Cambridge A Level Mathematics 9709 — 2009 Oct/Nov Paper 1 · Variant 2

5 (i) Prove the identity (sin x + cos x)(1 −sin x cos x) ≡sin3x + cos3x. [3] (ii) Solve the equation (sin x + cos x)(1 −sin x cos x) = 9 sin3x for 0◦≤x ≤360◦. [3]

Mark scheme: 5 (i) (sin x + cos x)(1 – sin x cos x) = sin x + cos x – sin2 x cos x – cos2 x sin x M1 Needs 4 terms from the product. sin2 x = 1 – cos2 x and cos2 x = 1 – sin2 x M1 Needs to be used once. → sin3 x + cos3 x A1 All ok. [3] (ii) (sin x + cos x)(1 – sin x cos x) = 9 sin3 x Uses part (i) → 8 sin3 x = cos3 x → tan3 x = 1 → tan x = ½ M1 Uses tan x = sin x ÷ cos x → tan3x = k. 8 → x = 26.6º and 206.6º A1 B1√ Co. √ for 180º + first answer and providing [3] there are no other answers in range.

More questions on Trigonometry

6 G F Q D E C B k j P O A i In the diagram, OABCDEFG is a cube in which each side has length 6. Unit vectors i, j and k are parallel to −−→OA, −−→OC and −−→OD respectively. The point P is such that −−→AP 1 −−→AB and the point Q is the = 3 mid-point of DF. (i) Express each of the vectors −−→OQ and −−→PQ in terms of i, j and k. [3] (ii) Find the angle OQP. [4]

Mark scheme: 6 (i) OQ = 3i + 3j + 6k B1 co PQ = −3i + j + 6k B2, 1 Loses one for each error. [3] (ii) (3i + 3j + 6k).( −3i + j + 6k) = −9 + 3 + 36 = 30 M1 Use of x1x2 + y1y2 + z1z2 co. 30 = √54√46cosθ M1 Correct method for modulus (once) and all θ = 53.0º M1A1 correctly linked. co. [4] Cosine rule M1 modulus nb QO .QP can gain 4/4. M1 attempt at 3 sides but OQ . PO can only gain 3/4. M1 A1 answer. Use of other vectors (e.g. OP .OQ ) M3 ok.

More questions on Vectors

7 P r cm q rad O Q A piece of wire of length 50 cm is bent to form the perimeter of a sector POQ of a circle. The radius of the circle is r cm and the angle POQ is θ radians (see diagram). (i) Express θ in terms of r and show that the area, A cm2, of the sector is given by A = 25r −r2. [4] (ii) Given that r can vary, find the stationary value of A and determine its nature. [4]

Mark scheme: 7 (i) 2r + rθ = 50 M1 Must use s = rθ and link with perimeter 1 θ = (50 – 2r) A1 co r A = 1 r2θ M1 Used with θ as f(r) 2 → A = 25r – r2 A1 co (answer given) [4] d A (ii) = 25 − 2 r B1 co dr = 0 when r = 12.5 M1 sets differential to 0 + solution A = 156¼ A1 co 2nd differential negative → Maximum B1 Could be quoted directly from quadratic. [4] GCE A/AS LEVEL – October/November 2009 9709 12 3

More questions on Circular measure

3 8 The function f is such that f(x) = 2x 5 for x ∈>, x ≠−2.5. + (i) Obtain an expression for f ′(x) and explain why f is a decreasing function. [3] (ii) Obtain an expression for f −1(x). [2] (iii) A curve has the equation y = f(x). Find the volume obtained when the region bounded by the curve, the coordinate axes and the line x = 2 is rotated through 360◦about the x-axis. [4]

