Cambridge A Level Mathematics 9709 — 2025 May/June Paper 3 · Variant 3

1 (a) Sketch the graph of y = 3 x - 2 a , where a is a positive constant. [1] (b) Hence or otherwise solve the inequality 3x - 2a 1 x + 5a . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) y B1 Symmetrical. In correct position. Lines intended to be straight. Must be in both first and second quadrants. 2a Key coordinates must be correct. Ignore y = x + 5 a if seen. x O 2a 3 1 1(b) 7 a B1 Allow if seen in an inequality. Obtain critical value from x + 5a = 3x − 2a 2 3 a B1 Allow if seen in an inequality. Obtain critical value − from x + 5a = 2a − 3x 4 3a 7 a B1 3a 7 a State final answer −  x  SC B1 only for −  x  with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7 a Allow −  x and x  . 4 2 Alternative Method for Question 1(b) Solve quadratic equation ( 3 x − 2a ) 2 = ( x + 5a ) 2 M1 8 x 2 − 22ax − 21a 2 = 0 3a 7 a A1 Obtain critical values − and 4 2 3a 7 a A1 3a 7 a State final answer −  x  SC B1 only for −  x  with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7 a Allow −  x and x  . 4 2 3

More questions on Quadratics

2 Solve the equation 2 ln ( 2x + 3) - ln ( 2x + 5) = ln ( 3x) . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Use the correct rule for logarithm of a power, product or quotient or an equivalent *M1 Use of any correct law applied to original terms. method using exponentials Obtain an equation free of logarithms A1 ( 2 x + 3 ) 2 E.g. = 3 x. 2 x + 5 Form a 3-term quadratic from completely correct use of logarithms and solve for x DM1 2 x 2 + 3x −=9 0 State final answer 32 only A1 OE Rejection of negative value if given must be clear. 4

More questions on Logarithmic and exponential functions

4 2 3 Find the exact value of ; 1 3 cos 5 x d x . [4] r 5 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 Use correct double angle formula M1* 1 2 (1 + cos10x ) Obtain 203 sin10x + 23 x A1 OE Use correct limits correctly DM1 3 15π 12π ( − 0 ) + − 20 40 40 Obtain 203 + 403 π A1 Or exact simplified equivalent. 4

More questions on Integration

4 (a) It is given that z = r e i i 1 and z = r e i i 2 . 1 1 2 2 Show that ( z z )* = z * z * . [3] 1 2 1 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 ri 2 4 (b) z = 3e is a root of the equation z + bz + c = 0 , where b and c are real. State the other root and hence find the values of b and c. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) i (1 +2 ) B1 Allow equivalent forms. State z1 z 2 = r1 r2e State correct conjugate of z1 z2 , z1 or z2 B1 Allow equivalent forms. Obtain given result from correct working B1 Clear demonstration that the product of the conjugates is identical to the conjugate of the product. Need to see a conclusion. 3 4(b) * - 14 πi B1 Allow equivalent forms. State z = 3e ( ) 7 Allow 3e 4πi . Complete method to find both b and c M1 E.g. find the product and sum of the roots, or 4 πi z − 3e 1 4 πi . expand z − 3e - 1 ( )( ) Obtain b = −3 2, c = 9 A1 OE Allow z 2 − 3 2 z + 9 = 0. 3

More questions on Complex numbers

5 The equation of a curve is xy + y 2 e -x = 4 . dy y 2 - ye x (a) Show that = x . [4] dx xe + 2y ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the gradients of the tangents to the curve when x = 0 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) dy B1 State or imply y + x as the derivative of xy dx − x d y 2 − x − x B1 State or imply 2 ye − y e as the derivative of y 2e d x d y M1 Need implicit differentiation and attempt at a Equate attempted derivative to zero and solve for product. d x dy y 2 − ye x A1 AG Obtain = from correct working Need to see sufficient correct detail. x dx xe + 2 y Alternative Method for Question 5(a) x dy x x B1 Using xye x + y 2 − 4e x = 0. State or imply xe + y xe + e as the derivative of xy ex ( ) dx dy x 2 B1 Using xye x + y 2 − 4e x = 0. State or imply 2 y − 4e as the derivative of y − 4ex dx d y M1 Must make use of 4 − xy = y 2 e − x Equate attempted derivative to zero and solve for d x dy y 2 − ye x A1 AG Obtain = from correct working Need to see sufficient correct detail. x dx xe + 2 y 4 5(b) Obtain one correct gradient B1 1 E.g. at ( 0, 2 ) . 2 Obtain second correct gradient B1 − 3 E.g. at ( 0, − 2 ) . 2 2

