Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 3 · Variant 2

1 Solve the inequality 2x −5 > 3 2x + 1 . [4]

Mark scheme: 1 EITHER: State or imply non-modular inequality ( 2 x − 5) 2 > ((3 2 x + 1)) 2 , or corresponding quadratic equation, or pair of linear equations ( 2 x − 5) = ±3( 2 x + )1 B1 Make reasonable solution attempt at a 3-term quadratic, or solve two linear equations for x M1 1 A1 Obtain critical values –2 and 4 1 A1 State final answer − 2 < x < 4 OR: Obtain critical value x = −2 from a graphical method, or by inspection, or by solving a linear equation or inequality B1 1 similarly B2 Obtain critical value x = 4 State final answer − 2 < x < 14 B1 [4] [Do not condone ⩽ for < ]

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2 Using the substitution u = 3x, solve the equation 3x + 32x = 33x giving your answer correct to 3 significant figures. [5]

Mark scheme: 2 State or imply 1 + u = u 2 B1 Solve for u M1 1 1( + 5 ) , or decimal in [1.61, 1.62] A1 Obtain root 2 Use correct method for finding x from a positive root M1 Obtain x = 0.438 and no other answer A1 [5]

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4 The equation x3 −x2 −6 = 0 has one real root, denoted by !. (i) Find by calculation the pair of consecutive integers between which ! lies. [2] (ii) Show that, if a sequence of values given by the iterative formula _P Q 6 xn+1 = xn + xn converges, then it converges to !. [2] (iii) Use this iterative formula to determine ! correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

Mark scheme: 4 (i) Evaluate, or consider the sign of, x 3 −x 2 − 6 for two integer values of x, or equivalent M1 Obtain the pair x = 2 and x = 3, with no errors seen A1 [2] (ii) State a suitable equation, e.g. x = ( x + ( 6 / x )) B1 Rearrange this as x 3 −x 2 − 6 = 0 , or work vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.219 A1 Show sufficient iterates to 5 d.p. to justify 2.219 to 3 d.p., or show there is a sign change in the interval (2.2185, 2.2195) A1 [3]

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9 The complex number 3 −i is denoted by u. Its complex conjugate is denoted by u*. (i) On an Argand diagram with origin O, show the points A, B and C representing the complex numbers u, u* and u* −u respectively. What type of quadrilateral is OABC? [4] u* (ii) Showing your working and without using a calculator, express in the form x + iy, where x u and y are real. [3] u* (iii) By considering the argument of , prove that u tan−1 3 = 2 tan−1 1 . 4 3

Mark scheme: 9 (i) Show u in a relatively correct position B1 Show u* in a relatively correct position B1 Show u* − u in a relatively correct position B1 State or imply that OABC is a parallelogram B1 [4] (ii) EITHER: Substitute for u and multiply numerator and denominator by 3 + i, or equivalent M1 Simplify the numerator to 8 + 6i or the denominator to 10 A1 3 4 + i , or equivalent A1 Obtain final answer 5 5 OR: Substitute for u, obtain two equations in x and y and solve for x or for y M1 3 4 or y = , or equivalent A1 Obtain x = 5 5 3 4 + i , or equivalent A1 [3] Obtain final answer 5 5 (iii) State or imply arg(u*/ u) = tan −1 ( 3 ) B1 4 ∗ ∗ / u ) = arg u − arg u M1 Substitute exact arguments in arg(u Fully justify the given statement using exact values A1 [3]

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10 y M R x O 1 p x2 The diagram shows the curve y = for x ≥0, and its maximum point M. The shaded region R 1 + x3 is enclosed by the curve, the x-axis and the lines x = 1 and x = p. (i) Find the exact value of the x-coordinate of M. [4] (ii) Calculate the value of p for which the area of R is equal to 1. Give your answer correct to 3 significant figures. [6]

Mark scheme: 10 (i) Use the quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 3 2 , or exact equivalent A1 [4] (ii) State or imply indefinite integral is of the form k ln(1 + 3x ) M1 1 ln(1 + x 3 ) A1 State indefinite integral 3 3 ) M1 Substitute limits correctly in an integral of the form k ln(1 + x 1 State or imply that the area of R is equal to ln(1 + p 3 ) − 1 ln 2 , or equivalent A1 3 3 Use a correct method for finding p from an equation of the form ln(1 + p 3 ) = a or ln((l + p 3 /) 2) = b M1 Obtain answer p = 3.40 A1 [2]

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