Cambridge A Level Mathematics 9709 — 2021 May/June Paper 3 · Variant 2

1 Solve the inequality 2x 3 x 1 . [4] −1 < + ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

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

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3 The variables x and y satisfy the equation x A 3−y , where A is a constant. = (a) Explain why the graph of y against ln x is a straight line and state the exact value of the gradient of the line. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that the line intersects the y-axis at the point where y 1.3. = (b) Calculate the value of A, giving your answer correct to 2 decimal places. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) = − B1 1 ln ln ln3 ln3   = − +     A y x State that the graph of y against ln x has an equation that is linear in y and ln x, or has an equation of the standard form ‘y = mx + c’ and is thus a straight line B1 Must be a correct statement. Accept if the 2 equations are written side by side with no comment. An equation with ln3 y should be compared with the form py + q ln x = c. State that the gradient is – 1 ln3 B1 OE. Exact answer required. ISW after a correct statement. 3 Question Answer Marks Guidance 3(b) Substitute ln x = 0, y = 1.3 and use correct method to solve for A M1 ( ) ln 1.3ln3 A = 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

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2 4 Using integration by parts, find the exact value of tan−1 1 dx. [5] 2x Ó0 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 Commence integration and reach 1 2 1 1 tan d 2 − +  + ax x b x x c x *M1 OE. Denominator might be 2 2 1 or 2 4 2 + + x x . Obtain 1 1 tan . 2 − −     x x x 2 2 4 + x dx A1 OE Complete integration and obtain ( ) 1 2 1 tan ln 4 2 − − +     x x x A1 OE e.g. with 2 1 ln 4 x +       Substitute limits correctly in an expression of the form ( ) 1 2 tan ln − + + px x q c x DM1 1 2tan 1 ln8 ln 4 − − + OE Obtain final answer 1 π ln 2 2 − A1 OE exact answer. Needs a value for 1 tan 1 − and a single log term Alternative method for Question 4 Use the substitution 1 tan 2 θ − = x to obtain 2 2 sec d λ θ θ θ  and reach tan tan d θ θ θ θ +  p q *M1 Question Answer Marks Guidance 4 Obtain 2 tan 2 tan d θ θ θ θ − A1 OE Complete integration and obtain ( ) 2 tan 2ln cos θ θ θ + A1 OE Substitute correct limits correctly in an expression of the form ( ) tan ln cos θ θ θ + r s DM1 Limits should be 4 π and 0. Limits must be in radians. Obtain final answer 1 π ln 2 2 − A1 OE exact answer. Need values for trig. functions and a single log term. 5

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5 The complex number u is given by u 10 6i. = −4 Find the two square roots of u, giving your answers in the form a ib, where a and b are real and + exact. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 5 Square a + ib , use 2i 1 = − and equate real and imaginary parts to 10 and 4 6 − respectively M1 Obtain 2 2 10 − = a b and 2 4 6 = − ab A1 Allow 2 i 4 6i = − ab Eliminate one unknown and find an equation in the other M1 Must be sensible algebra e.g. use of 2 2 − = − a b a b socres M0 Obtain [ ] 4 2 10 24 0 − − = a a , or [ ] 4 2 10 24 0 + − = b b , 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 1 1 tan 6 −− using an exact method A1 e.g. by using half angle formula which gives 2 2 6 10 2 6 0 − − = t t Convert to the required form M1 6 1 14 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

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18 The equation of a curve is y tan2x for x = e−5x −12π < < 2π. Find the x-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate. 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Mark scheme: 8 Use correct product (or quotient) rule M1 At least 3 of 4 terms correct Obtain 5 2 5 2 d 5e tan 2e tan sec d − − = − + x x y x x x x A1 OE. Equate their derivative to zero, use 2 2 sec 1 tan = + x x and obtain an equation in tan x M1 Obtain 2 2tan 5tan 2 0 − + = x x A1 Allow 3 2 2tan 5tan 2tan 0 − + = x x x 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 Obtain 5 2 5 2 d 5e tan 2e tan sec d − − = − + x x y x x x x A1 OE Equate their derivative to zero and obtain an equation in sin x and cos x M1 Obtain 5cos sin 2 = x x 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

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11 With respect to the origin O, the points A and B have position vectors given by −−¿OA 2i and = −j −−¿OB j = −2k. (a) Show that OA OB and use a scalar product to calculate angle AOB in degrees. 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The midpoint of AB is M. The point P on the line through O and M is such that PA : OA 7 : 1. = (b) Find the possible position vectors of P. 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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 evaluate the inverse cosine of the result M1 Much reach an angle. The question asks for the use of 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 Question Answer Marks Guidance 11(b) State or imply M has position vector i – k B1 OE Taking a general point of OM to have position vector λi – λk, express 7 = AP OA as an equation in λ *M1 ( ) λ  their OM State a correct equation in any form A1 e.g. ( ) ( ) 2 2 2 1 7 5 λ λ −+ + + − = Reduce to 2 2 15 0 λ λ − − = 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 2 γ = OP B1 State or imply that 1 2 cos 2 5 = AOB and use cosine rule to form an equation in γ *M1 Allow 1 cos 0.632... 2 = AOB State a correct equation in any form A1 e.g. 2 2 35 5 2 2 5. 2. 5 γ γ = + − Reduce to 2 2 15 0 γ γ − − = A1 OE Solve a quadratic and state a position vector DM1 Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates Question Answer Marks Guidance 11(b) Alternative method for Question 11(b) State or imply M has position vector i – k B1 OE State or imply that 3 = AM B1 Use Pythagoras to find MP *M1 ( ) 2 35 = − MP AM Obtain 4 2 = MP A1 Correct method to find a position vector DM1 ( ) ( ) 4 − ± − i k i k Obtain answers 5i – 5k and – 3i + 3k A1 Accept coordinates 6

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