Mathematics 9709/13 May/June 2024: question paper and mark scheme

Mark scheme

1 Correct second term 30x in expansion of (1 + 3 x)10 B1 WWW, may be implied later. Correct third term + 405 x 2 B1 Ignore subsequent terms, may be implied later. Multiply ( 2 − 5 x ) by their 30 x + 405x 2 to obtain two x 2 terms only M1 Expect −150 x 2 ,810 x 2 . Coefficient is 660 A1 Must be clearly identified. Allow final answer 660x 2 . 4

Mark scheme

2(a) State ( 53 π, 0) for point A B1 Or exact equivalent. 5 Allow x = π or exact equivalent. 3 19 B1 Or exact equivalent. x= π for point B May be implied in coordinate or vector form. 6 y = −k for point B B1 May be implied in coordinate or vector form. 3 2(b) ( Solve at least as far as sin −1 3t = kπ with correct value for cos −1 12 2 ) M1 Allow use of π = 3.14... . Allow sin −1 3t = 30 . sin −1 3t = 16 π and hence t = 16 A1 Or exact equivalent. Can use degrees if consistent. 2

Mark scheme

3(a) 1 2 B1 State 2r + rθ = 65 and r θ = 225 2 Form a 3-term quadratic or cubic in r or θ or r θ from correct arc and sector *M1 Condone sign errors. formula Solve their 3 term quadratic or cubic to obtain values of r or θ DM1 Expect 2r 2 − 65r + 450 = ( 2r − 45 )( r − 10 ) or 18θ 2 − 97θ + 72 = ( 9θ − 8 )( 2θ − 9 ) . 8 A1 B1 SC if no quadratic or cubic solution. r = 10 and θ = 4.5 ignore r = 22.5 and θ = , do not ignore r = 0 If r = 0 included A0 or B0 SC. 9 4 3(b) Use correct formula for area of triangle with clear use of angle being 2π − their θ M1 Expect 1.783 or 102.2o, their θ must be reflex. 48.9 A1 AWRT, WWW or a second answer. Or greater accuracy; condone absence of units. 2

Mark scheme

4(a) sin θ M1 Use identity tan θ = cos θ Use identity cos2 θ = 1 − sin 2 θ M1 ± ( 5sin 2 θ + 7sin θ − 6 = 0 ) A1 3 4(b) Attempt solution of their 3 term equation and correct process to find at least 1 M1 Expect ( 5s − 3)( s + 2 ) = 0, s = 3 / 5. value of sin x or sin 2x or sin θ x = 18.4 A1 Or greater accuracy. B1 SC if no solution to the quadratic. x = 71.6 or (90 – their 18.4) or greater accuracy; and no other solutions for A1FT WWW 0° < x < 180° B1 SC FT if no solution to the quadratic. B1 SC both correct in radians, 0.322, 1.25. 3

Mark scheme

5(a) Differentiate to obtain 4 x + 12 x −2 B1 OE Condone ‘+c’. K M1 Not given if ‘+c’ used. Equate first derivative to zero and solve 4 x + 2 = 0 as far as x3 = k , K and k non- x zero x = − 12 and y = 92 A1 OE B1 SC if no visible solution of the cubic. 3 5(b) Differentiate their first derivative, substitute their x value. Substitution may be M1 Must differentiate one term correctly. implied by a correct inequality or correct value, −1 Expect 4 − x −3 = 12 at x = 2 dy Alternative: substitute values of x into . dx 1 1 One value x < − and one value − < x < 0. 2 2 conclude minimum A1 Following correct work only 2 5(c) State increasing … B1 … with clear reference to first derivative always being positive [for x > 0] B1 Dependent on first derivative being correct. It is not sufficient to substitute values of x. 2

Mark scheme

6(a) Integrate to obtain form k (5 x − 3) −1 *M1 OE 4(5 x − 3) −1 A1 Or unsimplified equivalent. Condone absence of ... + c so far. Substitute x = 54 and y = −3 to attempt value of c DM1 4 DM0 for substituting −3, . 5 y = 4(5 x − 3) −1 − 7 allow f(x) or f = 4(5 x − 3) −1 − 7 A1 OE Condone c = –7 as the final answer providing 4 y = or f ( x ) = + c OE is seen earlier. ( − 3) 5 x Attempts to write equation in y = mx + c form scores A0. Do not ISW. Gains max 3/4. 4 6(b) Carry out stretch by replacing x by 2x in their equation M1 Award if given as the second transformation. Do not ignore sign errors. Carry out translation by replacing x by x − 2 and y by y − 10 M1 OE Award if given as the first transformation. Do not ignore sign errors. 4 A1 Or similarly simplified equivalent, WWW. y= +3 10 x − 23 3

Mark scheme

7(a) 5(3 + 9d ) = 127.5 B1 OE d = 2.5 B1 2 7(b) Attempt to find either the first term or the last term in the set by considering M1 Using their d. 1.5 + 2.5( n − 1) > 25 or 1.5 + 2.5( n − 1) < 100 or equivalent equations May be implied by correct answers. State or imply that 11th term or 26.5 is the first in the set A1 State or imply that 40th term or 99 is the last in the set A1 Either use S40 − S10 DM1 Their 40 and 10 from correct working with their d. Or use 12 n( a + l ) with correct results for their d Correct values 30, 26.5 and 99 respectively. Or use 12 n[2a + (n − 1)d ] with correct results for their d Correct values 30, 26.5 and 2.5 respectively. Obtain 1882.5 A1 OE 5

