Cambridge A Level Mathematics 9709 — 2012 May/June Paper 1 · Variant 1

1 Solve the equation sin 2x 2 cos 2x, for [4] = 0◦≤x ≤180◦.

Mark scheme: 1 tan 2x = 2 M1 2x = 63.4 or 243.4 A1 1 solution sufficient x = 31.7 or 121.7 (allow 122) A1A1 For 2nd A1 allow 90 + 1st soln prov. [4] only 2 solns in range. Alt methods possible

More questions on Trigonometry

7 2 Find the coefficient of x6 in the expansion of 2x3 . [4] −1x2

Mark scheme: 2 [7C3] × [(2x3)4] × [(–1/x2)3] seen soi B1B1 2 elements correct, 3rd element correct 35 × 24 × (–1)3 leading to their answer soi B1 2 elements correct. Identifying reqd ‒560(x6) as answer B1 term [4] SC B3 for [560(x)6] as answer

More questions on Series

3 A 2 cm 2 cm P R B C Q 2 cm In the diagram, ABC is an equilateral triangle of side 2 cm. The mid-point of BC is Q. An arc of a circle with centre A touches BC at Q, and meets AB at P and AC at R. Find the total area of the shaded regions, giving your answer in terms of π and √3. [5]

Mark scheme: 3 AQ (or r) = 3 B1 soi Allow 1.73 soi ft their 3 Allow 1.73 3 B1 Area ∆ = 3 (or area ∆AQC = ) 2 ft their 3 . Allow 1.57. SCA1 for π/4 1 2 π π 1 2 π 3 Area sector APR = ( 3 ) × = M1A1 from ( 3 ) × provided ∆ = 2 3 2 2 6 2 π Shaded region = 3 − oe cao A1 2 [5] 3 2 2

More questions on Trigonometry

4 A watermelon is assumed to be spherical in shape while it is growing. Its mass, M kg, and radius, r cm, are related by the formula M kr3, where k is a constant. It is also assumed that the radius is increasing at a constant rate of 0.1 centimetres= per day. On a particular day the radius is 10 cm and the mass is 3.2 kg. Find the value of k and the rate at which the mass is increasing on this day. [5]

Mark scheme: 2.3 2 4 1000k = 3.2 ⇒k = or or 0.0032 oe M1A1 1000 625  dM  2   = 3kr B1 dr   dM dM dr 2 = × used e.g. 3 × k × 10 × 1.0 M1 Must eventually make dM/dt subject dt dr dt cao. Non-calculus methods (e.g. → A1 0.096 0.09696) can score only 1st 2 marks [5] ( )2

More questions on Differentiation

5 y y = 6x + k y = 7Ö x B A x O The diagram shows the curve y and the line y 6x k, where k is a constant. The curve and 7√x the line intersect at the points A =and B. = + (i) For the case where k 2, find the x-coordinates of A and B. [4] = (ii) Find the value of k for which y 6x k is a tangent to the curve y [2] 7√x. = + =

Mark scheme: M1 Expressing as a clear quadratic soi5 (i) 6 x + 2 = 7 x ⇒ 6 ( x )2 − 7 x + 2 = 0 M1 oe e.g. (3t − 2 )(2t − 1) = 0 (3 x − 2 )(2 x − 1) = 0 2 1 x = or A1 1 solution sufficient. Accept e.g. t = 2/3 3 2 4 1 x = or (or 0.444, 0.25) A1 Both solutions required cao 9 4 OR (6 x + 2 )2 = 49 x → 36 x 2 − 25 x + 4 = 0 M1A1 Attempt to square both sides (9 x − 4 )(4 x − 1) = 0 M1 Attempt to solve (or formula etc.) 4 1 x = or (or 0.444, 0.25) oe A1 9 4 [4] (ii) 72 − 4 × 6 × k (= 0 ) M1 Apply b 2 −ac4 (= 0 ) 49 k = or 2.04 A1 Attempt to equate derivatives 24 d 1 d 7 −1 2 OR 7 x 2 = 6 M1 6 x + k ) → x ) = ( ( dx dx 2 49 49 49 x = , y = → k = or 2.04 A1 144 12 24 [2] GCE AS/A LEVEL – May/June 2012 9709 11

More questions on Quadratics

p 2 6 Two vectors u and v are such that u and v p where p is a constant. = −26 ! = 2p −11 !, + (i) Find the values of p for which u is perpendicular to v. [3] (ii) For the case where p 1, find the angle between the directions of u and v. [4] =

Mark scheme: 6 (i) 2 p 2 − 2 p + 2 + 12 p + 6 → 2 p 2 + 10 p + 8 M1 Correct method for scalar product u.v = 0 B1 Scalar product = 0 ( p + 1)( p + 4 ) = 0 → p = −1 or p = −4 A1 cao Both solutions required [3] (ii) u.v = 2 + 0 + 18 = 20 M1 Use of x1 x 2 + y1 y 2 + z1 z 2 │u│ = 41 or │v│ = 13 M1 Correct method for moduli 20 = 41 × 13 × cos θ oe M1 All connected correctly θ = 300.° or 0.523 rads A1 cao [4] 10

More questions on Vectors

7 (a) The first two terms of an arithmetic progression are 1 and cos2x respectively. Show that the sum of the first ten terms can be expressed in the form a where a and b are constants to be found. −bsin2x, [3] (b) The first two terms of a geometric progression are 1 and 1 tan2θ respectively, where 0 θ 1 3 2π. < < (i) Find the set of values of θ for which the progression is convergent. [2] (ii) Find the exact value of the sum to infinity when θ 16π. [2] =

Mark scheme: 10 7 (a) S10 = 2 M1 Correct formula with d = ± (cos2 x − )1 [2 2 + 9(cos x − 1)] S10 = [5 2 − 9 sin 2 x ] M1 Use of c 2 + s 2 = 1 in a correct S10 S10 = 10 − 45 sin 2 x A1 Or a = 10, b = 45 [3] 1 2 (b) (i) (0 < ) tan θ < 1 oe M1 Allow < 3 π (0 < ) θ < A1 cao Allow < 3 [2] 1 (ii) S ∞ = M1 1 2 π 1 − tan 3 6 9 S ∞ = or 1.125 A1 cao 8 [2]

More questions on Series

11 y 2 y = Ö(x + 1) y = 1 x O 2 The diagram shows the line y 1 and part of the curve y = = √(x + 1). 2 4 (i) Show that the equation y can be written in the form x [1] y2 −1. = √(x + 1) = 4 (ii) Find dy. Hence find the area of the shaded region. [5] ä y2 −1 (iii) The shaded region is rotated through 360◦about the y-axis. Find the exact value of the volume of revolution obtained. [5]

Mark scheme: B1 AG At least 1 step of working needed 4 11 (i) x = − 1 [1] 2 y  4   4  B1B1  (ii) ∫  − 1 dy = − − y  y 2   y  4 B1 For − , –y Upper limit = 2 y  4    − − 2  − (− 4 − 1 M1 Apply limits 1 and their 2 ‘correctly’  2  ) 2 d x − 3 → 1 SC B2 for 2( x + 1)− 1 1 A1 ∫ [5] 16 8   dy B1B1 − + 1 2   (iii) (π )∫ x 2 dy = (π )∫ y 4 y    − 16 8  (π ) 3 B1  3 y + y + y   − 16   − 16   (π )  + 4 + 2  −  + 8 + 1 M1 Apply limits 1 and their 2 ‘correctly’  24   3   5π A1 3 [5]

More questions on Integration