More questions on Differentiation

z + 4 6 Find the complex numbers z for which is real and z = 10 . Give your answers in the form z + 4i z = x + yi , where x and y are real. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................. ............................................................................................................................

Mark scheme: 6 ( x + 4 ) + iy x − i ( y + 4 ) *M1 Multiply numerator (and denominator) by the  conjugate of the denominator. x + i ( y + 4 ) x − i ( y + 4 ) Equate imaginary part of numerator to zero DM1 Numerator = x ( x + 4 ) + y ( y + 4 ) + i  xy − ( y + 4 )( x + 4 )  Obtain x + y = − 4 A1 OE Correct use of modulus and solve for x or y DM1 2 2 2 x + ( −−x 4 ) = 10  x + 4 x + 3 = 0 or ( −−y 4 ) 2 + y 2 = 10  y 2 + 4 y + 3 = 0 =z −−3 i A1 One correct solution A1. SC A1 A0 for both pairs of x and y correct but not in given form. or z = −−1 3i A1 Two solutions only. 6 Alternative Method for Question 6 z + 4 = c ( z + 4i ) , c or x + iy + 4 = c ( x + iy + 4i ) *M1 Equate to a real constant x + 4 = cx and y = cy + 4c DM1 Equate real and imaginary parts 4 4 c y y + 4 A1 x = , y = or = c − 1 1 − c x + 4 x Correct use of modulus and solve for a value of c, or x or y DM1 6c 2 + 20c + 6 = 0 OE x 2 + ( −−x 4 ) 2 = 10  x 2 + 4 x + 3 = 0 or ( −−y 4 ) 2 + y 2 = 10  y 2 + 4 y + 3 = 0. c = − 13 leading to z = −−3 i A1 One correct solution A1. SC A1 A0 for both pairs of x and y correct but not in given form. c = −3 leading to z = −−1 3i A1 Two solutions only. 6 Alternative Method 2 for Question 6 arg ( z + 4 ) = arg ( z + 4i ) *M1  z + 4 Use of arg  = 0.  z + 4i  y y + 4 DM1 = x + 4 x Obtain x + y = − 4 A1 Correct use of modulus and solve for x or y DM1 2 2 2 x + ( −−x 4 ) = 10  x + 4 x + 3 = 0 or ( −−y 4 ) 2 + y 2 = 10  y 2 + 4 y + 3 = 0. =z −−3 i A1 One correct solution A1. SC A1 A0 for both pairs of x and y correct but not in given form. or z = −−1 3i A1 Two solutions only. 6

More questions on Complex numbers

3a - 5x 7 Let f ( x) = , where a is a positive constant. ( 3 a + 2x)( 2a - x) (a) Express f ( )x in partial fractions. 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(b) Hence obtain the expansion of f ( )x in ascending powers of x, up to and including the term in x2. 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(c) State the set of values of x for which the expansion in part (b) is valid. 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Mark scheme: 7(a) A B M1 State or imply the form + and use a correct method to find a constant 3a + 2 x 2 a − x Obtain one of A = 3 and B = −1 A1 Allow M1 A1 if correct A or B found even if unwanted terms in the partial fractions expression. Obtain the second value A1 ISW 3 7(b) Use a correct method to obtain the first two terms in the expansion of M1 −1 −1 −1  2 x  −1  x  ( 3a + 2 x ) ,  1 +  , ( 2 a − x ) , or  1 −   3a   2 a  Obtain the correct unsimplified expansions in terms of a, up to the term in 2x . A2 FT A1 FT for each partial fraction. Follow their A, B 3  2 x  2 x  2  1  x  x  2   1 − +   ..  , −  1 + +   ..  . 3a  3a  3a   2 a  2 a  2 a       1 11 23 2 A1 Ignore terms in higher powers of x. Obtain final answer − x + x 2 3 Do not ISW. 2a 12a 72a Allow reverse order. 4 7(c) State x  32 a B1 OE 1