Mark scheme

8 Substitute x = 0 and attempt solution of 3-term quadratic equation in y M1 If y = 0 used can score a maximum of M0 A0 B1 M1 A0 A1FT DM1 A0, i.e. 4/8. –5 and 3 A1 B1 SC if no working to solve the quadratic. State or imply centre of circle is (3, − 1) B1 Condone errors which don’t affect finding centre. May be implied by the correct final y coordinate. Attempt gradient of AC or BC *M1 − 43 or 43 A1 State or imply gradient of tangent is 34 or − 34 A1FT Following their gradient of radius. Only FT when previous 2 marks are M1 A0. Either solve simultaneous equations (of 2 tangent equations) to find x- coordinate DM1 Or Substitute y-value of centre into either tangent equation x = − 163 , y = −1 A1 Alternative Method 1: for the 4th and 5th marks Rearrange and differentiate the circle equation or differentiate implicitly (M1) Replaces the second M1. dy 3 − x dy 3− x (A1) Replaces the second A1. = or = dx y + 1 dx 1 ( 25 − ( x − 3) ) 2 2 8 Alternative Method 2: for the last 5 marks = tan −1 4 or identifying similar triangles PMA and AMC (M1A1) C is the circle centre, P is intersection of the two ACP = MAP tangents, M is intersection of PC and the y-axis. 3 PM 4 PM 16 (M1A1) tan MAP = , = , PM = or use of similar triangles 4 3 4 3 −16 (A1) P is , −1 3 Alternative Method 3: for the last 5 marks Pythagoras on triangle PAC, PC 2 = PA2 + AC 2 , (M1) Identifies the required 3 sides and sets up formula. PC 2 = ( PM + 3) , PA2 = PM 2 + 42 , AC = radius = 5 (A1) Finds each side with two in terms of PM OE. 2 (M1A1) Sets up and solves equation. ( PM + 3)2 = PM 2 + 42 + 52 leads to 6 PM = 32, PM = 16 3 −16 (A1) P is , −1 3 8

Mark scheme

9(a) −1 Differentiate to obtain form kx 2 (2 x3 + 10) 2 M1 OE −1 3 x 2 (2 x 3 + 10) 2 A1 Or unsimplified equivalent. Substitute x = 3 in first derivative and evaluate to find gradient *M1 Expect 278 . −1 3 Allow if first derivative of forms k (2x + 10) 2 , −1 −1 kx(2x 3 + 10) 2 or kx 2 (2x 3 + 10) 2 . Attempt equation of tangent at ( 3,8 ) with numerical gradient DM1 Use of gradient of the normal is DM0. [±]( 27 x − 8 y − 17) = 0 or integer multiples A1 5 9(b) State or imply volume is π (2 x3 + 10) dx B1 Implied if π appears only at the end. Do not allow an unsimplified: (( π 2 x3 + 10 ) ). 1/2 2 Integrate to obtain k1 x 4 + k2 x and evaluate using limits 1 and 3 M1 Where k1k2 ≠ 0 . 60π A1 OE Allow from a correct integral and sight of limits. Allow numerical answers in the range 188-189. 3

Mark scheme

10(a) Substitute to obtain equation 9r 4 + 14r 2 − 8 = 0 B1 OE Attempt solution of quadratic equation in r 2 to obtain at least one value of r or r 2 ( )( M1 Expect 9r 2 − 4 r 2 + 2 . ) r = 23 only A1 SC B1 answer without working. 3 10(b) Substitute a = 2 and their r in correct formula and attempt to evaluate M1 2 20 2 20 2 1 − 2 − 1 3 Expect or 3 . 2 2 1 − − 1 3 3 5.998 A1 AWRT and no other value. 2 10(c) 4 8 B1 FT Following their r provided r < 1. Identify a2 = and common ratio as . 3 27 May be implied in the sum to infinity. 3 2 Allow . 3 Substitute their new a and r in correct formula for sum to infinity and evaluate M1 r < 1 otherwise M0. 36 A1 OE 19 Accept 1.89 or better from 1.894736….. 3

Mark scheme

11(a) Express f ( x) as: a − ( x − 3) 2 or a − ( 3 − x ) where a = ±19 or ± 1 M1 OE 2 ( ) If the form − f ( x ) = x 2 − 6 x − 10 is used the form must be returned to f ( x ) = Completed square form must give −x 2 . Answers must come from completion of the square (not calculus or graphs). 19 − (3 − x)2 or 19 − ( x − 3)2 A1 OE f ( x ) 19 or y 19 with ⩽, not < A1 FT Using their constant following the award of M1. SC B1 answer only or answer from a method not or –∞ < f(x) 19 or –∞ ⩽ f(x) 19 involving completion of the square. or (–∞, 19] or [–∞, 19] 3 11(b) g −1 ( x ) = 14 ( x − k ) B1 M1 OE g −1 f ( x ) = 1 4 ( ) 10 + 6 x − x 2 − k = 4 x + k May use their completed square form for f(x). Simplify the quadratic equation obtained from g −1f ( x) = g( x) provided k is *M1 Expect x 2 + 10 x − 10 + 5k = 0. present and apply b2 − 4ac = 0 to this quadratic equation Obtain 100 − 4 ( 5k − 10 ) = 0 and hence k = 7 A1 Use their k to form and solve a quadratic in x DM1 Allow if their quadratic has two solutions. ( −5, −13) only A1 SC B1 if no method seen. Alternative Method for first 4 marks State f ( x) = gg( x) (B1) gg( x ) = 16 x + 5k (M1) Apply b2 − 4ac = 0 to quadratic equation obtained from f ( x) = gg( x) (*M1) Provided k is present. 100 − 4(5k − 10) = 0 and hence k = 7 (A1) 6