Mathematics 9709/31 May/June 2021: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality 2 2 ( 3 x − 1) < ( x + 1) , or 2 2 B1 corresponding quadratic equation, or pair of linear equations Form and solve a 3-term quadratic, or solve two linear equations for x M1 e.g. 35x2 − 26 x + 3 = 0 3 1 A1 Allow 0.143 or better Obtain critical values x = and x = 5 7 1 3 A1 1 3 State final answer <x< Exact values required. Accept x > and x < 7 5 7 5 Do not condone ⩽ for < in the final answer. Fractions need not be in lowest terms. Alternative method for Question 1 3 B1 Obtain critical value x = from a graphical method, or by solving a linear 5 equation or linear inequality 1 B2 Allow 0.143 or better Obtain critical value x = similarly 7 1 3 B1 1 3 State final answer <x< OE. Exact values required. Accept x > and x < 7 5 7 5 Do not condone ⩽ for < in the final answer. Fractions need not be in lowest terms. 4

Mark scheme

2 Reduce to a 3-term quadratic u 2 + 6u − 1 = 0 OE B1 Allow ‘= 0’ implied Solve a 3-term quadratic for u M1 Obtain root 10 − 3 A1 Obtain answer x = – 1.818 only A1 The question asks for 3 d.p. Reject − 10 − 3 correctly ( ) B1 e.g. by stating that e x > 0 or ln −10 − 3 is impossible Not "math error". Alternative method for Question 2 Rearrange to obtain a correct iterative formula ( B1 e.g. x = − ln 6 + e xn n +1 ) Use the iterative process at least twice M1 Obtain answer x = – 1.818 A1 Show sufficient iterations to at least 4 d.p. to justify x = – 1.818 A1 1, −2.165..., −1.811..., −1.819..., −1.818..., −1.818... Clear explanation of why there is only one real root B1 5

Mark scheme

3(a) Use correct trig expansions and obtain an equation in sin x and cos x *M1 Use correct exact trig ratios for 30° in their expansion B1 FT e.g. cos x ( − 1) = sin x ( 3 − ) 2 3 1 2 Obtain an equation in tan x DM1 Allow if their error in line 1 was a sign error 2− 3 A1 AG Obtain tan x = from correct working 1− 2 3 4 3(b) Obtain answer in the given interval, e.g.173.8 ° B1 Accept 174°, 354° or better Obtain a second answer and no other in the given interval, e.g. 353.8 ° B1 Ignore answers outside the given interval. Treat answers in radians (3.03 and 6.17) as a misread. 2

Mark scheme

4(a) Use correct double angle formula or t-substitution twice M1 1 − cos 2θ A1 AG Obtain = tan 2 θ from correct working 1 + cos 2θ 2 4(b) Express tan 2θ in terms of sec2θ M1 π3 π ( sec θ ± 1) dθ 2 6 Integrate and obtain terms tan θ − θ A1 Accept with a mixture of x and θ Substitute limits correctly in an integral of the form M1 π 1 π a tan θ + bθ , where ab ≠ 0 3−3− + Allow if trig. not substituted 3 6 2 1 A1 or equivalent exact 2-term expression Obtain answer 3− π 3 6 4

Mark scheme

5(a) Use quadratic formula and i 2 = −1 M1 Obtain answers pi + q − p 2 and pi − q − p 2 A1 2 pi ± −4 p 2 + 4q Accept and ISW 2 2 5(b) State or imply that the discriminant must be negative M1 State condition q < p 2 A1 2 5(c) Carry out a correct method for finding a relation, e.g. use the fact that the M1 argument of one of the roots is ( ± ) 60° p A1 State a correct relation in any form, e.g. = (±) 3 q − p2 4 2 A1 Simplify to q = p 3 Alternative method for Question 5(c) Carry out a correct method for finding a relation, e.g. use the fact that the M1 sides have equal length ( ) State a correct relation in any form, e.g. 4 q − p 2 = p 2 + q − p 2 A1 4 2 A1 Simplify to q = p 3 3

Mark scheme

6(a) Use correct chain rule or correct quotient rule to differentiate x or y M1 dx 3 dy 2 A1 OE Obtain = or = dt 2 + 3t dt ( 2 + 3t )2 dy dy dx M1 Use = ÷ dx dt dt 2 A1 OE. Express as a simple fraction but not necessarily Obtain answer fully cancelled. 3 ( 2 + 3t ) Explain why this is always positive A1 For correct gradient. e.g. x is only defined for 2 + 3t > 0 hence gradient > 0 Alternative method for Question 6(a) Form equation in x and y only M1 ex − 2 1 2 −x A1 OE Obtain y = = − e 3e x 3 3 Differentiate M1 2 A1 OE Obtain y ′ = e − x 3 Explain why this is always positive A1 5 6(b) 1 B1 Obtain y = − when x = 0 3 Use a correct method to form the given tangent M1 y+1 3 = 2 x 3 Obtain answer 3 y = 2 x − 1 A1 OE 3

