180. − 5π − π SC B1 for , , 0.730c, 2.41c if use of 6 6 radians is clear. 4

More questions on Trigonometry

1 dy 3 22 The equation of a curve is such that = 4 ( 2x - 5) - 9x . The curve passes through the point dx A b,4 - 11 l. 2 (a) Find the gradient of the normal to the curve at the point A. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the equation of the curve. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 1 M1 dy 3 2 [Gradient of tangent] = 4 ( 2 −4 5 ) −9 4  = 90  Substitute x = 4 into . dx −11 3 1 2 is M0 unless they = 4 ( 2 −4 5 ) −9 4 2 1 reach − . 90 1 A1 AWRT −0.0111. [Gradient of normal] = − 90 2 2(b)  1 4   32  B1 B1 Accept unsimplified. y =  ( 2 x − 5 )  −6 x   + c   2   3 M1 11 11 1 Sub x = 4, y = − into an integrated expression − = 2 + c  ( 2  4 − 5 ) 4 −6 4 2 2 2 and attempt to find c. 3 A1 Condone c = 2 as final answer if ‘y = …’ seen 1 4 2 y = ( 2 x − 5 ) − 6 x + 2 previously. 2 Fractions must be simplified. Accept f(x) in place of y. 4

More questions on Integration

3 The third term of a geometric progression is 18 and the sum of the first three terms is 26. It is given that the common ratio is negative. (a) Find the tenth term of the progression. Give your answer correct to 3 significant figures. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact value of the sum to infinity of the progression. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) ar 2 = 18 , a + ar + ar 2 = 26 or a + ar + 18 = 26 B1 For first and second, or first and third equations. 18 18 2 M1 For expressing as a 3-term quadratic from two 2 + + 18 = 26 ⇒ 8r − 18r − 18  = 0 ( 4r + 3)( r − 3) = 0  expressions with sign errors only. Terms need r r not all be on one side. 3 A1 CAO r = − [or r = 3 ] Allow – 0.75. 4 a = 32 A1 CAO Ignore a = 2. May not find a, but instead use expression for a  18  in terms of r  e.g. a = 2  and then find a in  r  part (b). Tenth term = −2.40 A1 AWRT. Ignore other values. Alternative Method for first three marks in Question 3(a) 3 B1 a 1 − r 2 ( ) ar = 18 , 26 = 1 − r 18 3 M1 Expressing as a 3-term cubic from two 1 − r ( ) 3 2 expressions with sign errors only. Terms need r 2 26 = ⇒ 8 r − 26 r + 18  = 0  not all be on one side 1 − r 3 A1 CAO r = − [or r = 3, 1 ] Accept −0.75. 4 5 Note: SC B1 B1 B1 is possible following B1 3 M0 if no method is shown for finding r = − 4 [or r = 3 ]. 3(b) 32 M1 FT on values of a and r, provided −1 r 1. S=  3  1 −−   4  128 A1 FT FT on values of a and r, provided −1 r  0. 7 Condone extra answers. 2

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4 y x O 3 The diagram shows the curve with equation y = 5x 2 - 20x and the line with equation y = x - 16 . The x-coordinates of the points of intersection of the curve and line are 1 and 16. Find the area of the shaded region between the curve and the line. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4  3  3  *M1 Attempt to integrate both terms and subtract 2 − 20 x  = −5 x 2 + 21x − 16  areas. Accept subtraction either way round. Attempt to integrate ( x − 16 ) −  5 x       1 2   52 2  52 21 2  B1 B1 B1 for each integral.5 2 x − 10 x = −2 x + x − 16 x     x − 16 x  −   21   2  2  x 2 − 16 x.  B2,1, 0 for −2 x 2 +       2 Use of limits 1 and 16 DM1 Limits either way round. Minimum –7.5 – 384. 1 E.g. (( – 16) – (2 – 10)) – ((128 – 256) – 2 (2048 – 2560)) = (15.5 + 8) – (–128 + 512) = –7.5 – 384 = –391.5 Or (–2 + 10.5 – 16) – (–2048 + 2688 – 256) = –7.5 – 384 = –391.5 Or 225 –504 + = –391.5 2 783 A1 CAO 391.5 or SC B1 for correct answer if M1 B1 B1 DM0 2 scored. Alternative Method for Question 4 Height of triangle = 15 B1 1 M1 Area of triangle =  15  their height = 112.5 2  3   5  B1 Integrates   5 x 2 − 20 x  to   2 x 2 − 10 x 2          Use of limits 1 and 16 on their integral and subtracts area of triangle DM1 Limits either way round. 391.5 A1 CAO SC B1 for correct answer if M1 B1 B1 DM0 scored. 5

