Mathematics 9709/32 May/June 2021: question paper and mark scheme

Mark scheme

1 State or imply non-modular inequality ( 2 x − 1) < 32 ( x + 1) , or 2 2 B1 e.g. 5 x 2 + 22 x + 8 = 0 corresponding quadratic equation Allow recovery from ‘invisible brackets’ on RHS Form and solve a 3-term quadratic in x M1 2 A1 Obtain critical values x = – 4 and x = − 5 2 A1 Do not condone ⩽ for <, or ⩾ for > in the final answer. State final answer x < – 4, x > − Allow ‘or’ but not ‘and’. 5 − 2 < x < −4 scores A0. 5 Accept equivalent forms using brackets e.g. x ∈ ( −∞, −4 ) ∪ ( −0.4, ∞ ) Alternative method for Question 1 Obtain critical value x = – 4 from a graphical method, or by solving a linear B1 equation or linear inequality 2 B2 Obtain critical value x = − similarly 5 -1 1 2 2 B1 Do not condone ⩽ for <, or ⩾ for > in the final answer. State final answer x < – 4, x > − Allow ‘or’ but not ‘and’. 5 − 2 < x < −4 scores A0. 5 Accept equivalent forms using brackets e.g. x ∈ ( −∞, −4 ) ∪ ( −0.4, ∞ ) 4

Mark scheme

2 Show a circle with centre – 1 + i. B1 Need some indication of scale or a correct label. Could just be mark(s) on the axes Show a circle with radius 1 and centre not at the origin (or relevant part B1 thereof). Show correct half line from 1(or relevant part thereof) . B1 Shade the correct region on a correct diagram. B1 Im z 4 N.B. If they have very different scales on their 2 axes the diagram must match their scale - the ‘circle’ should be an ellipse. Allow freehand diagrams with clear correct intention. 1 -1 Re z

Mark scheme

3(a) State or imply ln x = ln A − y ln 3 B1 1 ln A y=− ln x + ln 3 ln 3 State that the graph of y against ln x has an equation that is linear in y and B1 Must be a correct statement. ln x, or has an equation of the standard form ‘y = mx + c’ and is thus a Accept if the 2 equations are written side by side with no straight line comment. An equation with y ln 3 should be compared with the form py + q ln x = c. 1 B1 OE. Exact answer required. State that the gradient is – ISW after a correct statement. ln 3 3 3(b) Substitute ln x = 0, y = 1.3 and use correct method to solve for A M1 ( ln A = 1.3ln 3) Follow their equation in y and ln x . Must be substituting lnx = 0, not x = 0. ln 0 ‘used’ in the solution scores M0A0. Obtain answer A = 4.17 only A1 Must be 2 d.p. as specified in question 2

Mark scheme

4 Commence integration and reach ax tan −1 1 x + b x 1 2 dx *M1 2 2 2 OE. Denominator might be 1 + x or 2 + x . c+x 4 2 1 2 A1 OE Obtain x tan −1 x − x. dx 2 4 + x2 1 A1 x2 ( Complete integration and obtain x tan −1 x − ln 4 + x 2 2 ) OE e.g. with ln 1 + 4 Substitute limits correctly in an expression of the form DM1 2 tan −1 1 − ln 8 + ln 4 OE ( ) px tan −1 x + q ln c + x 2 1 A1 OE exact answer. Obtain final answer π − ln 2 2 Needs a value for tan −1 1 and a single log term Alternative method for Question 4 Use the substitution θ = tan −1 x to obtain λ 2θ sec2 θ dθ and reach *M1 2 pθ tan θ + q tan θ dθ 4 Obtain 2θ tan θ − 2 tan θ dθ A1 OE Complete integration and obtain 2θ tan θ + 2ln ( cos θ ) A1 OE Substitute correct limits correctly in an expression of the form DM1 Limits should be π and 0. Limits must be in radians. rθ tan θ + s ln ( cosθ ) 4 1 A1 OE exact answer. Obtain final answer π − ln 2 Need values for trig. functions and a single log term. 2 5

