Cambridge A Level Mathematics 9709 — 2015 Oct/Nov Paper 1 · Variant 2

@ A2 1 1 1 −cosx 4 (i) Prove the identity −  [4] sin x tan x 1 + cosx. @ A2 1 1 2 (ii) Hence solve the equation − = for 0 ≤x ≤20. [3] sin x tan x 5

Mark scheme:  1 1   1 c  4 (i)  −  =  −  M1 Use of tan = sin/cos  sin x tan x   s s  1( − c ) 2 1( − c ) 2 = M1 Use of s² = 1 – c² s 2 1 − c 2 1( − c )(1 − c ) 1( − c ) 2 = or A1 1( − c )(1 + c ) 1( − c )(1 + c ) 1 − cos x ≡ A1 [4] ag 1 + cos x 2  1 1  2 (ii)  −  =  sin x tan x  5 1 − cos x 2 3 = → cos x M1 Making cosx the subject 1 + cos x 5 7 → x = 1.13 or 5.16 A1 A1 2π – 1st answer. [3] 6

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5 C B 0.6 rad O 6 cm A The diagram shows a metal plate OABC, consisting of a right-angled triangle OAB and a sector OBC of a circle with centre O. Angle AOB = 0.6 radians, OA = 6 cm and OA is perpendicular to OC. (i) Show that the length of OB is 7.270 cm, correct to 3 decimal places. [1] (ii) Find the perimeter of the metal plate. [3] (iii) Find the area of the metal plate. [3]

Mark scheme: 6 5 (i) Length of OB = = 7.270 M1 ag Any valid method cos 6.0 [1] (ii) AB = 6tan0.6 or 4.1 B1 Sight of in (ii) Arc length = 7.27 × (½π – 0.6) = (7.06) M1 Use of s= rθ with sector angle Perimeter = 6 + 7.27 + 7.06 + 6tan0.6 = 24.4 A1 [3] (iii) Area of AOB = ½ × 6 × 7.27 × sin0.6 M1 Use of any correct area method Area of OBC = ½ × 7.27² × (½π – 0.6) M1 Use of ½r²θ. → area = 12.31 + 25.65 = 38.0 A1 [3]

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6 Points A, B and C have coordinates A −3, 7 , B 5, 1 and C −1, k , where k is a constant. (i) Given that AB = BC, calculate the possible values of k. [3] The perpendicular bisector of AB intersects the x-axis at D. (ii) Calculate the coordinates of D. [5]

Mark scheme: 6 A(−3, 7), B(5, 1) and C(−1, k) (i) AB = 10 B1 6² + (k – 1)² = 10² M1 Use of Pythagoras k = −7 and 9 A1 [3] 4 (ii) m of AB = −¾ m perp = B1 M1 B1 M1 Use of m1m2 = −1 3 M = (1, 4) 4 Eqn y − 4 = ( x − )1 B1 3 Set y to 0, → x = –2 M1 A1 Complete method leading to D. [5] 0  2  3       

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