More questions on Integration

5 (a) Find the first three terms, in ascending powers of x, in the expansion of each of the following expressions. (i) ( 2 - px) 5 [2] .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... 4 (ii) b1 - 1 xl [2] 2 .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... 4 (b) Given that the coefficient of x2 in the expansion of ( 2 - px) 5 b1 - 1 xl is 93, find the possible 2 values of the constant p. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a)(i) 32 − 80 px + 80 p 2 x 2 B2,1,0 B2 for all correct. B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(a)(ii) 3 2 B2,1,0 OE 1 − 2 x + x B2 for all correct. 2 B1 for any two correct. May be in a list. Ignore terms with higher powers. 2 5(b) 3 2 *M1 3 terms FT their values. Coefficient of 2x = 32  + ( −−2 80 p ) + 80 p 2 48 + 160 p + 80 p 2 = 93 ⇒ 5 ( 4 p + 9 )( 4 p − 1) = 0  DM1 Set their 3-term coefficient to 93 and attempt to solve by factorising or other accepted method for solving their 3-term quadratic. 9 1 A1 SC B1 following M0 if method for solving p = − and p = quadratic is not shown. 4 4 3

More questions on Quadratics

6 The equation of a curve is 2x 2 - kxy + 2 = 0 and the equation of a line is y = px + 3 , where k and p are constants. (a) Given that k = 2 and p = 11, find the coordinates of the points of intersection of the curve and the line. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given instead that p = 4 , find the set of values of k for which the curve and the line do not intersect. 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Mark scheme: 6(a) 2 x 2 − 2 x (11x + 3 ) + 2 [ = 0] *M1 Substitutes k = 2, p = 11 and eliminates y or x. 2 x 2 + 2 Note: = 11x + 3. 2 x −20 x 2 − 6 x + 2  = 0  ⇒ [( 5 x − 1)( 2 x + 1) = 0 ] DM1 Simplifies to a 3-term quadratic. Terms need not all be on one side.  1 5   1 26  A1 A1 A1 for either both x-values correct or for both Coordinates  − , −  ,  ,  coordinates of one point correct.  2 2   5 5  Need not be written as coordinates. Fractions must be simplified. 4 6(b) 2 x 2 − kx ( 4 x + 3 ) + 2 = 0 ⇒ ( 2 − 4 k ) x 2 − 3kx + 2 [= 0] *M1 Substitute and reduce to 3-term quadratic. Terms need not all be on one side. Allow 2x2 – 4kx2 – 3kx + 2 [= 0]. ( 2 − 4 k ) x 2 − 3kx + 2 [= 0] A1 Correct quadratic. All terms to one side. Allow 2x2 – 4kx2 – 3kx + 2 [= 0]. b 2 − 4 ac = 9 k 2 − 4  ( 2 − 4 k )  2 DM1 Use of b 2 − 4ac. Must be correct for their a, b, c. a term must have two components. 9 k 2 + 32 k − 16 [  0] ⇒ ( k + 4 )( 9 k − 4 ) [  0] M1 Attempt to solve a 3-term quadratic in k by factorising or other accepted method for solving their 3-term quadratic. 4 A1 SC B1 following M0 if no method shown for −4 k  solving quadratic. 9 A0 for correct answer following incorrect quadratic. Must be k. Allow other correct notation. 5

More questions on Coordinate geometry

2 9 7 The equation of a curve is y = 4 x + - 8 . x 2 (a) A point P is moving along the curve in such a way that its y-coordinate is decreasing at 5 units per second. Find the rate at which the x-coordinate of point P is changing when x = 2 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the coordinates of the stationary points of the curve and determine their nature. 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Mark scheme: 7(a)  dy  18 B1 OE = 8 x −   3 Accept unsimplified.  dx  x d y 55 M1 OE At x = 2, = d x 4 d y dx For evaluating their or . d x dy dy dy dt 55 d M1 For correct use of chain rule with ±5 and their =  ⇒ = −t5 dx dt dx 4 d x d y (may be algebraic). d x Condone missing brackets.  d x  4 A1 4 = − Or decreasing at a rate of .    d t  11 11 AWRT – 0.364. 4 7(b) 18  4 9  M1 d y 8 x − = 0  x = Equating their 2-term to zero. 3   x  4  d x 3 6 A1 AWRT 1.22. x =  or ± 2 2 y = 4 (for both) A1 A0 A1 if one point correct. AWRT 4.00.  d 2 y  2 = 8 + 544 M1 ForAt leastdifferentiation.one correct term needed.  d x  x So both are minima A1 No need for reason. WWW on x-values. 5