More questions on Logarithmic and exponential functions

8 (a) Prove the identity cot 2 i - tan 2 i / 4 cot 2i cosec 2i . 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(b) Hence solve the equation cot 2 x - tan 2 x = 5 sec 2x for 0° 1 x 1 90° . 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Mark scheme: 8(a) Express the left hand side in terms of sin and cos M1 cos 2  sin 2  − sin 2  cos 2  Combine to a single term and factorise the numerator M1 2 2 2 2 cos − sin  cos + sin  ( )( ) cos 2 sin 2  Use correct double angle formulae in numerator and denominator M1 4cos2 sin 2 2 Obtain 4cot 2cosec2 from correct working A1 4cos2 1 4cos2  or . sin2 sin2 sin 2sin 2 Alternative Method for Question 8(a) Express the right hand side in terms of sin2 and cos2or tan2and sin2 M1 4cos2 1 4 1  or  sin 2 sin 2 tan 2 sin 2 4cos2 or . sin 2sin 2 Use correct double angle formulae in numerator and denominator M1 2 2 cos − sin  ( ) sin 2 cos 2  4 1 − tan 2  ( ) 1 or  OE 2tan 2sincos Split into 2 terms and simplify M1 cosec2− sec2  Obtain cot 2 − tan 2  from correct working A1 1+cot 2 − 1 + tan 2  ( ) ( ) 8(a) Alternative Method 2 for Question 8(a) Express the left hand side as a difference of 2 squares in terms of tan M1  1  1   − tan + tan  tan  tan  Use correct formula for tan2 M1  2  1    + tan  tan 2 tan  Use 1 + tan 2 = sec2  and simplify using correct double angle formula M1  2  sec 2      tan 2 tan  Obtain 4cot 2cosec2 from correct working A1  2  2      tan 2 sin 2 Alternative Method 3 for Question 8(a) Express the left hand side using appropriate identities M1 cot 2 = cosec2− 1 and tan 2 = sec2 − 1 leading to cosec2− sec2 . Combine to a single term in terms of sin and cos M1 2 2 cos − sin  ( ) sin 2 cos 2  Use correct double angle formulae in numerator and denominator M1 4cos2 sin 2 2 Obtain 4cot 2cosec2 from correct working A1 4cos2 1 4cos2  or . sin2 sin2 sin 2sin 2 4 8(b) Use the identity to obtain an expression in one trigonometric function *M1 Obtain tan 2 2x = 54 A1 OE Obtain one solution e.g. 20.9 DM1 AWRT Obtain a second value e.g. 69.1 and no extras in range A1 AWRT 4

More questions on Trigonometry

9 With respect to the origin O, the points A, B and C have position vectors given by OA = i + 2j , O B = i + 3j - 2k and O C = 2i - j + 3k . The line l passes through B and C. (a) Find a vector equation for l. 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(b) The point P is the foot of the perpendicular from A to l. Find the position vector of P. 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(c) The point D is the reflection of A in l. Find the position vector of D. 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Mark scheme: 9(a) Use a correct method to form an equation for the line through B and C M1 E.g. r = OB + BC. Obtain r = i + 3 j − 2k + ( i − 4 j + 5k ) A1 OE Must have r or component or column vector form. E.g. r = 2i −+j 3k + ( −+i 4 j − 5k ) . l = ... scores A0. 2 9(b) Find AP for a general point P on l B1 Allow unsimplified. E.g. j − 2 k + ( i − 4 j + 5k ) or i − 3 j + 3k + ( i − 4 j + 5k ) . Calculate the scalar product of AP (not OP ) and a direction vector for l and equate M1 E.g. ( j − 2 k + ( i − 4 j + 5k ) )  ( i − 4 j + 5k ) = 0 the result to zero.  − 4 (1 − 4) + 5 ( −+2 5) = 0. Obtain = 13 or = − 23 A1 Or correct equivalents. Obtain 43 i + 53 j − 13 k A1 Or equivalent column vector. 4 9(c) Use a correct method to find their position vector of D M1 E.g. OD = OA + 2 AP Allow a slip in one component. Obtain 53 i + 34 j − 32 k A1 Or equivalent column vector. 2

