Mathematics 9709/32 May/June 2020: question paper and mark scheme

Mark scheme

1 Commence division and reach partial quotient 3x2 + kx M1 Obtain quotient 3x2 + 2x –1 A1 Obtain remainder 2x – 5 A1 4

Mark scheme

2 State or imply 2 ln y = ln A + kx B1 Substitute values of ln and x, or equate gradient of line to k, and solve for k M1 Obtain k = 0.80 A1 Solve for ln A M1 Obtain A = 3.31 A1 Alternative method for question 2 Obtain two correct equations in y and x by substituting y- and x- values in the given equation B1 Solve for k M1 Obtain k = 0.80 A1 Solve for A M1 Obtain A = 3.31 A1 5

Mark scheme

3 1 5 5 M1* Commence integration and reach ax 2 ln x + b x 2 . dx x 2 2 5 2 2 1 5 A1 Obtain 5 x ln x − 5 x . dx x 2 2 5 4 2 5 A1 Complete the integration and obtain x ln x − x , or equivalent 5 25 Use limits correctly, having integrated twice DM1 e.g 2 × 32 ln 4 − 4 2 × 32 − × 0 + 4 5 25 5 25 128 124 A1 Obtain answer ln 2 − , or exact equivalent 5 25 5

Mark scheme

4 Use correct product rule M1 Obtain correct derivative in any form, e.g. –sin x sin 2x + 2cos x cos 2x A1 Use double angle formula to express derivative in terms of sin x and cos x M1 Equate derivative to zero and obtain an equation in one trig function M1 Obtain 3 sin 2x = 1, or 3 cos 2x = 2 or 2 tan 2x = 1 A1 Solve and obtain x = 0.615 A1 6

Mark scheme

5(a) State R = 7 B1 Use trig formulae to find α M1 Obtain α = 57.688° A1 3 5(b) 1 B1 FT Evaluate cos − 1 to at least 3 d.p. (67.792°) 7 (FT is on their R) Use correct method to find a value of in the interval M1 Obtain answer, e.g. 5.1° A1 Obtain second answer, e.g.117.3°, only A1 4

Mark scheme

6(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 e.g. (1 + 3x ) − x × 12 x 4 3 (1 + 3 x ) 4 2 Equate derivative to zero and solve for x M1 Obtain answer 0.577 A1 4 6(b) State or imply du = 2 3x dx , or equivalent B1 Substitute for x and dx M1 1 A1 Obtain integrand , or equivalent ( 2 3 1+ u 2 ) State integral of the form tan and use limits u = 0 and u = 3 (or x = 0 and x = 1) correctly M1 3 A1 Obtain answer π , or exact equivalent 18 5

Mark scheme

7 Separate variables correctly and integrate at least one side B1 Obtain term ln(y – 1) B1 1 A B M1 Carry out a relevant method to determine A and B such that ≡ + ( x + 1)( x + 3) x +1 x + 3 1 1 A1 Obtain A = and B = − 2 2 1 1 A1 FT Integrate and obtain terms ln ( x + 1) − ln ( x + 3) ln + 1 − ln( + 3), or equivalent + A1 FT 2 2 (FT is on A and B) Use x = 0, y = 2 to evaluate a constant, or as limits in a solution containing terms of the form a ln(y – 1), b ln(x + 1) and M1 c ln(x + 3), where abc ≠ 0 Obtain correct answer in any form A1 3x + 3 A1 Obtain final answer y = 1 + , or equivalent x+3 9

Mark scheme

8(a) Substitute and obtain a correct equation in x and y B1 Use i2 = –1 and equate real and imaginary parts M1 Obtain two correct equations in x and y, e.g. x – y = 3 and 3x + y = 5 A1 Solve and obtain answer z = 2 – i A1 4 8(b)(i) Show a point representing 2 + 2i B1 Show a circle with radius 1 and centre not at the origin B1 FT (FT is on the point representing the centre) Show the correct half line from 4i B1 Shade the correct region B1 4 8(b)(ii) Carry out a complete method for finding the least value of Im z M1 1 A1 Obtain answer 2 − 2 , or exact equivalent 2 2

Mark scheme

9(a) k B1 State cos p = 1+ p Differentiate both equations and equate derivatives at x = p M1 k A1 Obtain a correct equation in any form, e.g. − sin p = − (1 + p ) 2 Eliminate k M1 Obtain the given answer showing sufficient working A1 5 9(b) Use the iterative formula correctly at least once M1 Obtain final answer p = 0.568 A1 Show sufficient iterations to justify 0.568 to 3 d.p., or show there is a sign change in the interval (0.5675, 0.5685) A1 3 9(c) Use a correct method to find k M1 Obtain answer k = 1.32 A1 2

Mark scheme

10(a) State that the position vector of M is 3i + j B1 Use a correct method to find the position vector of N M1 10 A1 Obtain answer i + 2 j + 2k 3 Use a correct method to form an equation for MN M1 1 A1 Obtain correct answer in any form, e.g. r = 3i + j + λ i + j + 2k 3 5 10(b) State or imply r = µ(2i + 2j + 3k) as equation for OB B1 Equate sufficient components of MN and OB and solve for λ or for µ M1 Obtain λ = 3 or µ = 2 and position vector 4i + 4j + 6k for P A1 3 10(c) Carry out correct process for evaluating the scalar product of direction vectors for OP and MP, or equivalent M1 Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse M1 cosine of the result Obtain answer 21.6° A1 3