Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 3 · Variant 2

1 Solve the equation ln 1 2x 2, giving your answer correct to 3 decimal places. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: Question Answer Marks 1 Remove logarithm and obtain 1 + 2 x = e 2 B1 Use correct method to solve an equation of the form 2 x = a , where a > 0 M1 Obtain answer x = 2.676 A1 Total: 3

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

Mark scheme: 2 EITHER: State or imply non-modular inequality ( x − 4) 2 < (2(3 x + 1)) 2 , or corresponding (B1 quadratic equation, or pair of linear equations x − 4 = ±2(3 x + 1) Make reasonable solution attempt at a 3-term quadratic, or solve two linear M1 equations for x Obtain critical values x = − 65 and x = 72 A1 State final answer x < − 65 , x > 72 A1) OR: Obtain critical value x = − 65 from a graphical method, or by inspection, or by (B1 solving a linear equation or inequality Obtain critical value x = 72 similarly B2 State final answer x < − 65 , x > 72 B1) Total: 4

More questions on Algebra

2x3 (i) By sketching 4 has one positive root and one suitable graphs, show that the equation e−1 = −x2 negative root. [2] (ii) Verify by calculation that the negative root lies between and [2] −1 −1.5. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 2xn(iii) Use the iterative formula xn+1 = − 4 −e−1 to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 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Mark scheme: 3(i) − 12 x B1 Sketch a relevant graph, e.g. y = e Sketch a second relevant graph, e.g. y = 4 − x 2 , and justify the given statement B1 Total: 2 3(ii) Calculate the value of a relevant expression or values of a pair of expressions at M1 x = – 1 and x = – 1.5 complete the argument correctly with correct calculated values A1 Total: 2 3(iii) Use the iterative formula correctly at least once M1 Obtain final answer – 1.41 A1 Show sufficient iterations to 4 d.p. to justify – 1.41 to 2 d.p., or show there is a sign A1 change in the interval ( – 1.415, – 1.405) Total: 3

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5 The curve with equation y tan x, where a is a positive constant, has only one point in the interval e−ax 0 x 1 at which the tangent= is parallel to the x-axis. Find the value of a and state the exact value of<the <x-coordinate20 of this point. 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Mark scheme: 5 Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero, use Pythagoras and obtain a quadratic equation in tan x M1 Obtain tan 2 x − a tan x + 1 = 0 , or equivalent A1 Use the condition for a quadratic to have only one root M1 Obtain answer a = 2 A1 Obtain answer x = 14π A1 Total: 7

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6 The line l has equation r i 2j 2i k . The plane p has equation 3x y 20. = + −3k + , −j + + −5z = (i) Show that the line l lies in the plane p. 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(ii) A second plane is parallel to l, perpendicular to p and contains the point with position vector 3i 2k. Find the equation of this plane, giving your answer in the form ax by cz d. [5] −j + + + = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(i) Verify that the point with position vector i + 2 j − 3k lies in the plane B1 EITHER: Find a second point on l and substitute its coordinates in the equation of p (M1 Verify that the second point, e.g. (3, 1, – 2), lies in the plane A1) OR: Expand scalar product of a normal to p and the direction vector of l (M1 Verify scalar product is zero A1) Total: 3 6(ii) EITHER: Use scalar product to obtain a relevant equation in a , b and c, e.g. 2 a − b + c = 0 (B1 Obtain a second relevant equation, e.g. 3a + b − 5c = 0 , and solve for one ratio M1 e.g. a : b Obtain a : b : c = 4 : 13 : 5 , or equivalent A1 Substitute (3, – 1, 2) and the values of a, b and c in the general equation and find d M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR1: Attempt to calculate vector product of relevant vectors, e.g. (M1 (2i −+j k ) × (3i + j − 5k ) Obtain two correct components A1 Obtain correct answer, e.g. 4i + 13 j + 5k A1 Substitute (3, – 1, 2) in 4 x + 13 y + 5 z = d , or equivalent, and find d M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR2: Using the relevant point and relevant vectors form a 2-parameter equation for the (M1 plane State a correct equation, e.g. r = 3i −+j 2k + λ(2 i −+j k ) + µ(3i + j − 5k ) A1 State three correct equations in x, y, z, λand µ A1 Eliminate λand µ M1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) OR3: Using the relevant point and relevant vectors form a determinant equation for the (M1 plane x − 3 y + 1 z − 2 A1 State a correct equation, e.g. 2 −1 1 = 0 3 1 −5 Attempt to expand the determinant M1 Obtain or imply two correct cofactors A1 Obtain answer 4 x + 13 y + 5 z = 9 , or equivalent A1) Total: 5

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7 h m A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is 2 m2. At time t 0 the tank is empty and water begins to flow into it at a rate of 1 m3 per hour. At the same time water= begins to flow out from the base at a rate of 0.2 h m3 per hour, where h m is the depth of water in the tank at time t hours. (i) Form a differential equation satisfied by h and t, and show that the time T hours taken for the depth of water to reach 4 m is given by 4 10 T dh. 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(ii) Using the substitution u 5 h, find the value of T. 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Mark scheme: 7(i) dV dh B1 State or imply = 2 dt dt dV B1 State or imply = 1 − 0.2 h dt Obtain the given answer correctly B1 Total: 3 7(ii) 1 B1 State or imply d u = − d h , or equivalent 2 h Substitute for h and dh throughout M1 5 A1 20(5 − u ) Obtain T d u , or equivalent =∫ u 3 Integrate and obtain terms 100ln u − 20u , or equivalent A1 Substitute limits u = 3 and u = 5 correctly M1 Obtain answer 11.1, with no errors seen A1 Total: 6

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10 y P Q x e O The diagram shows the curve y ln x 2. The x-coordinate of the point P is equal to e, and the normal to the curve at P meets the x-axis= at Q. (i) Find the x-coordinate of Q. 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(ii) Show that ln x dx x ln x c, where c is a constant. [1] Ó = −x + ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the x-axis and the normal PQ. 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Mark scheme: 10(i) ln x B1 State or imply derivative is 2 x State or imply gradient of the normal at x = e is − 12 e , or equivalent B1 Carry out a complete method for finding the x-coordinate of Q M1 2 A1 Obtain answer x = e + , or exact equivalent e Total: 4 10(ii) Justify the given statement by integration or by differentiation B1 Total: 1 10(iii) 2 ln x M1* x. dx Integrate by parts and reach ax (ln x ) + b ∫ x Complete the integration and obtain x (ln x ) 2 − 2 x ln x + 2 x , or equivalent A1 Use limits x = 1 and x = e correctly, having integrated twice DM1 Obtain exact value e – 2 A1 1 B1 Use x- coordinate of Q found in part (i) and obtain final answer e − 2 + e Total: 5

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