Cambridge A Level Mathematics 9709 — 2014 May/June Paper 1 · Variant 2

6 The 1st, 2nd and 3rd terms of a geometric progression are the 1st, 9th and 21st terms respectively of an arithmetic progression. The 1st term of each progression is 8 and the common ratio of the geometric progression is r, where r Find ≠1. (i) the value of r, [4] (ii) the 4th term of each progression. [3]

Mark scheme: 6 (i) GP 8 8 r 8r² AP 8 8 + 8d 8 + 20d 8r = 8 + 8d and 8 r 2 = 8 + 20 d B1 B1 B1 for each equation. Eliminates d → 2 r 2 −r5 + 3 = 0 M1 Correct elimination. → r = 1.5 ( or 1) A1 co (no penalty for including r = 1) [4] (ii) 4th term of GP = ar³ = 8 × 27/8 = 27 B1 co If r = 1.5, d = 0.5 4th term of AP = a + 3d = 9½ M1A1 needs a +3d and correct method for d [3] − 2   3 

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7 C D B A The diagram shows a trapezium ABCD in which BA is parallel to CD. The position vectors of A, B and C relative to an origin O are given by ` a ` a ` a −−→OA 34 , −−→OB 13 and −−→OC 45 . = = = 0 2 6 (i) Use a scalar product to show that AB is perpendicular to BC. [3] (ii) Given that the length of CD is 12 units, find the position vector of D. [4]

Mark scheme:     7 (i) (b − a).(b − c) =  − 1  .  2  M1 AB = b − a once (a – b is ok)     M1 Use of x1x2... with AB and CB 2 4     → −6 − 2 + 8 = 0 → 90° A1 All correct [3]  2    (ii) Unit vector = ⅓  1  M1 Method for unit vector.   −2    8    CD = 12 × unit vector = ±  4  M1 Knows to multiply by 12 or ±4BA   −8    12    OD = OC + CD =  9  M1 A1 Correct method. co   [4] −2   d2 y

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9 y y = 8 −ï 4 −x P 3, 7 x O The diagram shows part of the curve y 8 4 and the tangent to the curve at P 3, 7 . = − −x dy (i) Find expressions for and y dx. [5] dx Ó (ii) Find the equation of the tangent to the curve at P in the form y mx c. [2] = + (iii) Find, showing all necessary working, the area of the shaded region. [4]

Mark scheme: 9 y = 8 − 4 − x d y 1 − 1 (i) = − (4 − x ) 2 × −1 B1 B1 Without (−1). For (×−1). d x 2 3 (4 − x ) 2 ∫y dx = 8 x − ÷ −1 3 × B1 B1 for "8x" and +c". B1 for all except 3 2 ÷(−1). B1 for ÷(−1). [5] (n.b. these 5 marks can be gained in(ii) or (ii) Eqn y − 7 = 1 2 ( x − 3 ) (iii)) → y = ½x + 5½ M1A1 M1 unsimplified. A1 as y=mx+c [2] (iii) Area under curve = ∫ from 0 to 3 (58/3) M1 Use of limits – needs use of “0” Area under line = ½(5½ + 7)×3 M1 Correct method  1 2 11x  Or 4 x + from 0 to 3 M1 A1 M1 Subtraction. A1 co  2  [4] 58 75 7 → − = 3 4 12

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