More questions on Differentiation

8 The circle with equation x 2 + y 2 - 6x + 10y - 27 = 0 intersects the line x =-2 at the points P and Q. Find the area of the triangle formed by the tangents to the circle at P and Q, and the line x =-2 . 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Mark scheme: B1 Seen or implied.8 Centre of circle is ( 3, −5 ) x = −2 ⇒ y 2 + 10 y − 11 = 0   ( y − 1)( y + 11) = 0  M1 3-term quadratic. Terms need not all be on one side. or (y + 5)2 = 36. ( 3, −5 ) must be correct if it is used. y = 1 and y = − 11 A1 No method needed for solving quadratic. −−5 1  −6  −+5 11  6  M1 At least one correct. Gradient of line PC is = or gradient of line QC is =     3 −−( 2 )  5  3 −−( 2 )  5  5 5 A1 At least one correct. Gradient of tangent at P is or gradient of tangent at Q is − 6 6 5 46 M1 Distance from point of intersection to line Equation of tangent is y −=1 ( x + 2 ) . Crosses y = −5 at x = − , so 6 5 6  36  x = −2 is 6  = .   36 5  5  distance = 5 5 8 Note y = x + and y = −x5 – 38. 5 5 6 3 6 3 or equations of two tangents are y −=1 ( x + 2 ) and y + 11 = − ( x + 2 ) and these 6 6 46 36 meet when x = − so distance = 5 5 1 36 M1 Area =  12 their Condone use of 46. 2 5 5 8 432 216 A1 or or 43.2 10 5 8

More questions on Coordinate geometry

9 C r cm a rad A B r cm The diagram shows a sector ABC of a circle with centre A and radius r cm. The angle BAC is a radians, where 0 1 a 1 1 r . 2 (a) It is given that the area of the triangle ABC is 4 cm 2 and the area of the sector ABC is 8 a cm 2 . Find the exact area of the shaded segment. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) It is given instead that the length of the chord BC is r cm but the area of the triangle ABC is still 2 4 cm 2. Find the area of the shaded segment. Give your answer correct to 3 significant figures. 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Mark scheme: 9(a) 1 2 B1 r = 8 ⇒ r = 4 2 1 2 π B1 r sin= 4 ⇒ = 2 6 1 2 π M1 Using their r and their . Area of segment =  4  − 4 2 6 1 2 π Condone 4 –  4  . 2 6 Allow use of 8 × their  – 4 4 A1 Fraction must be simplified. = π − 4 3 4π − 12. Allow 3 4 9(b) 1 3 *M1 -1  2  -1  1  r 2 + r 2 − 2 r 2 cos= r 2  2cos= Or  = 2 sin   or 2 sin   2 2    4   2 2  π Using  = or r = 4 implies 0/4. 6 1 2 DM1 1 2 = 0.723  0.72273  r sin ( their 0.723 ) = 4  Or = 41.4  r sin ( their 41.4 ) = 4  2 2 r = 3.48  3.4777  A1 Accept 2r = 12.1 AWRT. 1 2 A1 1 2 π Area of segment =  3.48  0.723 − 4 = 0.371 [0.37068…] Or  3.48   41.4 − 4 = 0.371 AWRT. 2 2 180 4

More questions on Circular measure

10 The functions f and g are defined by f ( )x = x for x H 0 , g ( )x = 3 x + 2 - 5 for x H -2 . (a) Describe fully a sequence of transformations which transforms the graph of y = f ( x) to the graph of y = g ( x) . You should make clear the order in which the transformations are applied. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ y x O The diagram shows the graph of y = g ( x) . (b) On the diagram sketch the graph of y = g -1 ( x) together with any relevant mirror line. [2] (c) Find an expression for g -1 ( )x . 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(d) State the range of g -1 . [1] ............................................................................................................................................................ ............................................................................................................................................................ The function h is defined by h ( )x = x - 2 for x H 0 . (e) Find the value of g -1 h ( 4 ) . 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(f) Explain why the composite function hg -1 cannot be formed. 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Mark scheme: 10(a) {Stretch}{factor 3} {parallel to y-axis/in y-direction/vertically} B2,1,0  −2  B2,1,0 Translation    −5  B1 Transformations correct and in the correct order. Alternative solution for Question 10(a)  − 2  B2,1,0   Translation 5   −   3  {Stretch} {factor 3} {parallel to y-axis/in y-direction/vertically} B2,1,0 B1 Transformations correct and in the correct order. 5 The translation parallel to the x-axis can be made anywhere in the sequence. Note: If 3 or more transformations are given then maximum 2/5 for any correct one. 10(b) y B1 For line or curve in correct quadrants only. B1 Must not pass through (0, 0). Fully correct including the line y = x . No label needed. Approximate reflection of y = f(x). Curve must not come back on itself. 2 10(c) y + 5 M1 Allow x/y swap. y = 3 x + 2 − 5  = x + 2 3 2 A1 Must be in terms of x. −1  x + 5  [g ( x )] =   − 2 Not ‘x = …’.  3  2 10(d) [Range of g− 1 is g − 1 ( x ) ] − 2 B1 FT 1  x + a  2 where a, Following their g− ( x ) =   − c  b  b, c are non-zero. Not x −2. Not −2. Accept other notations, e.g. [ −2,  ]. 1 10(e) − 1 −1 31 B1 AWRT 3.44. [g h ( 4 ) = g ( 2 )] = 9 1 10(f) hg−1 is impossible since the range of g − 1 is x − 2 is not within the domain of h, B1 Minimum acceptable: ‘The range of g−is1 not which is x 0 within the domain of h’. 1

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