Cambridge A Level Mathematics 9709 — 2009 May/June Paper 1 · Variant 1

sin x sin x 1 Prove the identity − ≡2 tan2x. [3] 1 −sin x 1 + sin x

Mark scheme: s s 2 s 1 − = 2 B1 Correct algebra 1 − s 1 + s 1 − s Use of 1 − s² = c² M1 Use of this formula. 2 s 2 → 2 A1 Evidence of tan=sin/cos and everything c [3] completed accurately. → 2t 2

More questions on Trigonometry

2 Find the set of values of k for which the line y = kx −4 intersects the curve y = x2 −2x at two distinct points. [4]

Mark scheme: 2 kx − 4 = x² − 2x → x² − (2 + k)x + 4 = 0 M1 Complete elimination of y (or x) Use of b² − 4ac M1 Any use ( =0, <0, >0) (2 + k)² = 16 k = 2 or −6 A1 For the values of k, however used k > 2 or k < −6 A1 Correct only. [4] 5

More questions on Quadratics

3 (i) Find the first 3 terms in the expansion of (2 + 3x)5 in ascending powers of x. [3] (ii) Hence find the value of the constant a for which there is no term in x2 in the expansion of (1 + ax)(2 + 3x)5. [2]

Mark scheme: 3 (i) (2 + 3x)5 = 32 + 240x + 720x² 3 × B1 All co. [3] (ii) (1 + ax) (2 + 3x)5 → ( 1 × 720) + (a × 240) = 0 M1 Must be considering exactly 2 terms. → a = −3 A1√ √ for ( − coeff x² ÷ coeff x). [2]

More questions on Series

4 y 9 3 x O p 2p –3 The diagram shows the graph of y = a sin(bx) + c for 0 ≤x ≤2π. (i) Find the values of a, b and c. [3] (ii) Find the smallest value of x in the interval 0 ≤x ≤2π for which y = 0. [3]

Mark scheme: 4 (i) a = 6 B1 co b = 2 B1 co c = 3 B1 co [3] (ii) 6sin2x + 3 = 0 M1 Setting to 0 and attempt at making sinbx the → sin2x = −½ subject. Works with “2x” first M1 Must be evidence of ÷b 7π x = or 1.83. A1 Co (radians only) 12 [3]

More questions on Trigonometry

5 R1 A B q rad O R2 The diagram shows a circle with centre O. The circle is divided into two regions, R1 and R2, by the radii OA and OB, where angle AOB = θ radians. The perimeter of the region R1 is equal to the length of the major arc AB. (i) Show that θ = π −1. [3] (ii) Given that the area of region R1 is 30 cm2, find the area of region R2, correct to 3 significant figures. [4]

Mark scheme: 5 (i) Perimeter of R1 = r + r + rθ B1 Major arc length = 2πr − rθ B1 Equated and solved → θ = π − 1 B1 co answer was given [3] (ii) ½r²θ with θ = π − 1, equated to 30 M1 Use of correct formula once → r ² = 60/(π − 1) ( r = 5.29) A1 Any correct form for r or r² R2 = ½r ² (π + 1) → 58.0 M1A1 Attempt at r (or r²) and at area of R2 or [4] co (could be full circle − sector) [Reflex = π + 1. R2 = 30(π + 1) ÷ (π − 1)] [ M1 A1 M1 A1] GCE A/AS LEVEL – May/June 2009 9709 01

More questions on Circular measure

6 Relative to an origin O, the position vectors of the points A and B are given by −−→ −−→ OA = 2i −8j + 4k and OB = 7i + 2j −k. −−→ (i) Find the value of OA.−−→OB and hence state whether angle AOB is acute, obtuse or a right angle. [3] −−→ 2 −−→ (ii) The point X is such that AX = AB. Find the unit vector in the direction of OX. [4] 5

Mark scheme: 6 (i) OA.OB = 14 − 16 − 4 = −6 M1 A1 Must be scalar from correct method. This is −ve → Obtuse angle. B1√ co. Correct deduction from his scalar. [3] (ii) AB = 5i + 10j − 5k AX = 2 (AB) 5 OX = OA + AX M1 Needs AB and OX attempting. OX = 4i − 4j + 2k A1 co Divides by the modulus M1 Must finish with a vector, not a scalar. Unit vector = 1 (4i − 4j + 2k) A1√ Correct for his OX. 6 [4]

More questions on Vectors

7 (a) Find the sum to infinity of the geometric progression with first three terms 0.5, 0.53 and 0.55. [3] (b) The first two terms in an arithmetic progression are 5 and 9. The last term in the progression is the only term which is greater than 200. Find the sum of all the terms in the progression. [4]

Mark scheme: 7 (a) a = 0.5, r = 0.5² B1 For both a and r. Uses correct formula = 0.5÷ 0.75 M1 Uses correct formula with some a, r. → S∞ = ⅔ (or 0.667) A1 co. [3] (b) a = 5, d = 4 Uses 200 = a + (n − 1)d or T.I. M1 Attempt at finding the number of terms. 50 terms in the progression A1 co. Use of correct Sum formula M1 Correct formula (could use the last term → 5150 A1 (201)). [4] co.

