Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 3 · Variant 3

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

Mark scheme: 1 2 2 2 3 1 + > − x x , or corresponding quadratic equation, or pair of linear equations 2(x + 2) = ± (3x – 1) B1 Make reasonable attempt at solving a 3-term quadratic, or solve two linear equations for x M1 Obtain critical values 3 5 = − x and x = 5 A1 State final answer 3 5 5 − < < x A1 Alternative method for question 1 Obtain critical value x = 5 from a graphical method, or by inspection, or by solving a linear equation or an inequality B1 Obtain critical value 3 5 = − x similarly B2 State final answer 3 5 5 − < < x B1 4

More questions on Quadratics

32x 3 Showing all necessary working, solve the equation + 3−x 4. Give your answer correct to 32x = −3−x 3 decimal places. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Reduce the equation to a horizontal equation in M1 Simplify and reach 3( 33 ) 5 = x , ( ) 3 27 5 = x , or equivalent A1 Use correct method for finding x from a positive value of 33 x , 3 1 3 + x or 27x M1 Obtain answer x = 0.155 A1 4

More questions on Logarithmic and exponential functions

6 Throughout this question the use of a calculator is not permitted. The complex number with modulus 1 and argument 1 is denoted by w. 30 (i) Express w in the form x iy, where x and y are real and exact. 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The complex number 1 2i is denoted by u. The complex number v is such that v 2 u and 1 + = arg v arg u = + 30. (ii) Sketch an Argand diagram showing the points representing u and v. [2] (iii) Explain why v can be expressed as 2uw. Hence find v, giving your answer in the form a ib, + where a and b are real and exact. 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Mark scheme: 6(i) Obtain answer 1 3 i 2 2 = + w B1 1 6(ii) Show point representing u B1 Show point representing v in relatively correct position B1 2 6(iii) Explain why the moduli are equal B1 Explain why the arguments are equal B1 Use 2i 1 =− and obtain 2uw in the given form M1 Obtain answer ( ) 1 2 3 2 3 i − + + A1 4

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7 The plane m has equation x 4y 2. The plane n is parallel to m and passes through the point + −8z = P with coordinates 5, 2, . −2 (i) Find the equation of n, giving your answer in the form ax by cz d. 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(ii) Calculate the perpendicular distance between m and n. 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(iii) The line l lies in the plane n, passes through the point P and is perpendicular to OP, where O is the origin. Find a vector equation for l. 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Mark scheme: 7(i) Substitute coordinates (5, 2, – 2) in x + 4y – 8z = d M1 Obtain plane equation x + 4y – 8z = 29, or equivalent A1 2 7(ii) Attempt to use perpendicular formula to find perpendicular from (5, 2, – 2) to m M1 Obtain a correct unsimplified expression, e.g. ( ) 5 8 16 2 1 16 64 + + − + + A1 Obtain answer 3 A1 Alternative method 1 for question 7(ii) State or imply perpendicular from O to m is 2 9 or from O to n is 29 9 B1 Find difference in perpendiculars M1 Obtain answer 3 A1 Alternative method 2 for question 7(ii) Obtain correct parameter value, or position vector or coordinates of the foot of the perpendicular from (5, 2, – 2) to m, e.g. µ = 1 3 ± ; 14 2 2 , , 3 3 3       B1 Calculate the length of the perpendicular M1 Obtain answer 3 B1 3 Question Answer Marks Guidance 7(iii) Calling the direction vector ai + bj + ck, use a scalar product to form a relevant equation in a, b and c, e.g. a + 4b – 8c = 0 or 5a + 2b – 2z = 0 B1 Solve two relevant equations for the ratio a : b : c M1 Obtain a : b : c = 4 : – 19: – 9 A1 OE State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE Alternative method for question 7(iii) Attempt to calculate vector product of two relevant vectors, e.g. (i + 4j – 8k)×(5i + 2j – 2k) M1 Obtain two correct components A1 Obtain 8i – 38j – 18k A1 OE State answer r = 5i + 2j – 2k + λ(4i – 19j – 9k) A1 OE 4

More questions on Vectors

8 y x O 120 The diagram shows the graph of y sec x for 0 1 = ≤x < 20. 1.2 (i) Use the trapezium rule with 2 intervals to estimate the value of sec x dx, giving your answer Ó 0 correct to 2 decimal places. 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(ii) Explain, with reference to the diagram, whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i). 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(iii) P is the point on the part of the curve y sec x for 0 1 at which the gradient is 2. By first = ≤x < 20 1 differentiating cosx, find the x-coordinate of P, giving your answer correct to 3 decimal places. 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Mark scheme: 8(i) State or imply ordinates 1, 1.2116…, 2.7597... B1 Use correct formula, or equivalent, with h = 0.6 M1 Obtain answer 1.85 A1 3 8(ii) Explain why the rule gives an overestimate B1 1 8(iii) Differentiate using quotient or chain rule M1 Obtain correct derivative in terms of sin x and cos x A1 Equate derivative to 2, use Pythagoras and obtain an equation in sin x M1 Obtain 2 2sin sin 2 0 + − = x x A1 OE Solve a 3-term quadratic for x M1 Obtain answer x = 0.896 only A1 6

More questions on Differentiation

20 40 109 The variables x and t satisfy the differential equation 5dx . It is given that x −x −x = dt = when t 0. = (i) Using partial fractions, solve the differential equation, obtaining an expression for x in terms of t. 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(ii) State what happens to the value of x when t becomes large. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 9(i) Separate variables correctly and integrate one side B1 Obtain term 0.2t, or equivalent B1 Carry out a relevant method to obtain A and B such that ( )( ) 1 20 40 − − x x ≡ 20 − A x + 40 − B x *M1 OE Obtain A = 1 20 and B = 1 20 − A1 Integrate and obtain terms ( ) ( ) 1 1 ln 20 ln 40 20 20 − − + − x x OE A1FT +A1FT The FT is on A and B Use x = 10, t = 0 to evaluate a constant, or as limits DM1 Obtain correct answer in any form A1 Obtain final answer = x 4 4 60e 40 3e 1 − − t t A1 OE 9 9(ii) State that x approaches 20 B1 1

More questions on Differential equations

10 y M R x 0 O The diagram shows the graph of y ecos x sin3x for 0 and its maximum point M. The shaded = ≤x ≤0, region R is bounded by the curve and the x-axis. (i) Find the x-coordinate of M. Show all necessary working and give your answer correct to 2 decimal places. 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(ii) By first using the substitution u cosx, find the exact value of the area of R. 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Mark scheme: 10(i) Use product rule and chain rule at least once M1 Obtain correct derivative in any form A1 Equate derivative to zero, use Pythagoras and obtain an equation in cos x M1 Obtain 2 cos 3cos 1 0 + −= x x , or 3-term equivalent A1 Obtain answer x = 1.26 A1 5 10(ii) Using du = ± sin x dx express integrand in terms of u and du M1 Obtain integrand ( ) 2 e 1 − u u A1 OE Commence integration by parts and reach ( ) 2 e 1 e d − + ∫ u u a u b u u *M1 Obtain ( ) 2 e 1 2 e d − − ∫ u u u u u A1 OE Complete integration, obtaining ( ) 2 e 2 1 − + u u u A1 OE Substitute limits u = 1 and u = – 1 (or x = 0 and x = π ), having integrated completely DM1 Obtain answer 4 e , or exact equivalent A1 7

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