Cambridge A Level Mathematics 9709 — 2015 May/June Paper 1 · Variant 3

4 (i) Express the equation 3 sin 1 = cos 1 in the form tan 1 = k and solve the equation for 0Å < 1 < 180Å. [2] (ii) Solve the equation 3 sin22x = cos22x for 0Å < x < 180Å. [4]

Mark scheme: 4 (i) tan θ = 1 / 3 M1 θ = 184.° only A1 Ignore solns. outside range 0→180 [2] (ii) tan 2 x = ( ± 1) / 3 Must be sq. root soi M1 sin 2 x = ( ± ) 1 / 2 or cos 2 x = ( ± ) 3/2 2 2 1 using c + s = .1 Not tan x = ( ± ) etc. 3 ( x ) = 15 A1 ft for (90 their 15) or (180 – their 15) (x ) = any correct second value (75, 105, 165) A1 All four correct. Extra solns in range 1 (x ) = cao A1 [4]  5   3   2       

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5 Relative to an origin O, the position vectors of the points A, B and C are given by ` a ` a ` a 3 5 6 −−→ −−→ −−→ OA = 2 , OB = −1 and OC = 1 . −3 −2 2 (i) Show that angle ABC is 90Å. [4] (ii) Find the area of triangle ABC, giving your answer correct to 1 decimal place. [3]

Mark scheme:       5 (i) AB =  − 1  −  2  =  − 3  B1 Or BA, CB . Allow any combination. Ignore       labels. 2 3 1 −  −     6   5   1        BC =  1  −  − 1  =  2  B1       2 − 2 4       AB. BC = 2 − 6 + 4 oe must be seen = 0 M1 Could be part of calculation for angle ABC hence ABC = 90º A1 AG Alt methods Pythag, Cosine Rule [4] (ii) AB = 14 , BC = 21 oe B1 At least one correct 1 Area = 14 21 M1 Reasonable attempt at vectors and their 2 magnitudes 8.6 oe A1 7 6 [3] Allow 2 1 1

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7 The point A has coordinates p, 1 and the point B has coordinates 9, 3p + 1 , where p is a constant. (i) For the case where the distance AB is 13 units, find the possible values of p. [3] (ii) For the case in which the line with equation 2x + 3y = 9 is perpendicular to AB, find the value of p. [4]

Mark scheme: 7 (i) (9 − p ) 2 + (3 p ) 2 = 169 M1 Or = 13 10 p 2 − 18 p − 88 ( = 0 ) oe A1 3-term quad p = 4 or − 11 / 5 oe A1 [3] 2 (ii) Gradient of given line = − B1 3 3 Hence gradient of AB = M1 Attempt using m1m2 = −1 2 3 3 p − 2   3 p  −9 p   3  = oe eg     = 1 M1 Or vectors   .   2 9 − p  3   9 − p   3 p   − 2  (includes previous M1) p = 3 A1 [4] 2 3

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9 (a) The first term of an arithmetic progression is −2222 and the common difference is 17. Find the value of the first positive term. [3] (b) The first term of a geometric progression is ï3 and the second term is 2 cos 1, where 0 < 1 < 0. Find the set of values of 1 for which the progression is convergent. [5]

Mark scheme: 9 (a) 2222/17 (=131 or 130.7) M1 Ignore signs. Allow 2239/17→131.7 or 132 131 × 17 (=2227) M1 Ignore signs. Use 131. −2222 + 2227 = 5 A1 5 www gets 3/3 [3] 2 cos θ (b) r = soi oe B1 3 2 cos θ 2 cos θ ( − 1 < ) < 1 or (0 <) < 1 soi M1 Ft on their r. Ignore a 2nd inequality on 3 3 LHS π / ,6 5π / 6 soi (but dep. on M1) A1A1 Allow 30º, 150º. π / 6 < θ < 5π / 6 cao A1 Accept Y [5] d

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10 y A 2, 9 y = 9 + 6x −3x2 x O B C 3, 0 Points A 2, 9 and B 3, 0 lie on the curve y = 9 + 6x −3x2, as shown in the diagram. The tangent at A intersects the x-axis at C. Showing all necessary working, (i) find the equation of the tangent AC and hence find the x-coordinate of C, [4] (ii) find the area of the shaded region ABC. [5] [Question 11 is printed on the next page.]

Mark scheme: d y10 (i) = 6 − 6 x B1 d x At x = 2 , gradient = −6 soi B1 y − 9 = −6 ( x − 2 ) oe Expect y = −6 x + 21 M1 Line through (2, 9) and with gradient their −6 When y = ,0 x = 3 12 cao A1 [4] (ii) Area under curve: ∫ 9 + 6 x − 3 x 2 dx = 9 x + 3 x 2 − x 3 B2,1,0 Allow unsimplified terms ( 27 + 27 − 27 ) − (18 + 12 − 8) M1 Apply limits 2,3. Expect 5 27 3 27 1 × × 9 ( = Area under tangent: 2 ( −6 x + 21) dx ( → ). Ft on their ) B1 OR ∫ 2 7 2 2 4 4 −x6 + 21 and/or their 7/2. 27 7 Area required − 5 = A1 4 4 [5]

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11 A r O ! rad C B In the diagram, OAB is a sector of a circle with centre O and radius r. The point C on OB is such that angle ACO is a right angle. Angle AOB is ! radians and is such that AC divides the sector into two regions of equal area. (i) Show that sin ! cos ! = 12!. [4] It is given that the solution of the equation in part (i) is ! = 0.9477, correct to 4 decimal places. (ii) Find the ratio perimeter of region OAC : perimeter of region ACB, giving your answer in the form k : 1, where k is given correct to 1 decimal place. [5] (iii) Find angle AOB in degrees. [1]

Mark scheme: 11 (i) OC = r cos α or AC = r sin α or oe soi M1 (Area ∆OAC = ) 12 r 2 sin α cos α A1 1 r 2 sin α cos α = 1 × 1 r 2α oe M1 Or e.g. 2 2 2 1 2 r 2α − 1 2 r 2 cos α sin α = 1 4 r 2α 1 2 r 2α − 1 2 r 2 cos α sin α = 1 2 r 2 cos α sin α sin α cos α = 12 α A1 AG [4] (ii) Perimeter ∆OAC = r + r sin α + r cos α = 4.2 ( 0 ) r M1A1 Allow with r a number. 2.0164 gets M1A0 Perim. ACB = rα + r sin α + r − r cos α = 2.18r or 2.17r M1A1 Allow with r a number. 0.9644 gets M1A0 Allow 2.2 www. 4.2 ( 0 ) Ratio = : 1 = 1.1 : 1 A1 Use of cos = 0.6, sin = 0.8, α = 9.0 is PA 1 .2 18 or .2 17 [5] (iii) 54.3º cao B1 [1]

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