More questions on Series

8 y C B A x O D (10, –3) The diagram shows points A, B and C lying on the line 2y = x + 4. The point A lies on the y-axis and AB = BC. The line from D (10, −3) to B is perpendicular to AC. Calculate the coordinates of B and C. [7]

Mark scheme: 8 m of AC = ½ Perpendicular gradient = −2 M1 Use of m1m2 = −1 Eqn BD y + 3 = −2(x − 10) M1 Correct method for eqn of line ( or y + 2x = 17) A1 In any form. Sim. eqns BD with given eqn. M1 Correct method of solution. → B (6,5) A1 co. Vector move (step) → C (12, 8) M1 A1√ Any valid method. √ for his B. [7] 2

More questions on Coordinate geometry

9 y 6 y = 3x – 2 x O 1 2 6 The diagram shows part of the curve y = 3x −2. (i) Find the gradient of the curve at the point where x = 2. [3] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis, giving your answer in terms of π. [5]

Mark scheme: 9 (i) dy/dx = − 6(3x − 2)−2 × 3 B1 M1 B1 (without the ×3). Use of chain rule. If x = 2, m = −1⅛ (−1.125) A1 co. [3] 36 d x B1 2 Attempt at π ∫ y 2 - even if π missing. (ii) Vol = ∫ (3 x − 2 ) − 36 [ ÷ 3] B1 B1 (3 x − 2) No need for π here. Use of limits [2] − [1] → 9π M1 Correct use of correct limits. A1 co. [5] GCE A/AS LEVEL – May/June 2009 9709 01

More questions on Integration

10 The function f is defined by f : x →2x2 −12x + 13 for 0 ≤x ≤A, where A is a constant. (i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (ii) State the value of A for which the graph of y = f(x) has a line of symmetry. [1] (iii) When A has this value, find the range of f. [2] The function g is defined by g : x →2x2 −12x + 13 for x ≥4. (iv) Explain why g has an inverse. [1] (v) Obtain an expression, in terms of x, for g−1(x). [3]

Mark scheme: 10 (i) 2x² − 12x + 13 = 2(x − 3)² − 5 3 × B1 Allow even if a, b, c not specifically quoted. [3] (ii) Symmetrical about x = 3. A = 6. B1√ For 2 × his (−b). [1] (iii) One limit is −5 B1√ For his c. Other limit is 13 B1 co. [2] (iv) Inverse since 1:1 (4 > 3). B1 Valid argument. [1] (v) Makes x the subject of the equation M1 Attempts to change the formula. Order of operations correct DM1 “+5”, ÷2, √, +3. Allow for simple algebraic slips such as − 5 for +5 etc. x + 5 → + 3 A1 co – as a function of x, not y. 2 [3] condone ±. dy 2

More questions on Functions

11 y A C D y = x 3 – 6x 2 + 9x x O B The diagram shows the curve y = x3 −6x2 + 9x for x ≥0. The curve has a maximum point at A and a minimum point on the x-axis at B. The normal to the curve at C (2, 2) meets the normal to the curve at B at the point D. (i) Find the coordinates of A and B. [3] (ii) Find the equation of the normal to the curve at C. [3] (iii) Find the area of the shaded region. [5]

Mark scheme: dy 11 (i) = 3x2 – 12x + 9 B1 co (can be given in part (ii)) dx dy Solves = 0 M1 Attempt to solve dy/dx = 0. dx → A (1, 4), B (3, 0). A1 Both needed. [3] (ii) If x = 2, m = −3 Normal has m = 1 M1 Use of m1m2 = −1. needs calculus. 3 Eqn y −2 = 1 (x − 2) or 3y = x + 4. M1 A1 Correct form of equation – needs calculus. 3 [3] A1 any form. (iii) area under curve – integrate y. x2 → 1 x4 − 2x3 + 9 B2,1 For the 3 terms. −1 for each error. 4 2 Limits 2 to “his 3” → ¾ (0.75) M1 Using 2 to “his 3” with integration. Area of trapezium = ½ × 1 × (2 + 2⅓) M1 Any correct method for trapezium. = 2 1 6 Subtract → shaded area of 1 5 A1 co 12 [5]

More questions on Integration