Mathematics 9709/22 May/June 2022: question paper and mark scheme

Mark scheme

1 Differentiate using quotient rule (or product rule) M1 x − 2 x ln x A1 OE Obtain x4 1 A1 Substitute x = e to obtain − 3 or exact equivalent e 3

Mark scheme

2(a) Draw V-shaped graph with vertex on positive x-axis *B1 Draw (more or less) correct graph of y = 5 x − 3 with greater gradient DB1 crossing x-axis between origin and vertex of first graph 2 2(b) Attempt solution of linear equation where signs of 2x and 5x are M1 different 12 A1 and no second answer Solve −2 x + 9 = 5 x − 3 to obtain , 1.71 or better 7 Alternative method for question 2(b) Attempt solution of 3-term quadratic equation (2 x − 9) 2 = (5 x − 3) 2 to M1 7 x 2 + 2 x − 24 = 0 obtain at least one value of x 12 A1 and no second answer Obtain , 1,71 or better 7 2

Mark scheme

3 Use product rule to differentiate e 2 x cos 2 y M1 Must be evidence of implicit differentiation dy A1 Obtain 2e 2 x cos 2 y − 2e2 x sin 2 y dx dy dy B1 Obtain 2e2 x cos 2 y − 2e 2 x sin 2 y + cos y =0 dx dx Substitute x- and y-values to find value of first derivative M1 dy Dependent at least two terms, with at least one involving dx 2 A1 Obtain or 23 3 or exact equivalent 3 5

Mark scheme

4(a) Use y-values [ ln1] , ln 2, ln 3, ln 4 B1 but not their decimal equivalents Use correct formula, or equivalent, with h = 1 M1 allow with decimal equivalents Use both relevant logarithm properties correctly M1 Obtain 12 [ln1 + 2ln 2 + 2ln 3 + ln 4] and hence ln12 A1 AG – necessary detail needed 4 4(b) Sketch correct graph of y = ln x for at least y ⩾ 0 *B1 Indicate that top of each trapezium is below curve or clear equivalent DB1 but B0 if only one chord shown from (1, 0) to (4, ln 4) allow ‘concave down’ 2

Mark scheme

5(a) Substitute x = 4 , equate to zero and attempt solution M1 Obtain a = −7 A1 Divide by x − 4 at least as far as the x term M1 or use of identity or by inspection Obtain 2 x 2 + x + 1 and conclude ( x − 4)(2 x 2 + x + 1) A1 Alternative method for question 5(a) Divide by x − 4 at least as far as the x term M1 Equate the remainder to zero M1 Obtain a = −7 A1 Obtain 2 x 2 + x + 1 and conclude ( x − 4)(2 x 2 + x + 1) A1 4 5(b) Apply logarithms and use power law for e3 y = 4 M1 Obtain 13 ln 4 or exact equivalent A1 Use discriminant [ = 1 − 8 = −7 ] or equivalent to show no other roots B1 AG – necessary detail needed 3

Mark scheme

6(a) Integrate to obtain form k e 2 x −1 M1 any constant k, k ≠ 6 Obtain correct 32 e 2 x −1 A1 Apply limits correctly to k e 2 x −1 and equate to 120 M1 Allow one slip Rearrange as far as a = ... M1 Obtain a = 12 ln(80 + e 2 a −1 ) − 12 A1 AG – necessary detail needed 5 6(b) Use iteration process correctly at least once M1 Obtain final answer 1.76 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 s.f. to justify answer or show sign change in A1 the interval [1.755, 1.765] 3

Mark scheme

7 Integrate 2π − 2 x to obtain form k1 (2π − 2 x) 2 3 M1 any constant k1 Obtain correct − 13 (2π − 2 x) 2 3 A1 OE 3 Apply limits to obtain 13 (2π) 2 or exact equivalent A1 at this stage or later Use identity − sin 2 x = − 12 + 12 cos 2 x B1 OE Integrate to obtain form k 2 x + k3 sin 2 x M1 any constants k2 , k3 Obtain correct − 12 x + 14 sin 2 x A1 OE if area taken as positive Substituting correct limits into their 12 x − 14 sin 2 x M1 Can be SOI by ± 12 π Allow one slip 3 Obtain 13 (2π) 2 − 12 π or exact equivalent A1 CWO 8

Mark scheme

8(a) 1 B1 Use sin 2θ = 2sin θ cos θ and secθ = to obtain 6sin θ cos θ Expand second term to obtain 5 3 cosθ + 5sin θ B1 Simplify to obtain 11sin θ + 5 3 cosθ B1 State R = 14 B1 FT FT their k1 cos θ + k 2 sin θ Use appropriate trigonometry to find α M1 Obtain α = 38.21 A1 AWRT 6 8(b) State or imply 14sin(2β + 38.21) = 2 B1 FT FT their R and α Carry out correct process to find value of β between 0° and 90° M1 Obtain 66.8 A1 AWRT 3