Cambridge A Level Mathematics 9709 — 2014 Oct/Nov Paper 1 · Variant 2

1 y 2, 5 y = x2 + 1 0, 1 x O The diagram shows part of the curve y = x2 + 1. Find the volume obtained when the shaded region is rotated through 360 about the y-axis. [4]

Mark scheme: 1 Vol = ( π) ∫x²dy = (π) ∫ (y − 1) dy M1 Use of ∫x ² – not ∫y² – ignore π 2 A1 co 1 2 ( y − 1) Integral is y − y or B1 Sight of an integral sign with 1 and 5 2 2 Limits for y are 1 to 5 → 8π or 25.1(AWRT) A1 co [4] (no π max 3/4) 5

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2 P B Q 5 cm A O 12 cm The diagram shows a triangle AOB in which OA is 12 cm, OB is 5 cm and angle AOB is a right angle. Point P lies on AB and OP is an arc of a circle with centre A. Point Q lies on AB and OQ is an arc of a circle with centre B. (i) Show that angle BAO is 0.3948 radians, correct to 4 decimal places. [1] (ii) Calculate the area of the shaded region. [5]

Mark scheme: 5 2 (i) tanθ = M1 Any valid trig method ag 12 → ( θ = 0.3948 ) [1] (ii) Other angle in triangle = − π – 0.3948 B1 Unsimplified OK Area of triangle AOB = ×12×5 (= 30) B1 co Use of r²θ once M1 With θ in radians and r = 5 or 12 Shaded area = sector + sector – triangle = ×12²×0.3948 + 5²θ – 30 DM1 Sum of 2 sectors – triangle or any other valid method using the given angle and a different one. = 28.43 + 14.70 – 30 = 13.1 A1 co [5] 5

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3 (i) Find the first 3 terms, in ascending powers of x, in the expansion of 1 + x 5. [2] 5 The coefficient of x2 in the expansion of 1 + px + x2 is 95. (ii) Use the answer to part (i) to find the value of the positive constant p. [3]

Mark scheme: 3 (i) (1 + x 5) = 1 + 5x + 10x² B2,1 Loses 1 for each error [2] (ii) 1( + px + x 2) 5 (1+) 5(px + x²) + 10(px + x²)2 M1 Replace x by (px + x²) in their expansion Coeff of x² = 5 + 10p² DM1 Considers 2 terms = 95 → p = 3 A1 co – no penalty for ±3 [3] 12

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12 4 A curve has equation y = 3 −2x. dy (i) Find dx. [2] A point moves along this curve. As the point passes through A, the x-coordinate is increasing at a rate of 0.15 units per second and the y-coordinate is increasing at a rate of 0.4 units per second. (ii) Find the possible x-coordinates of A. [4]

Mark scheme: 12 4 y = 3 − 2 x (i) Differential = −12(3 – 2x)−2 × −2 B1 B1 co co (even if 1st B mark lost) [2] d y d y d x (ii) = ÷ = 0.4 ÷ 0.15 M1 Chain rule used correctly (AEF) d x d t d t 24 8 dy 8 3 → = M1 Equates their with their or 2 (3 − 2 x ) 3 dx 3 8 → x = 0 or 3 A1 A1 co co [4]

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5 (i) Show that the equation 1 + sin x tan x = 5 cosx can be expressed as 6 cos2x −cosx −1 = 0. (ii) Hence solve the equation 1 + sin x tan x = 5 cosx for 0 ≤x ≤180 . [3]

Mark scheme: 5 1 + sinxtanx = 5cosx (i) Replaces t by s/c M1 Correct formula 2s 1 + = 5c c Replace s² by 1 − c² M1 Correct formula used in appropriate place → 6c² − c − 1 (= 0) A1 AG [3] (ii) Soln of quadratic → (c = −⅓ or ½) M1 Correct method → x = 60° or 109.5° A1 A1 co co [3] 3 2

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6 The equation of a curve is y = x3 + ax2 + bx, where a and b are constants. (i) In the case where the curve has no stationary point, show that a2 < 3b. [3] (ii) In the case where a = −6 and b = 9, find the set of values of x for which y is a decreasing function of x. [3]

Mark scheme: 6 y = x3 + ax2 + bx dy (i) = 3x² + 2ax + b B1 co dx dy (ii) b² − 4ac = 4a² − 12b (I 0) M1 Use of discriminant on their quadratic dx or other valid method → a² I= 3b A1 co – answer given [3] (iii) y = x³ − 6x² + 9x dy = 3x² − 12x + 9 I 0 M1 Attempt at differentiation dx = 0 when x = 1 and 3 A1 co → 1 I x I 3 A1 condone < [3]

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7 X M 10 C B k D j 8 O i A The diagram shows a pyramid OABCX. The horizontal square base OABC has side 8 units and the centre of the base is D. The top of the pyramid, X, is vertically above D and XD = 10 units. The −−→ −−→ mid-point of OX is M. The unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. −−→ −−→ (i) Express the vectors AM and AC in terms of i, j and k. [3] (ii) Use a scalar product to find angle MAC. [4]

Mark scheme: 7 (i) AM = −6i + 2j + 5k B2,1 co −1 each error AC = −8i + 8j B1 co [3] (ii) AM.AC = 48 + 16 = 64 M1 Use of x1y1 + etc. with suitable vectors 64 = √128√65cosθ M1 M1 Product of moduli. Correct link. → θ = 45.4° A1 co [4]

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8 (a) The sum, Sn, of the first n terms of an arithmetic progression is given by Sn = 32n −n2. Find the first term and the common difference. [3] (b) A geometric progression in which all the terms are positive has sum to infinity 20. The sum of the first two terms is 12.8. Find the first term of the progression. [5] [Questions 9, 10 and 11 are printed on the next page.]

Mark scheme: 8 (a) Sn = 32n − n². Set n to 1, a or S1 = 31 B1 co Set n to 2 or other value S2 = 60 → 2nd term = 29 → d = − 2 M1 A1 Correct method. (or equates formulae – compares co coeffs n², n) [3] [M1 comparing, A1 d A1 a] [M1 only when coeffs compared] a a 1( − r ) 2 (b) = 20 , , or a + ar = 12.8 B1 B1 co co 1 −r 1 − r a Elimination of or a or r M1 ‘Correct’ elimination to form equation in a 1 − r or r → (r = 0.6) → a = 8 DM1 A1 Complete method leading to a = [5] Condone a = 8 and 32 − 9

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9 y D A 2, 6 C 8, 3 x O B 5, −3 The diagram shows a trapezium ABCD in which AB is parallel to DC and angle BAD is 90 . The coordinates of A, B and C are 2, 6 , 5, −3 and 8, 3 respectively. (i) Find the equation of AD. [3] (ii) Find, by calculation, the coordinates of D. [3] The point E is such that ABCE is a parallelogram. (iii) Find the length of BE. [2]

Mark scheme: 9 9 (i) mAB = −3 or B1 oe 3 1 mAD = M1 use of m1m2 = −1 with grad AB 3 1 Eqn AD y – 6 = (x – 2) or 3y = x + 16 A1 co – OK unsimplified 3 [3] (ii) Eqn CD y – 3 = −3(x – 8) or y = −3x + 27 B1 OK unsimplified. on m of AB. Sim Eqns M1 Reasonable algebra leading to x = or y = with AD and CD → D (6½, 7½) A1 [3] (iii) Use of vectors or mid-point → E (5, 12) or mid-point (5,4.5) B1 May be implied Length of BE = 15 B1 co [2] d 2 y 24

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