Cambridge A Level Mathematics 9709 — 2010 May/June Paper 2 · Variant 1

1 Solve the inequality 5. [3] |2x −3| >

Mark scheme: 1 EITHER: State or imply non-modular inequality (2x – 3)2 > 52 , or corresponding equation or pair of linear equations M1 Obtain critical values –1 and 4 A1 State correct answer x < –1, x > 4 A1 OR: State one critical value, e.g. x = 4, having solved a linear equation (or inequality) or from a graphical method or by inspection B1 State the other critical value correctly B1 State correct answer x < –1, x > 4 B1 [3]

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6 1 2 Show that dx 2 ln 2. [4] x 2 = ä 0 +

Mark scheme: 2 Obtain integral ln(x + 2) B1 Substitute correct limits correctly M1 Use law for the logarithm of a product, a quotient or a power M1 Obtain given answer following full and correct working A1 [4]

More questions on Integration

3 (i) Show that the equation 6 tan x can be written in the form tan(x + 45◦) = 6 tan2x tan x 1 0. −5 + = [3] (ii) Hence solve the equation [3] 6 tan x, for x 0◦< tan(x + 45◦) = < 180◦.

Mark scheme: 3 (i) Use tan(A ± B) formula to obtain an equation in tan x M1 Use tan 45° = 1 and obtain a correct equation in any form A1 Obtain the given equation correctly A1 [3] (ii) Solve the given quadratic in tan x and evaluate an inverse tangent M1 Obtain a correct answer, e.g. 18.4° A1 Obtain second answer, e.g. 26.6°, and no others in the given interval A1 [3] [Treat the giving of answers in radians as a misread. Ignore answers outside the given interval.] 2

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4 The polynomial x3 3x2 4x 2 is denoted by + + + f(x). (i) Find the quotient and remainder when is divided by x2 x [4] f(x) + −1. (ii) Use the factor theorem to show that is a factor of [2] (x + 1) f(x).

Mark scheme: 4 (i) Commence division by x2 + x – 1 obtaining quotient of the form x + k M1 Obtain quotient x + 2 A1 Obtain remainder 3x + 4 A1 Identify the quotient and remainder correctly A1√ [4] (ii) Substitute x = –1 and evaluate expression M1 Obtain answer 0 A1 [2] 1 1

More questions on Algebra

5 (i) Given that y 2x, show that the equation = 2x 4 + 3(2−x) = can be written in the form y2 3 0. −4y + = [3] (ii) Hence solve the equation 2x 4, + 3(2−x) = giving the values of x correct to 3 significant figures where appropriate. [3]

Mark scheme: 1 5 (i) State or imply 2–x = , or 2–x = y–1 B1 y Substitute and obtain a 3-term quadratic in y M1 Obtain the given answer correctly A1 [3] (ii) Solve the given quadratic and carry out correct method for solving an equation of the form 2x = a, where a > 0 M1 Obtain answer x = 1.58 or 1.585 A1 Obtain answer x = 0 B1 [3] GCE AS/A LEVEL – May/June 2010 9709 21 2 dy 2

More questions on Logarithmic and exponential functions

6 The equation of a curve is x2y y2 6x. + = dy 6 (i) Show that . [4] −2xy dx = x2 2y + (ii) Find the equation of the tangent to the curve at the point with coordinates giving your answer in the form ax by c 0. (1, 2), [3] + + =

Mark scheme: dy 6 (i) State 2xy + x2 as derivative of x2y B1 dx dy State 2y as derivative of y2 B1 dx dy Equate derivatives of LHS and RHS, and solve for M1 dx Obtain given answer A1 [4] (ii) Substitute and obtain gradient 2 , or equivalent B1 5 Form equation of tangent at the given point (1, 2) M1 Obtain answer 2x – 5y + 8 = 0, or equivalent A1 [3] [The M1 is dependent on at least one of the B marks being obtained.]

More questions on Differentiation

7 (i) By sketching a suitable pair of graphs, show that the equation e2x 2 = −x has only one root. [2] (ii) Verify by calculation that this root lies between x 0 and x 0.5. [2] = = (iii) Show that, if a sequence of values given by the iterative formula 1 2 xn+1 = ln(2 −xn) converges, then it converges to the root of the equation in part (i). [1] (iv) Use this iterative formula, with initial value x1 0.25, to determine the root correct to 2 decimal = places. Give the result of each iteration to 4 decimal places. [3]

Mark scheme: 7 (i) Make a recognisable sketch of a relevant graph, e.g. y = 2 – x B1 Sketch an appropriate second graph, e.g. y = e2x, and justify the given statement B1 [2] (ii) Consider sign of e2x – (2 – x) at x = 0 and x = 0.5, or equivalent M1 Complete the argument correctly with correct calculations A1 [2] (iii) Show that e2x = 2 – x is equivalent to x = 1 ln(2 – x), or vice versa B1 [1] 2 (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.27 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.265, 0.275) A1 [3]

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cos x dy 8 (i) By differentiating , show that if y cot x then [3] sin x dx = = −cosec2x. (ii) By expressing cot2x in terms of cosec2x and using the result of part (i), show that 12π cot2x dx 1 4π. 1 ã 4π = −1 [4] 1 1 (iii) Express cos 2x in terms of sin2x and hence show that can be expressed as 2 cosec2x. 1 2x Hence, using the result of part (i), find −cos 1 dx. ä 1 2x [3] −cos

Mark scheme: 8 (i) Use quotient rule M1 Obtain correct derivative in any form A1 Obtain given result correctly A1 [3] (ii) State cot2 x ≡ –1 + cos ec2x , or equivalent B1 Obtain integral –x – cotx (f.t. on signs in the identity) B1√ Substitute correct limits correctly M1 Obtain given answer A1 [4] 1 (iii) Use trig formulae to convert integrand to 2 where k = ±2, or ±1 M1 k sin x Obtain given answer 1 cos ec2x correctly A1 2 Obtain answer – 1 cot x + c, or equivalent B1 [3] 2

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