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

2 y y = 2x2 Q x X −2, 0 O P p, 0 The diagram shows the curve y = 2x2 and the points X −2, 0 and P p, 0 . The point Q lies on the curve and PQ is parallel to the y-axis. (i) Express the area, A, of triangle XPQ in terms of p. [2] The point P moves along the x-axis at a constant rate of 0.02 units per second and Q moves along the curve so that PQ remains parallel to the y-axis. (ii) Find the rate at which A is increasing when p = 2. [3]

Mark scheme:  1  2   cos  x  = B1  2  3  2  3 1 x = 0.84 x = 1.68 only, aef M1A1 Looks up cos−1 first, then ×2 2 [3] (in given range) (ii) B1 y always +ve, m always –ve. B1 from (0, 8) to (2π, 2) (may be [2] implied) 2 (iii) No turning point on graph or 1:1 B1 cao, independent of graph in (ii) [1] 1   M1 Tries to make x subject. x (iv) y = 5 + 3cos    2  Order; −5, ÷3, cos−1, ×2 M1 Correct order of operations  x − 5  A1 cao x = 2cos−1    3  [3] 9 y = x 3 + px 2 dy (i) = 3x² + 2px B1 cao dx 2p Sets to 0 → x = 0 or − M1 Sets differential to 0 3 2 p 4 p 3  → (0, 0) or − , A1 A1 cao cao, first A1 for any correct    3 27  [4] turning point or any correct pair of x values. 2nd A1 for 2 complete TPs d 2 y (ii) 2 = 6x + 2p M1 Other methods include; clear dx demonstration of sign change of gradient, clear reference to the shape of the curve At (0, 0) → 2p +ve Minimum A1 www  

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4 Relative to the origin O, the position vectors of points A and B are given by ` a ` a 3 6 −−→ −−→ OA = 0 and OB = −3 . −4 2 (i) Find the cosine of angle AOB. [3] ` a k −−→ The position vector of C is given by OC = −2k . 2k −3 (ii) Given that AB and OC have the same length, find the possible values of k. [4]

Mark scheme: 2 p 4 p 3 At , →−2p –ve Maximum A1 − 

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6 The line with gradient −2 passing through the point P 3t, 2t intersects the x-axis at A and the y-axis at B. (i) Find the area of triangle AOB in terms of t. [3] The line through P perpendicular to AB intersects the x-axis at C. (ii) Show that the mid-point of PC lies on the line y = x. [4]

Mark scheme: 6 (i) y − 2t = −2( x − 3t )( y + 2 x = 8t ) M1 Unsimplified or equivalent forms Set x to 0 → B(0, 8t) Set y to 0 → A(4t, 0) M1 Attempt at both A and B, then using → Area = 16t² A1 cao [3] (ii) 1 cao m = B1 Unsimplified or equivalent forms 2 co 1 → y − 2t = ( x − 3t )(2 y = x + t ) M1 2 A1 correctly shown. Set y to 0 → C (−t, 0) Midpoint of CP is (t, t) This lies on the line y = x. A1 [4] 1 2

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7 (a) The third and fourth terms of a geometric progression are 1 and 2 respectively. Find the sum to 3 9 infinity of the progression. [4] (b) A circle is divided into 5 sectors in such a way that the angles of the sectors are in arithmetic progression. Given that the angle of the largest sector is 4 times the angle of the smallest sector, find the angle of the largest sector. [4]

Mark scheme: 1 2 7 (a) ar² = , ar³ = 3 9 2 → r = aef M1 Any valid method, seen or implied. 3 Could be answers only. 3 Substituting → a = A1 Both a and r 4 3 → S∞ = 4 = 2 14 aef M1 A1 Correct formula with r < 1 , cao 1 3 [4] (b) 4 a = a + 4 d → 3a = 4d B1 May be implied in 360 = 5 / 2( a + 4 a ) 5 360 = S5 = ( 2 a + 4 d ) or 12.5a M1 Correct Sn formula or sum of 5 2 terms → a = 28.8º aef A1 cao, may be implied Largest = a + 4d or 4a = 115.2º aef B1 (may use degrees or radians) [4]  1  8 f : x⟼5 + 3cos  x  for 0 ø x ø 2π.  2   1  (i) 5 + 3cos  x  = 7  2  1 2 Makes cos  x   

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