Mathematics 9709/33 May/June 2025: question paper and mark scheme

Mark scheme

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 + 5a if seen. x O 2a 3 1 1(b) 7a B1 Allow if seen in an inequality. Obtain critical value from x + 5a = 3x − 2a 2 3a B1 Allow if seen in an inequality. Obtain critical value − from x + 5a = 2a − 3x 4 3a 7a B1 3a 7a State final answer − <x< SC B1 only for − <x< with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7a Allow − < x and x < . 4 2 Alternative Method for Question 1(b) Solve quadratic equation ( 3x − 2a ) = ( x + 5a ) M1 2 2 8x2 − 22ax − 21a2 = 0 3a 7a A1 Obtain critical values − and 4 2 3a 7a A1 3a 7a State final answer − <x< SC B1 only for − <x< with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7a Allow − < x and x < . 4 2 3

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 = 3x. E.g. 2x + 5 Form a 3-term quadratic from completely correct use of logarithms and solve for x DM1 2x2 + 3x − 9 = 0 State final answer 32 only A1 OE Rejection of negative value if given must be clear. 4

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 State z1 z2 = r1r2 e ( 1

Mark scheme

4(a) i θ +θ 2 ) B1 Allow equivalent forms. State correct conjugate of z1z2 , 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) ( ) State z * = 3e 4 - 1 πi B1 Allow equivalent forms. 7 πi Allow 3e 4 . Complete method to find both b and c M1 E.g. find the product and sum of the roots, or ( - 1 πi expand z − 3e 4 )( z − 3e ). 1 πi 4 Obtain b = −3 2, c = 9 A1 OE Allow z 2 − 3 2z + 9 = 0. 3

Mark scheme

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

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 x2 + ( − x − 4) = 10 x2 + 4 x + 3 = 0 or 2 ( − 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 4c 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 6c2 + 20c + 6 = 0 OE x2 + ( − x − 4) = 10 x2 + 4 x + 3 = 0 or 2 ( − y − 4)2 + y2 = 10 y2 + 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 x2 + ( − x − 4) = 10 x2 + 4 x + 3 = 0 or 2 ( − y − 4)2 + y2 = 10 y2 + 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

Mark scheme

7(a) A B M1 State or imply the form + and use a correct method to find a constant 3a + 2 x 2a − 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 ( 3a + 2 x )−1 , 1 + 2x x , ( 2a − x ) , or 1 − −1 3a 2a Obtain the correct unsimplified expansions in terms of a, up to the term in x 2 . A2 FT A1 FT for each partial fraction. Follow their A, B 3 2x 2x 1 x x 2 2 1 − + .. , − 1 + + .. . 3a 3a 3a 2a 2a 2a 1 11 23 2 A1 Ignore terms in higher powers of x. Obtain final answer − 2 x+ x Do not ISW. 2a 12a 72a3 Allow reverse order. 4 7(c) State x < 32 a B1 OE 1

Mark scheme

8(a) Express the left hand side in terms of sin θ and cosθ M1 cos 2 θ sin 2 θ − sin 2 θ cos 2 θ Combine to a single term and factorise the numerator M1 ( cos θ − sin θ )( cos θ + sin θ ) 2 2 2 2 cos 2 θ sin 2 θ Use correct double angle formulae in numerator and denominator M1 4cos 2θ sin 2 2θ Obtain 4cot 2θ cosec2θ from correct working A1 4cos 2θ 1 4cos 2θ × or . sin 2θ sin 2θ sin 2θ sin 2θ Alternative Method for Question 8(a) Express the right hand side in terms of sin 2θ and cos 2θ or tan 2θ and sin 2θ M1 4cos 2θ 1 4 1 × or × sin 2θ sin 2θ tan 2θ sin 2θ 4cos 2θ or . sin 2θ sin 2θ Use correct double angle formulae in numerator and denominator M1 ( cos θ − sin θ ) 2 2 sin 2 θ cos 2 θ or ( 4 1 − tan 2 θ )× 1 OE 2 tan θ 2sin θ cos θ Split into 2 terms and simplify M1 cosec2θ − sec2 θ Obtain cot 2 θ − tan 2 θ from correct working A1 (1+cot θ ) − (1 + tan θ ) 2 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 θ M1 1 1 − tan θ × + tan θ tan θ tan θ Use correct formula for tan 2θ M1 2 1 × + tan θ tan 2θ tan θ Use 1 + tan2 θ = sec2 θ and simplify using correct double angle formula M1 2 sec θ 2 × tan 2θ tan θ Obtain 4cot 2θ cosec2θ from correct working A1 2 2 × tan 2θ sin 2θ Alternative Method 3 for Question 8(a) Express the left hand side using appropriate identities M1 cot 2 θ = cosec2θ −1 and tan2 θ = sec2 θ − 1 leading to cosec2θ − sec2 θ . Combine to a single term in terms of sin θ and cosθ M1 ( cos θ − sin θ ) 2 2 sin 2 θ cos 2 θ Use correct double angle formulae in numerator and denominator M1 4cos 2θ sin 2 2θ Obtain 4cot 2θ cosec2θ from correct working A1 4cos 2θ 1 4cos 2θ × or . sin 2θ sin 2θ sin 2θ sin 2θ 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

Mark scheme

9(a) Use a correct method to form an equation for the line through B and C M1 E.g. r = OB + λ BC. Obtain r = i + 3j − 2k + λ ( i − 4 j + 5k ) A1 OE Must have r or component or column vector form. E.g. r = 2i − j + 3k + λ ( −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 − 2k + λ ( i − 4 j + 5k ) or i − 3j + 3k + μ ( i − 4 j + 5k ) . Calculate the scalar product of AP (not OP) and a direction vector for l and equate M1 E.g. ( j − 2k + λ ( i − 4 j + 5k ) ) · ( i − 4 j + 5k ) = 0 the result to zero. λ − 4 (1 − 4λ ) + 5 ( −2 + 5λ ) = 0. Obtain λ = 13 or μ = − 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

Mark scheme

10(a) Separate variables correctly B1 Use correct double angle formula to simplify integral in y *M1 sin 4 y sin 2 y dy = 2cos 2 y dy Obtain sin 2 y A1 *M1 Commence integration by parts and obtain px cos3x + q cos3x dx A1 Obtain − 13 x cos3x + 13 cos3x dx Complete integration and obtain − 13 x cos3x + 19 sin 3x A1 Use y = 121 π when x = 12 π in an expression with sin 2 y, x cos3x and sin 3x to obtain DM1 1 =0− 1 +c 2 9 the constant of integration Obtain sin 2 y = − 13 x cos3x + 19 sin 3x + 18 11 A1 OE ISW 8 10(b) Solve sin 2 y = 18 11 to obtain one solution, e.g. 0.329 M1 AWRT Allow for their constant from a solution involving sin 2 y, x cos3x and sin 3x. M0 for answers in degrees. Obtain a second solution, e.g. 1.24, and no others in range A1 AWRT 2

Mark scheme

11(a) Use the correct product rule to differentiate *M1 Could be working in terms of a. p Obtain the form sin 2 x + q x cos 2 x. x dy 1 A1 Obtain = sin 2 x + 2 x cos 2 x dx 2 x Equate the derivative to zero and form an equation without surds DM1 E.g. sin 2a + 4a cos2a = 0. Could be working in terms of x. Obtain tan 2a = −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) ( ) State a = 12 π − tan −1 4a and rearrange to π − 2a = tan −1 4a B1 Or work from right to left. Allow working in x or a. State tan ( π − 2a ) = 4a and rearrange to tan 2a = −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