Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 2 · Variant 3

1 Find the gradient of the curve y at the point where x 4. [3] = ln(5x + 1) =

Mark scheme: k 1 1 Obtain derivative of the form , where k = 1, 5 or M1 5 x + 1 5 5 Obtain correct derivative A1 5 x + 1 5 Substitute x = 4 into expression for derivative and obtain A1√ [3] 21 2 2

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2 Solve the inequality [4] |2x −3| ≤|3x|.

Mark scheme: 2 EITHER State or imply non-modular inequality (2x – 3)2 Y (3x)2, or corresponding equation or pair of linear equations M1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations M1 3 Obtain critical values –3 and A1 5 3 State correct answer x Y –3 or x [ A1 5 OR State one critical value, e.g. x = –3, by solving a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B2 3 State correct answer x Y –3 or x [ B1 [4] 5 2

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3 Solve the equation 2 x [5] ln(x + 3) −ln = ln(2x −2).

Mark scheme: 3 Use 2 ln(x + 3) = ln(x + 3)2 M1 Use law for addition or subtraction of logarithms M1 Obtain correct quadratic expression in x A1 Make reasonable solution attempt at a 3-term quadratic M1 State x = 9 and no other solutions (condone x = –1 not deleted) A1 [5] 1 1

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4 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 16π 1 1 sin dx 1 √3 8 12π 4. + + [5] ã 0 (cos2x + 2x) =

Mark scheme: 1 1 4 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2x, where ab ≠ 0, correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 1 Obtain correct integral (for sin 2x term) of − cos 2 x B1 2 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [5]

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5 Solve the equation 5 sec22θ tan 2θ 9, giving all solutions in the interval [6] = + 0◦≤θ ≤180◦.

Mark scheme: 5 Use trig identity correctly to obtain a quadratic in tan 2θ M1 Solve the quadratic correctly M1 4 Obtain tan 2θ = 1 or – A1 5 Obtain one correct answer A1 Carry out correct method for second answer from either root M1 Obtain remaining 3 answers from 22.5°, 112.5°, 70.7°, 160.7° and no others in the range A1 [Ignore answers outside the given range] [6] GCE AS/A LEVEL – October/November 2011 9709 23

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6 (i) The polynomial x4 ax3 bx 2, where a and b are constants, is denoted by It is + −x2 + + p(x). given that and are factors of Find the values of a and b. [5] (x −1) (x + 2) p(x). (ii) When a and b have these values, find the quotient when is divided by x2 x [3] p(x) + −2.

Mark scheme: 6 (i) Substitute x = 1 or x = –2 and equate to zero M1 Obtain a correct equation in any form with powers of x values calculated A1 Obtain a second correct equation in any form A1 Solve a relevant pair of equations for a or for b M1 Obtain a = 3 and b = –5 A1 [5] (ii) Attempt division by x2 + x – 2, or equivalent, and reach a partial quotient of x2 + kx M1 Obtain partial quotient x2 + 2x A1 Obtain x2 + 2x – 1 with no errors seen A1 S.C. M1A1√ if ‘a’ and/or ‘b’ incorrect [3] 1 x

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7 y x O P 1 The diagram shows the curve y The curve has a gradient of 3 at the point P. = (x −4)e 2x. (i) Show that the x-coordinate of P satisfies the equation 2 x = + 6e −12x. [4] (ii) Verify that the equation in part (i) has a root between x 3.1 and x 3.3. [2] = = 2xn 2 6e−1 to determine this root correct to 2 decimal places. + (iii) Use the iterative formula xn+1 = Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) At any stage, state the correct derivative of e 2 B1 Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to 3 and obtain given equation correctly A1 [4] 1 − x (ii) Consider sign of 2 + 6e 2 – x, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.21 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (3.205, 3.215) B1 [3] dy 2

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8 The equation of a curve is 2x2 y2 6. −3x −3y + = dy 4x (i) Show that −3 [3] dx = 3 −2y. (ii) Find the coordinates of the two points on the curve at which the gradient is [6] −1.

Mark scheme: dy 8 (i) State 2 y as derivative of y2, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [3] (ii) Equate gradient expression to –1 and rearrange M1 Obtain y = 2x A1 Substitute into original equation to obtain an equation in x2 (or y2) M1 Obtain 2x2 – 3x – 2 = 0 (or y2 – 3y – 4 = 0) A1 Correct method to solve their quadratic equation M1 State answers (– 1 2 , –1) and (2, 4) A1 [6]

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