Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 1 · Variant 1

6 (a) The fifth term of an arithmetic progression is 18 and the sum of the first 5 terms is 75. Find the first term and the common difference. [4] (b) The first term of a geometric progression is 16 and the fourth term is 274 . Find the sum to infinity of the progression. [3]

Mark scheme: 6 (a) a + 4d = 18 B1 co or 75 = 5/2(a + 18) → a = 12 etc 5 ( 2 a + 4 d ) = 75 B1 co 2 Solution M1 Solution of sim equations → a = 12, d = 1½ A1 co for both [4] 27 (b) a = 16 and ar3 = B1 Needs both of these 4 3 r = 4 Sum to infinity = 64 M1 A1 Correct formula and r < 1 [3] 1

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8 x cm y cm x cm The diagram shows a metal plate consisting of a rectangle with sides x cm and y cm and a quarter-circle of radius x cm. The perimeter of the plate is 60 cm. (i) Express y in terms of x. [2] (ii) Show that the area of the plate, A cm2, is given by A 30x [2] = −x2. Given that x can vary, (iii) find the value of x at which A is stationary, [2] (iv) find this stationary value of A, and determine whether it is a maximum or a minimum value. [2] [Questions 9, 10 and 11 are printed on the next page.]

Mark scheme: πx 8 (i) 2 x + 2 y + = 60 M1 Linking 60 with sum of at least 4 sides 2 and use of radians πx → y = 30 − x − A1 co 4 [2] πx 2 (ii) A = xy + 4 πx πx 2 1 2 = x (30 − x − ) + M1 Subs “y” into area eqn and use r θ 4 4 2 = 30x – x2 A1 co. [2] dA (iii) = 30 − 2 x Knowing to differentiate dx = 0 when x = 15 cm M1 A1 Sets differential to 0 + solution. co. [2] (iv) Max. M1 A1 Any valid method. co. [2] GCE AS/A LEVEL – October/November 2010 9709 11

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9 C1 P 8 cm T C2 Q 2 cm R S The diagram shows two circles, C1 and C2, touching at the point T. Circle C1 has centre P and radius 8 cm; circle C2 has centre Q and radius 2 cm. Points R and S lie on C1 and C2 respectively, and RS is a tangent to both circles. (i) Show that RS 8 cm. [2] = (ii) Find angle RPQ in radians correct to 4 significant figures. [2] (iii) Find the area of the shaded region. [4]

Mark scheme: 9 (i) RS² = 10² – 6² M1 Use of Pythagoras (or other) → RS = 8 cm. A1 Answer given. [2] (ii) sin θ = 8/10 oe M1 Use of trig – even if with degrees. → angle RPQ = 0.9273 radians A1 co in radians. (Accept 0.927) [2] (iii) Region = trapezium − 2 sectors Area of trapezium = 40 cm² B1 co 1 1 Large sector = × 8² × 0.9273 M1 Use of r²θ. 2 2 Small sector angle = (π − 0.9273) 1 1 Small sector = × 2² × 2.214 M1 Use of r²θ with angle = π − (ii) 2 2 → 5.90 cm2 A1 [4] co 2

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10 The equation of a curve is y 3 4x = + −x2. (i) Show that the equation of the normal to the curve at the point is 2y x 9. [4] (3, 6) = + (ii) Given that the normal meets the coordinate axes at points A and B, find the coordinates of the mid-point of AB. [2] (iii) Find the coordinates of the point at which the normal meets the curve again. [4]

Mark scheme: 10 y = 4x – x2 + 3 dy (i) = 4 − 2 x B1 co dx At x = 3, m = − 2 1 Gradient of normal = M1 Use of m1m2 = −1 2 Eqn of normal y − 6 = 12 ( x − 3) M1 A1 Use of y – k = m(x – h) or y = mx + c → 2y = x + 9 (where m is gradient of normal) [4] 9 (ii) Meets axes at (0, ) and (−9, 0) M1 Sets x and y to 0 + midpoint formula. 2 − 9 9  Mid-point is  ,  A1 co.  2 4  [2] (iii) 2y = x + 9, y = 4x – x2 + 3 → 2x2 – 7x + 3 = 0 oe M1 A1 Eliminates x completely. Correct eqn. → (½, 4¾) M1 A1 Solution of quadratic. co [4] GCE AS/A LEVEL – October/November 2010 9709 11 9 11 y = 2 − x dy 2

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