Mark scheme

5 Square a + ib , use i 2 = −1 and equate real and imaginary parts to 10 and M1 −4 6 respectively Obtain a 2 − b 2 = 10 and 2ab = −4 6 A1 Allow 2abi = −4 6i Eliminate one unknown and find an equation in the other M1 Must be sensible algebra e.g. use of a2 − b2 = a − b socres M0 Obtain a 4 − 10a 2 − 24 [ = 0] , or b 4 + 10b 2 − 24 [ = 0] , or 3-term equivalent A1 Or equivalent horizontal equation from correct work ( ) Obtain final answers ± 2 3 − 2i , or exact equivalents A1 e.g. ± ( 12 − 2i ) from correct work Alternative method for Question 5 Use the correct method to find the modulus and argument of u M1 Obtain modulus 14 A1 Obtain argument tan −1 −1 using an exact method A1 e.g. by using half angle formula which gives 6 2 6t 2 − 10t − 2 6 = 0 Convert to the required form M1 ± 14 6 − 1 i 7 7 This mark is available if working in decimals ( ) Obtain answers ± 2 3 − 2i , or exact equivalents A1 e.g. ± ( 12 − 2i ) 5

Mark scheme

6(a) Express the LHS in terms of cos 2θ and sin 2θ B1 e.g. 1 − cos 2θ sin 2θ sin 2θ Use correct double angle formulae to express the LHS in terms of cos θ and sin θ M1 e.g. ( 1 − 1 − 2sin 2 θ ) 2sin θ cos θ Obtain tan θ from correct working A1 AG Alternative method for Question 6(a) Express the LHS in terms of sin 2θ and tan 2θ B1 Use correct double angle formulae to express the LHS in terms of cos θ and M1 2 1 − sin 2 θ sin θ e.g. 1 − cos θ = 4sin 2 θ 2sin θ cos θ 2 sin θ 4sin θ cos θ cos θ Obtain tan θ from correct working A1 AG Alternative method for Question 6(a) Express the LHS in terms of sin 2θ and tan 2θ B1 Use correct t substitution or rearrangement of sin 2θ in terms of sec2 2θ and M1 sec 2 θ 1 − tan 2 θ 1 + tan 2 1 − tan 2 tan θ to express the LHS in terms of tan θ. − = − 2 tan θ 2 tan θ 2 tan 2 tan Obtain tan θ from correct working A1 AG 3 6(b) State integral of the form μ ln cosθ or ± ln sec θ *M1 π [ − ln cosθ ] 3π4 OE Use correct limits correctly and insert exact values for the trigonometric DM1 Need to see evidence of the substitution ratios 1 1 A1 Obtain a correct expression, e.g. − ln + ln 2 2 Obtain 1 ln 2 from correct working A1 AG (must see an intermediate step) 2 4

Mark scheme

7 dy y B1 OE. Must be a differential equation. State equation = k dx x +1 Separate variables correctly for their differential equation and integrate at *M1 k least one side 1y dy = x + 1 dx Obtain ln y A1 Allow M1A1A1 if they have assumed k = 1 or are working with an incorrect value for k Obtain 2 [ k ] x + 1 A1 Use (0, 1) and (3, e) in an expression containing ln y and x + 1 and a DM1 If remove logs before finding the constant of integration constant of integration to determine k and/or a constant of integration c then the constant must be of the correct form. (or use (0, 1), (3, e) and (x, y) as limits on definite integrals) 1 A1 OE. ( ln y = x + 1 − 1) Obtain k = and c = – 1 2 Their value of c will depend on where c is in their equation and whether they are working with 1 ln y . k The value of k must be consistent with what they integrated. Obtain y = exp ( x + 1 − 1) A1 NFWW, OE, ISW. 7

Mark scheme

8 Use correct product (or quotient) rule M1 At least 3 of 4 terms correct dy A1 OE. Obtain = −5e −5 x tan 2 x + 2e −5 x tan x sec 2 x dx Equate their derivative to zero, use sec 2 x = 1 + tan 2 x and obtain an equation M1 in tan x Obtain 2 tan 2 x − 5 tan x + 2 = 0 A1 Allow 2 tan 3 x − 5 tan 2 x + 2 tan x = 0 State answer x = 0 B1 From correct derivative. Solve a 3 term quadratic in tan x and obtain a value of x M1 Must be in radians Obtain answer, e.g. 0.464 A1 Must be 3 d.p. as specified in the question. Allow A1A0 if both values given to 2 d.p. or > 3 d.p. Obtain second non-zero answer, e.g. 1.107 and no other in the given interval A1 Alternative method for Question 8 Use correct product (or quotient) rule M1 At least 3 of 4 terms correct dy A1 OE Obtain = −5e −5 x tan 2 x + 2e −5 x tan x sec 2 x dx Equate their derivative to zero and obtain an equation in sin x and cos x M1 Obtain 5cos x sin x = 2 A1 Or simplified equivalent (i.e. cancelled) State answer x = 0 B1 From correct derivative. Use double angle formula or square both sides and solve for x M1 Or equivalent method. Must be in radians. Obtain answer, e.g. 0.464 A1 Must be 3 d.p. as specified in the question. Allow A1A0 if both values given to 2 d.p. or > 3 d.p. Obtain second non-zero answer, e.g. 1.107 and no other in the given interval A1 8