Mark scheme: 3 8 x a 2 x + 5 (i) fV(x) = –3(2x + 5)–2 × 2 B1 B1 B1 for –3(2x + 5)–2. B1 for ×2 fV(x) is negative → decreasing B1√ √ providing bracket is squared. [3] (using value or values only B0) 3 3 (ii) y = → 2 x + 5 = M1 Attempt at making x the subject. 2 x + 5 y –1 1  3  3− 5 x → f (x) =  −5  or A1 co including f(x) not f(y) 2  x  2 x [2] 9 2 (iii) ∫ π ( 2 x + 5) dx B1 For –9(2x + 5)–1 = (–9π(2x + 5)–1 ÷ 2) B1 For ÷ 2 in ∫ of y2 Limits 0 to 2 → π (−½ − −0.9) M1 Use of correct limits with ∫ of y2. → = 0.4π (or 1.26) A1 co [4]

More questions on Integration

9 y C (12, 14) B D O x A (0, –2) The diagram shows a rectangle ABCD. The point A is (0, −2) and C is (12, 14). The diagonal BD is parallel to the x-axis. (i) Explain why the y-coordinate of D is 6. [1] The x-coordinate of D is h. (ii) Express the gradients of AD and CD in terms of h. [3] (iii) Calculate the x-coordinates of D and B. [4] (iv) Calculate the area of the rectangle ABCD. [3]

Mark scheme: 9 (i) y-coordinate same as the B1 co y-coordinate of the mid-point of [1] AC. 8 h − 12 (ii) m of AD = or M1 A1 any use of y-step ÷ x-step for M mark h 8 8 − h m of CD = or A1 co 12 − h 8 [3] nb AC = 20, M(6, 6) MD = 10 → D(16, 6) and B(−4, 6) (iii) Product of gradients = −1 M1 Used correctly with the two gradients → h2 – 12h – 64 = 0 M1 Forming a quadratic equation → h = 16 or − 4 DM1A1 Solution of equation. co so xD = 16 and xB = –4 [4] or Pyth h2 + 82 + 82 + (12 – h)2 = 400 (iv) Area = √320 × √80 M1 M1 M1 for method for one of the lengths → 160 A1 M1 for base × height. co [3] ( or Area = 2 × area of a triangle with base = BD, → 2 × ½ × 20 × 8 = 160) (or matrix method) GCE A/AS LEVEL – October/November 2009 9709 12 2

More questions on Coordinate geometry

10 y y = x 2 – 4x + 7 2y = x + 5 B A x O (i) The diagram shows the line 2y x 5 and the curve y x2 7, which intersect at the points = + = −4x + A and B. Find (a) the x-coordinates of A and B, [3] (b) the equation of the tangent to the curve at B, [3] (c) the acute angle, in degrees correct to 1 decimal place, between this tangent and the line 2y = x + 5. [3] (ii) Determine the set of values of k for which the line 2y = x + k does not intersect the curve y x2 7. [4] = −4x +

Mark scheme: 10 (i) (a) 2y = x + 5, y = x2 – 4x + 7 Sim equations → 2x2 – 9x + 9 = 0 M1 Complete elimination of x or y → x = 3 or x = 1½. DM1 A1 Correct method for quadratic. co. [3] dy (b) = 2 x − 4 B1 co dx → y – 4 = 2(x – 3) M1 A1 Correct form of eqn with m numeric. co [3] nb use of y + 4 or x, y interchanged M1 A0 (c) m = 2 → angle of 63.4º m = ½ → angle of 26.6º M1 Finds angle with x-axis once. → angle between = 37º M1A1 Subtracts two angles. co. [3] (i+2j).(2i+j) → 4=√5√5cosθ M1M1A1 or use of tan(A–B) M2A1 or Cosine rule with 3 sides found. (ii) y = x2 – 4x + 7 2y = x + k Sim eqns → 2x2 – 9x + 14 – k = 0 M1 A1 Eliminates y or x completely. Co (= 0) Uses b2 – 4ac, 81 − 8(14 − k) M1 Uses b2 – 4ac = 0, or < 0 or > 0 Key value is k = 3.875 or 31/8. k < 3.875 A1 Co condone Y. [4]

More questions on Quadratics