Cambridge A Level Mathematics 9709 — 2010 May/June Paper 1 · Variant 3

dy 6 5 The equation of a curve is such that dx = √(3x −2). Given that the curve passes through the point P (2, 11), find (i) the equation of the normal to the curve at P, [3] (ii) the equation of the curve. [4]

Mark scheme: dy 6 5 = dx 3 x − 2 (i) x = 2, tangent has gradient 3 M1 Use of mlm2 = –1 with dy/dx 1 → normal has gradient − M1 A1 Correct form of line eqn. for normal 3 1 → y − 11 = − ( x − 2 ) [3] 3 3 x − 2 B1 Without the ÷3 ÷ 3 (ii) Integrate → 6 B1 For ÷3, even if B0 above 1 2 → y = 4 3 x − 2 + c through (2,11) M1 Using (2, 11) for c A1 co → y = 4 3 x − 2 + 3 [4]

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6 Relative to an origin O, the position vectors of the points A, B and C are given by −−→OA i 4k, −−→OB 3i 2j 8k, −−→OC 10k. = −2j + = + + = −i −2j + (i) Use a scalar product to find angle ABC. [6] (ii) Find the perimeter of triangle ABC, giving your answer correct to 2 decimal places. [2]

Mark scheme: 6 OA = i – 2j + 4k, OB = 3i + 2j + 8k, OC = –i – 2j + 10k (i) (±) 2i + 4j + 4k B1 co (±) 4i + 4j – 2k B1 co M1 Needs to be scalar. AB.CB = 16 M1 For product of 2 moduli and cosine AB.CB = 36 36 cos θ θ = 63 6. ° M1 A1 All correct. [6] (ii) Perimeter = 6 + 6 + 40 M1 Correct overall method for perimeter. or 6 + 6 + 6 sin 318.° × 2 → 18.32 A1 co [2] 6

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7 C B D 10 cm O 12 cm A E F The diagram shows a metal plate ABCDEF which has been made by removing the two shaded regions from a circle of radius 10 cm and centre O. The parallel edges AB and ED are both of length 12 cm. (i) Show that angle DOE is 1.287 radians, correct to 4 significant figures. [2] (ii) Find the perimeter of the metal plate. [3] (iii) Find the area of the metal plate. [3]

Mark scheme: 6 7 (i) sin 1 2 θ = M1 Use of trig with/without radians 10 Angle DOE = 1.287 radians. A1 co – answer given. [2] (ii) P = 12 + 12 + 2 × 10 × angle BOD M1 Use of s = rθ for arc length. Angle BOD = (π – 1.287) M1 Correct angle → 61.1 A1 co [3] (iii) Sector DOE = ½ × 102 × 1.287 M1 Correct formula used with radians. Triangle DOE = ½ × 102 × sin 1.287 M1 Correct formula used with radians. Area = π × 102 – (2 sectors – 2 triangles) (or 48 + 48 + 2×½×102×(π – 1.287) M1 M1 → 281 or 282 A1 co [3] GCE AS/A LEVEL – May/June 2010 9709 13

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8 y B C (5, 4) A (–1, 2) x O D The diagram shows a rhombus ABCD in which the point A is (−1, 2), the point C is (5, 4) and the point B lies on the y-axis. Find (i) the equation of the perpendicular bisector of AC, [3] (ii) the coordinates of B and D, [3] (iii) the area of the rhombus. [3]

Mark scheme: 8 (i) Mid-point of AC = (2, 3) B1 Co Gradient of AC = 1/3 Gradient of BD = –3 M1 Use of mlm2 = –1 Equation y – 3 = –3(x – 2) A1 Co [3] (ii) If x = 0, y = 9, B (0, 9) B1√ √ on his equation. Vector move D (4, –3) M1 A1 Valid method. co. [3] (iii) AC = 40 BD = 160 M1 Correct use on either AC or BD, Area = 40 M1 A1 Full and correct method. co (or by matrix method M2 A1) [3] 4

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9 y 4 y = x + x y = 5 A B M x O 4 The diagram shows part of the curve y = x + x which has a minimum point at M. The line y = 5 intersects the curve at the points A and B. (i) Find the coordinates of A, B and M. [5] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis. [6]

Mark scheme: 4 9 y = x + x 4 (i) x + = 5 → A (1, 5), B(4, 5) B1 B1 co. co. x dy 4 = 1 − 2 M1 Differentiates. dx x = 0 when x = 2, M (2, 4). DM1 A1 Setting to 0. co. [5] (ii) Vol of cylinder = π52.3 B1 Any valid method. Vol under curve = π y 2 dx M1 Attempt at integrating y2 ∫ x 3 16 Integral = − + 8 x A2, 1, 0 Allow if no π present. 3 x Uses his limits “1 to 4” DM1 Using his limits. → 75π − 57π = 18π A1 co. [6] 2

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10 The function f : x →2x2 −8x + 14 is defined for x ∈>. (i) Find the values of the constant k for which the line y + kx = 12 is a tangent to the curve y = f(x). [4] (ii) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (iii) Find the range of f. [1] The function g : x →2x2 −8x + 14 is defined for x ≥A. (iv) Find the smallest value of A for which g has an inverse. [1] (v) For this value of A, find an expression for g−1(x) in terms of x. [3]

Mark scheme: 10 f : x a 2 x 2 −x8 + 14 (i) y + kx = 12, Sim Eqns. M1 Complete elimination of y (or x) → 2x2 – 8x + kx + 2 = 0 A1 Use of b2 – 4ac M1 Uses b2 – 4ac on eqn = 0, no “x” in a, b, c. → (k – 8)2 =16 → k = 12 or 4. A1 co.co [4] (ii) 2x2 – 8x + 14 = 2(x – 2)2 + 6 B1×3 [3] (iii) Range of f [ 6. B1√ √ for c or from calculus. [1] (iv) Smallest A = 2 B1√ √ to answer to (ii). [1] (v) Makes x the subject M1 Could interchange x, y first. Order of operations correct. M1 Order must be correct. −1 x − 6 g ( x ) = + 2 A1 co 2 [3]

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