Mark scheme

9(a) A B + Cx B1 State or imply the form + 2+ x 3 + x2 Use a correct method for finding a constant M1 SOI Obtain one of A = 4, B = 1 and C = – 2 A1 Obtain a second value A1 Obtain the third value A1 ISW 5 9(b) Use correct method to find the first two terms of the expansion of ( 2 + x ) , −1 M1 Allow unsimplified but not if still including n C r −1 −1 1 1 ( 1 + x , 3 + x 2 2 −1 ) or 1 + x 2 3 ( ) Obtain correct unsimplified expansions up to the term in x 2 of each partial A1 FT 2 A1 FT 2 1 − 1 x + 1 x .... fraction 2 2 1 + 1 (1 − 2 x ) 1 − x 2 .... 3 3 The FT is on their A, B and C Multiply out, up to the terms in x 2 , by B + Cx, where BC ≠ 0 M1 Allow with B and C as implied in part (b) 7 5 7 A1 Or equivalent in form p + qx + rx 2 . Obtain final answer − x + x2 3 3 18 A0 if they multiply through by 18. 5

Mark scheme

10(a) State or imply CD = 2r − 2r cos x B1 Using correct formulae for area of sector and trapezium, or equivalent, form M1 e.g. 2 × 1 r 2 x = 0.9 ( 2r + 2r − 2r cos x ) r sin x an equation in r and x 2 2 Obtain x = 0.9 ( 2 − cos x ) sin x A1 AG, NFWW 3 10(b) Calculate the values of a relevant expression or pair of expressions at x = 0.5 M1 Calculated for both values and correct for one value is and x = 0.7 sufficient for M1. Must be working in radians. Complete the argument correctly with correct values A1 Must have sufficient accuracy to support the answer 0.5 > 0.484 0.016 > 0 0.96... < 1 e.g. or or 0.7 < 0.716 −0.016 < 0 1.02... > 1 2 10(c) x B1 If working in reverse, the first B1 is for State a suitable equation, e.g. cosx = 2 − x 0.9sin x = 2 − cos x 0.9sin x Rearrange this as x = 0.9sin x ( 2 − cos x ) B1 Need to see the complete sequence of changes. 2 10(d) Use the iterative process correctly at least once M1 Must be working in radians Obtain answer 0.62 A1 Show sufficient iterations to at least 4 d.p.to justify 0.62 to 2 d.p., or show A1 Allow recovery. there is a sign change in the interval (0.615, 0.625) N.B. A candidate who starts with 0.5 and stops at 0.61 or starts at 0.7 and stops at 0.63 can score M1A0A1 if they have worked to the required accuracy. 3

Mark scheme

11(a) Show that OA = OB = 5 B1 CWO Evaluate the scalar product of the correct position vectors M1 e.g. ( 0 − 1 + 0 ) Condone of using AO and/or BO Divide their scalar product by the product of the moduli of their vectors and M1 Much reach an angle. The question asks for the use of evaluate the inverse cosine of the result scalar product, so alternative methods (e.g. cosine rule) are not accepted. Obtain answer 101.5° A1 The question asks for an answer in degrees. Accept 102° or better. Mark radians (1.77) as a misread. Do not ISW: 78.5° as final answer scores A0. 4 11(b) State or imply M has position vector i – k B1 OE ρ Taking a general point of OM to have position vector λi – λk, express *M1 ( λ their OM ) AP = 7 OA as an equation in λ State a correct equation in any form A1 e.g. ( −2 + λ )2 + 1 + ( −λ )2 = 7 5 Reduce to λ 2 − 2λ − 15 = 0 A1 OE Solve a quadratic and state a position vector DM1 Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates Alternative method for Question 11(b) State or imply that OP = γ 2 B1 *M1 Allow cos 1 AOB = 0.632... State or imply that cos 1 AOB = 2 and use cosine rule to form an equation 2 2 5 in γ State a correct equation in any form A1 2 e.g. 35 = 5 + 2γ 2 − 2 5.γ 2. 5 Reduce to γ 2 − 2γ − 15 = 0 A1 OE Solve a quadratic and state a position vector DM1 Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates 11(b) Alternative method for Question 11(b) State or imply M has position vector i – k B1 OE State or imply that AM = 3 B1 Use Pythagoras to find MP *M1 MP = 35 − ( AM ) 2 Obtain MP = 4 2 A1 Correct method to find a position vector DM1 (i − k ) ± 4(i − k ) Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates 6