Mark scheme

7(a) Use correct quotient rule or correct product rule M1 1 1 x. 2 − tan −1 x. dy 1+ x 2 x e.g. = dx x Obtain correct derivative in any form A1 Equate derivative to zero and remove inverse tangent M1 2a A1 AG. Accept with x in place of a. Obtain a = tan 2 from correct working 1+ a 4 7(b) Calculate the value of a relevant expression or pair of expressions at a = 1.3 M1 Must be using radians and a = 1.5 Complete the argument correctly with correct calculated values A1 e.g. 1.3 < 1.448, 1.5 > 1.322 ( 0.148, − 0.178 ) 2 7(c) 2 an M1 Use the iterative process an +1 = tan 2 correctly at least twice 1 + an Obtain final answer 1.39 A1 Show sufficient iterations to at least 4 d.p. to justify 1.39 to 2 d.p. or show A1 Allow recovery there is a sign change in the interval (1.385, 1.395) 3

Mark scheme

8(a) B1 OE. Allow ± ρ 2 State or imply AB = −1 −3 Use the correct process to calculate the scalar product of a pair of relevant ρ M1 ( 2 + 2 − 3 = 1) vectors, e.g. their AB and a direction vector for l Using the correct process for the moduli, divide the scalar product by the M1 1 product of the moduli of the two vectors and evaluate the inverse cosine of cos −1 6 14 the result Obtain answer 83.7° or 1.46 radians A1 Or answers rounding to 83.7° or 1.46 radians 4 ρ ρ 8(b) State or imply ± AP and ± BP in component form, i.e. B1 (1 + λ, 1 – 2λ, λ) and (– 1 + λ, 2 – 2λ, 3 + λ), or equivalent Form an equation in λ by equating moduli or by using cos BAP = cos ABP *M1 Obtain a correct equation in any form A1 Or (1 + λ ) 14 − 4λ + 6λ 2 = (13 − λ ) 2 − 2λ + 6λ 2 (1 + λ ) + (1 − 2λ ) + λ = ( λ − 1) + ( 2 − 2λ ) + ( λ + 3) 2 2 2 2 2 2 (83λ − 528λ + 207λ − 162 = 0 ) 3 2 Solve for λ and obtain position vector DM1 [λ = 6] Obtain correct position vector for P in any form, e.g. (8, – 9, 7) or A1 Accept coordinates 8i – 9j + 7k 5

Mark scheme

9(a) Use correct product rule or correct quotient rule M1 Obtain correct derivative in any form A1 − 2 5 x 3 2 −3 y' = − x ln x x 3 Equate 2 term derivative to zero and solve for x M1 3 A1 Or exact equivalent Obtain answer x = e 2 3 A1 Or exact equivalent Obtain answer y = 2e 5 9(b) 1 1 1 *M1 Commence integration and reach ax 3 ln x + b x 3 . dx x 1 1 1 A1 Obtain 3x 3 ln x − 3 x 3 . dx x 1 1 A1 OE Complete the integration and obtain 3x 3 ln x − 9 x 3 1 1 DM1 6 ln 8 − 9 × 2 − 0 + 9 Use limits correctly in an expression of the form px ln x + qx 3 3 ( pq ≠ 0 ) Obtain 18ln 2 − 9 from full and correct working A1 AG need to see ln 8 = 3ln 2 5

Mark scheme

10 1 B1 A B C Ax + B C State a suitable form of partial fractions for e.g. + 2 + or + x (1 + 2 x ) 2 x x 1 + 2x x 2 1 + 2x Use a relevant method to determine a constant M1 Obtain one of A = – 2, B = 1 and C = 4 A1 Obtain a second value A1 Obtain the third value A1 Separate variables correctly and integrate at least one term M1 1 B3 FT The FT is on A, B and C. Obtain terms −2ln x − + 2ln (1 + 2 x ) and t Withhold B1 for each error or omission. x Evaluate a constant, or use limits x = 1, t = 0 in a solution containing terms t, M1 a ln x and b ln (1 + 2 x ) , where ab ≠ 0 1 1 + 2x A1 Obtain a correct expression for t in any form, e.g. t = − + 2ln +1 x 3x 11