Cambridge A Level Mathematics 9709 — 2021 May/June Paper 3 · Variant 1

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

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

More questions on Quadratics

2ex e−x2 Find the real root of the equation 3, giving your answer correct to 3 decimal places. + 2 ex = Your working should show clearly that +the equation has only one real root. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 Reduce to a 3-term quadratic 2 6 1 0 + −= u u 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 0 > x or ( ) ln 10 3 − − is impossible Not "math error". Alternative method for Question 2 Rearrange to obtain a correct iterative formula B1 e.g. ( ) 1 ln 6 e + = − + n x nx 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

More questions on Logarithmic and exponential functions

1 4 (a) Prove that [2] −cos 21 1 cos  tan21. + 21 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 3π 1 (b) Hence find the exact value of [4] −cos 21 Ô 1 1 cos d1. 6π + 21 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) Use correct double angle formula or t-substitution twice M1 Obtain 2 1 cos2 tan 1 cos2 θ θ θ − = + from correct working A1 AG 2 Question Answer Marks Guidance 4(b) Express 2 tan θ in terms of 2 sec θ M1 ( ) 2 π 3 π 6 sec 1 d θ θ ±          Integrate and obtain terms tanθ θ − A1 Accept with a mixture of x and θ Substitute limits correctly in an integral of the form tanθ θ + a b , where 0 ≠ ab M1 3 π 1 π 3 6 3 − − +       Allow if trig. not substituted Obtain answer 2 1 3 π 3 6 − A1 or equivalent exact 2-term expression 4

More questions on Integration

5 (a) Solve the equation z2 0, where p and q are real constants. [2] −2piz −q = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ In an Argand diagram with origin O, the roots of this equation are represented by the distinct points A and B. (b) Given that A and B lie on the imaginary axis, find a relation between p and q. 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(c) Given instead that triangle OAB is equilateral, express q in terms of p. 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Mark scheme: 5(a) Use quadratic formula and 2i 1 =− M1 Obtain answers 2 i + − p q p and 2 i − − p q p A1 Accept 2 2 i 4 4 2 p p q ± − + and ISW 2 5(b) State or imply that the discriminant must be negative M1 State condition 2 < q p A1 2 Question Answer Marks Guidance 5(c) Carry out a correct method for finding a relation, e.g. use the fact that the argument of one of the roots is ( )60 ± ° M1 State a correct relation in any form, e.g. 2 − p q p ( ) 3 = ± A1 Simplify to 2 4 3 = q p A1 Alternative method for Question 5(c) Carry out a correct method for finding a relation, e.g. use the fact that the sides have equal length M1 State a correct relation in any form, e.g. ( ) 2 2 2 4 − = + − q p p q p A1 Simplify to 2 4 3 = q p A1 3

More questions on Complex numbers

6 The parametric equations of a curve are t x ln 2 3t , y . 2 3t = + = + (a) Show that the gradient of the curve is always positive. 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(b) Find the equation of the tangent to the curve at the point where it intersects the y-axis. 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Mark scheme: 6(a) Use correct chain rule or correct quotient rule to differentiate x or y M1 Obtain d d x t = 3 2 3 + t or d d y t = ( ) 2 2 2 3 + t A1 OE Use d d y x = d d y t ÷ d d x t M1 Obtain answer ( ) 2 3 2 3 + t A1 OE. Express as a simple fraction but not necessarily fully cancelled. Explain why this is always positive A1 For correct gradient. e.g. x is only defined for 2 3 0 + > t hence gradient > 0 Alternative method for Question 6(a) Form equation in x and y only M1 Obtain e 2 1 2 e 3 3 3e − −   = = −     x x x y A1 OE Differentiate M1 Obtain 2 e 3 x y − ′ = A1 OE Explain why this is always positive A1 5 Question Answer Marks Guidance 6(b) Obtain 1 3 y = − when x = 0 B1 Use a correct method to form the given tangent M1 2 3 1 3 y x + =           Obtain answer 3 2 1 = − y x A1 OE 3

More questions on Differentiation

7 y M x O a tan−1x The diagram shows the curve y and its maximum point M where x a. x = = (a) Show that a satisfies the equation @ A 2a a tan . 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(b) Verify by calculation that a lies between l.3 and 1.5. 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(c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 7(a) Use correct quotient rule or correct product rule M1 e.g. 1 2 1 1 . tan . d 1 2 d x x y x x x x − − + = Obtain correct derivative in any form A1 Equate derivative to zero and remove inverse tangent M1 Obtain 2 2 tan 1 a a a   =   +   from correct working A1 AG. Accept with x in place of a. 4 Question Answer Marks Guidance 7(b) Calculate the value of a relevant expression or pair of expressions at a = 1.3 and a = 1.5 M1 Must be using radians 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) Use the iterative process 1 + = na tan 2 2 1     +   n n a a correctly at least twice M1 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 there is a sign change in the interval (1.385, 1.395) A1 Allow recovery 3

More questions on Numerical solution of equations

1 8 With respect to the origin O, the points A and B have position vectors given by OA 2 and −−¿ = 1 ` a ` a ` a 3 2 1 OB 1 . The line l has equation r 3 . −−¿ = = 1 + , −21 −2 (a) Find the acute angle between the directions of AB and l. 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(b) Find the position vector of the point P on l such that AP BP. 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Mark scheme: 8(a) State or imply  AB = 2 1 3 − −           B1 OE. Allow ± Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. their  AB and a direction vector for l M1 ( ) 2 2 3 1 + − = Using the correct process for the moduli, divide the scalar product by the product of the moduli of the two vectors and evaluate the inverse cosine of the result M1 1 1 cos 6 14 −      Obtain answer 83.7° or 1.46 radians A1 Or answers rounding to 83.7° or 1.46 radians 4 Question Answer Marks Guidance 8(b) State or imply ±  AP and ±  BP in component form, i.e. (1 + λ, 1 – 2λ, λ) and (– 1 + λ, 2 – 2λ, 3 + λ), or equivalent B1 Form an equation in λ by equating moduli or by using cos cos = BAP ABP *M1 Obtain a correct equation in any form ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 1 1 2 1 2 2 3 λ λ λ λ λ λ + + − + = − + − + + A1 Or ( ) ( ) 2 2 1 14 4 6 13 2 2 6 λ λ λ λ λ λ + − + = − − + ( ) 3 2 83 528 207 162 0 λ λ λ − + − = 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 8i – 9j + 7k A1 Accept coordinates 5

More questions on Vectors

0. The curve has one stationary point.9 The equation of a curve is y 3 ln x for x = x−2 > (a) Find the exact coordinates of the stationary point. 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........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 8 (b) Show that y dx 18 ln 2 [5] Ó 1 = −9. 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Mark scheme: 9(a) Use correct product rule or correct quotient rule M1 Obtain correct derivative in any form A1 2 5 3 3 2 ln 3 ' x x x x y − − − = Equate 2 term derivative to zero and solve for x M1 Obtain answer 3 2 e = x A1 Or exact equivalent Obtain answer = y 3 2e A1 Or exact equivalent 5 Question Answer Marks Guidance 9(b) Commence integration and reach 1 1 3 3 1 ln . d ax x b x x x +  *M1 Obtain 1 1 3 3 1 3 ln 3 . d − x x x x x A1 Complete the integration and obtain 1 1 3 3 3 ln 9 − x x x A1 OE Use limits correctly in an expression of the form 1 1 3 3 ln + px x qx ( ) 0 pq ≠ DM1 6ln8 9 2 0 9 −× − + Obtain 18ln 2 9 − from full and correct working A1 AG need to see ln8 3ln 2 = 5

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