More questions on Vectors

10 The variables x and y satisfy the differential equation dy sin 4 y = x sin 2y sin 3x . dx It is given that y = 1 r when x = 1 r . 12 2 (a) Solve the differential equation, obtaining a relation between x and y. 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(b) Given that 0 1 y 1 1 r, find the values of y when x = 0 . 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Mark scheme: 10(a) Separate variables correctly B1 Use correct double angle formula to simplify integral in y *M1  sin4 y  dy 2cos2 y dy =  sin2 y Obtain sin 2y A1 *M1 cos3 x dx Commence integration by parts and obtain px cos3 x + q  1 1 A1 cos3 x dx Obtain − 3 x cos3 x + 3  Complete integration and obtain − 13 x cos3 x + 19 sin3 x A1 Use y = 121 π when x = 12 π in an expression with sin2 y , x cos3 x and sin3 x to obtain DM1 12 = 0 − 19 + c the constant of integration Obtain sin2 y = − 13 x cos3 x + 19 sin3 x + 1811 A1 OE ISW 8 10(b) Solve sin2y = 1811 to obtain one solution, e.g. 0.329 M1 AWRT Allow for their constant from a solution involving sin2 y , x cos3 x and sin3 x. M0 for answers in degrees. Obtain a second solution, e.g. 1.24, and no others in range A1 AWRT 2

More questions on Differential equations

11 y M O a 1 x r 2 The diagram shows the curve y = x sin 2x for 0 G x G 1 r. The curve has a maximum point at M, 2 where x = a . (a) Show that tan 2a =-4a [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show by calculation that 0.9 1 a 1 0. 95 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Show that if a sequence of values given by the iterative formula = x 1 -1 n + 1 2 br - tan `4x njl converges, then it converges to a. 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(d) Use the iterative formula in part (c) to calculate a correct to 4 decimal places. Give the result of each iteration to 6 decimal places. 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Mark scheme: 11(a) Use the correct product rule to differentiate *M1 Could be working in terms of a. p Obtain the form sin2 x + q x cos2 x. x dy 1 A1 Obtain = sin2 x + 2 x cos2 x dx 2 x Equate the derivative to zero and form an equation without surds DM1 E.g. sin2a + 4a cos2a = 0. Could be working in terms of x. Obtain tan2a = −4a from correct work A1 AG 4 11(b) Calculate the values of a relevant expression or pair of expressions at x = 0.9 and M1 Allow smaller interval, provided it contains the x = 0.95 root. Must be working in radians. Complete the argument correctly with correct calculated values A1 E.g.  tan1.8 + 3.6 = −0.686...  0   tan1.9 + 3.8 = 0.873...  0 2 11(c) 1 −1 −1 B1 Or work from right to left. π − tan 4 a and rearrange to π − 2a = tan 4a State a = 2 ( ) Allow working in x or a. State tan ( π − 2 a ) = 4 a and rearrange to tan2a = −4a B1 Allow working in x or a. 2 11(d) Use the iterative process correctly at least once M1 M0 if working in degrees. Obtain final answer 0.9183 A1 Show sufficient iterations to at least 6 d.p. to justify 0.9183 to 4 d.p. A1 E.g. or show there is a sign change in the interval ( 0.91825, 0.91835 ) 0.9,0.920872,0.917944,0.918347,0.918292... 3

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