Cambridge A Level Mathematics 9709 — 2011 Oct/Nov Paper 1 · Variant 2

2 The functions f and g are defined for x ∈> by f : x →3x + a, g : x →b −2x, where a and b are constants. Given that ff(2) = 10 and g−1(2) = 3, find (i) the values of a and b, [4] (ii) an expression for fg(x). [2]

Mark scheme: 2 f : x 3 x + a , g : x b − 2 x (i) f2(x) = 3(3x + a) + a B1 Must be correct – unsimplified ok f²(2) = 18 + 4a = 10 → a = −2 B1 co b − x b − 2 g–1(x) = → = 3 b = 8 M1 Correct method leading to a value for b 2 2 co or g(3) = 2 → b − 6 = 2 b = 8 A1 [4] (ii) fg(x) = 3(b − 2 x ) + a M1 Must be fg not gf. = 22 − 6x A1√ √ on a and b (3b + a − 6x) must be two [2] term answer.

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3 Relative to an origin O, the position vectors of points A and B are given by −−→OA 5i j 2k and −−→OB 2i 7j pk, = + + = + + where p is a constant. (i) Find the value of p for which angle AOB is 90◦. [3] (ii) In the case where p = 4, find the vector which has magnitude 28 and is in the same direction as −−→AB. [4]

Mark scheme: 3 OA = 5i + j + 2k , OB = 2 i + 7 j + pk M1 Use of x1x2 + y1y2 + z1z2 (i) OA. OB = 10 + 7 + 2p DM1 ....=0 = 0 → p = − 8½ A1 co [3] (ii) AB = −3i + 6j + 2k B1 co (accept negative) Modulus = √(9+36+4) M1 For modulus Magnitude 28 → 28 ×unit vector M1 Scales by ×28 ÷ modulus. → −12i + 24j + 8k. A1 Co – could leave as “4 × …”. [4]

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4 The equation of a curve is y2 + 2x = 13 and the equation of a line is 2y + x = k, where k is a constant. (i) In the case where k = 8, find the coordinates of the points of intersection of the line and the curve. [4] (ii) Find the value of k for which the line is a tangent to the curve. [3]

Mark scheme: 4 (i) y 2 + 2 x = 13 , 2 y + x = 8 M1 Complete elimination of x or y 2 2 A1 co (allow multiples) – needs 3 terms → y − 4 y + 3 = 0 , x −x8 + 12 = 0 DM1 Solution of quadratic = 0 → (2, 3) and (6, 1) A1 Needs all 4 coordinates. [4] (ii) Removes x → y 2 + 2( k − 2 y ) = 13 M1 Complete elimination of x or y. Uses b 2 − 4 ac on “quadratic = 0) DM1 Use of discriminant =0, <0 or >0 → k = 8½ A1 Co dy 1 [3] (M1 equating m of line and curve or = −½ = −→ y=2, x=4½, k= 8½ dx y M1 x to y A1 for k) GCE AS/A LEVEL – October/November 2011 9709 12

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5 (i) Sketch, on the same diagram, the graphs of y = sin x and y = cos 2x for 0◦≤x ≤180◦. [3] (ii) Verify that x = 30◦is a root of the equation sin x = cos 2x, and state the other root of this equation for which 0◦≤x ≤180◦. [2] (iii) Hence state the set of values of x, for 0◦≤x ≤180◦, for which sin x < cos 2x. [2]

Mark scheme: 5 (i) B1 y = sinx (0,0). (π,0) + curve B1 y = cos2x One full cycle. B1 y = cos2x starts and finishes at (0, 1) and oscillates between −1 and +1. [3] Do not penalise graphs from 0 to 360. B1 co (ii) Evidence of sin 30 = cos 60 = 0.5 B1 co Other root is 150º [2] (iii) 0 ≤ x < 30 and 150 < x ≤ 180 B1 B1√ Condone < or ≤ throughout (x < 30 or x > 150 ok) [2] √ B1 N d √3 2 j 3√3

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6 C2 D E C1 6 cm 10 cm 3p1 q A B X The diagram shows a circle C1 touching a circle C2 at a point X. Circle C1 has centre A and radius 6 cm, and circle C2 has centre B and radius 10 cm. Points D and E lie on C1 and C2 respectively and DE is parallel to AB. Angle DAX = 13π radians and angle EBX = θ radians. (i) By considering the perpendicular distances of D and E from AB, show that the exact value of θ 3 √3 is sin−1 . [3] 10 (ii) Find the perimeter of the shaded region, correct to 4 significant figures. [5]

Mark scheme: 6 (i) D to AX = 6 sin π3 = 6√3÷2 B1 co Needs –√3÷2 not just 3√3. E to AX = 10sinθ B1 co Correct method. ag. 3 3 B1 Use of decimals loses this B mark. . Equate these → θ = sin −1 10 [3] (ii) Arc DX = 6.⅓π = 2π B1 co Arc EX = 10×0.5464 =5.464 M1 Use of s=rθ radians. Horizontal steps = 6cos⅓π and 10cosθ M1 Attempt at both steps needed DE = 10 + 6 − 6cos⅓π − 10cosθ M1 Full method for DE. Perimeter = arc DX + arc BX + DE → 16.20 A1 Co – must be exactly 16.20, not more or [5] less places. dy 8

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dy 7 A curve is such that . The line 3y + x = 17 is the normal to the curve at the point P on the dx = 5 −8x2 curve. Given that the x-coordinate of P is positive, find (i) the coordinates of P, [4] (ii) the equation of the curve. [4]

Mark scheme: dy 8 7 = 5 − 2 , Normal 3 y + x = 17 dx x (i) Gradient of line = −⅓ B1 co dy M1 Use of m1m2 = − 1 = 3 → x = 2, y = 5 DM1 DM1 solution. A1 co. dx A1 [4] (ii) y = 5 x + 8 x −1 (+ c ) B1 B1 co.co. doesn’t need +c. Uses (2, 5) → c = −9 M1 A1 Use of +c following integration. co. [4] GCE AS/A LEVEL – October/November 2011 9709 12 2

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Find8 The equation of a curve is y = √(8x −x2). dy (i) an expression for dx, and the coordinates of the stationary point on the curve, [4] (ii) the volume obtained when the region bounded by the curve and the x-axis is rotated through 360◦about the x-axis. [4] [Questions 9 and 10 are printed on the next page.]

Mark scheme: 8 y = 8 x − x 2 dy (i) 2 × (8 − 2 x ) B1 B1 for everything but ×(8-2x) = 12 (8 x − x 2 ) − 1 dx B1 B1 for × (8−2x), even if B0 = 0 when x = 4. M1 Sets to 0 + attempt at solution. → (4, 4) A1 Co – A0 if fortuitous because of B0 [4] earlier. (ii) y = 0 when x = 0 or 8 B1 Vol = π ∫ (8 x − x 2 d)x Anywhere  2 x 3  B2,1 = π 4 x −1 for each error (not including π)  −3  256π B1 → 3 [4] co

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9 y B (3, 6) C (9, 4) M O x A (–1, –1) D The diagram shows a quadrilateral ABCD in which the point A is (−1, −1), the point B is (3, 6) and the point C is (9, 4). The diagonals AC and BD intersect at M. Angle BMA = 90◦and BM = MD. Calculate (i) the coordinates of M and D, [7] (ii) the ratio AM : MC. [2]

Mark scheme: 9 (i) Gradient of AC = ½ B1 co Gradient of BD = − 2 M1 Use of m1m2 = − 1 with AC Eqn of BD is y − 6 = − 2( x − 3) M1 Correct formula for straight line Eqn of AC is y + 1 = 12 ( x + )1 M1 Solution. Sim eqns → M (5, 2) A1 co Vector move – or midpoint back → D (7, − 2) M1 A1√ Correct method. √ on M. [7] (ii) Ratio of AM : MC = √45 : √20 M1 Correct distance formula. or Vector step → 3 : 2 A1 Looks at the two x or y steps. [2] Must be numerical, 1.5 ok, not as roots

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10 (a) An arithmetic progression contains 25 terms and the first term is −15. The sum of all the terms in the progression is 525. Calculate (i) the common difference of the progression, [2] (ii) the last term in the progression, [2] (iii) the sum of all the positive terms in the progression. [2] (b) A college agrees a sponsorship deal in which grants will be received each year for sports equipment. This grant will be $4000 in 2012 and will increase by 5% each year. Calculate (i) the value of the grant in 2022, [2] (ii) the total amount the college will receive in the years 2012 to 2022 inclusive. [2]

Mark scheme: 10 (a) a = −15, n = 25 (i) Use of Sn → d = 3. M1 A1 Must be correct formula. co [2] (ii) Last term = a + 24d M1 Must be a + 24d → 57 A1√ √ for his d. (or 525 = ½ × 25 × (−15 + l) → l = 57) [2] (iii) Positive terms are 3,6, ....57 Either a = 0 or 3, n = 19 or 20 M1 Correct use of formula for Sn. Use of S19 or S20 → 570 A1 co [2] (b) r = 1.05 B1 In either part (i) or (ii). (i) 11th term = ar10 = $6516 or $6520 B1 co [2] 4000 × .1( 0511 − )1 (ii) S11 = M1 Correct sum formula with their r. . 05 A1 co = $56800 or (56827) [2]

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