Cambridge A Level Mathematics 9709 — 2010 Oct/Nov Paper 1 · Variant 2

1 (i) Find the first 3 terms in the expansion, in ascending powers of x, of (1 −2x2)8. [2] (ii) Find the coefficient of x4 in the expansion of (2 −x2)(1 −2x2)8. [2]

Mark scheme: 1 (i) 1 + 8(–2x2) + 8C2(–2x2)2 B2, 1 Loses 1 for each error 1 – 16x2 + 112x4 [2] (ii) (2 – x2) × their (1 – 16x2 + 112x4) M1 Must consider exactly 2 terms (2 × their 112) – their (–16) 240 A1√ [2] 2 2 2

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2 Prove the identity tan2x −sin2x ≡tan2x sin2x. [4]

Mark scheme: 2 LHS = sin2x/cos2x – sin2x M1 Replace t² by s²/c² or sec² − 1 sin2x(1 – cos2x)/cos2x M1 Use of 1 – cos2x = sin2x sin 2 x sin 2 x oe M1 Valid overall method cos 2 x tan2xsin2x A1 AG sin 2 x 2 OR RHS = . sin x M1 Replace t² by s²/c² 2 cos x sin2x(1 – cos2x)/cos2x M1 Use of 1 – cos2x = sin2x (sin2x/cos2x) – sin2x M1 Valid overall method tan2x – sin2x A1 AG [4] 1/2

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3 The length, x metres, of a Green Anaconda snake which is t years old is given approximately by the formula x = 0.7 √(2t −1), where 1 ≤t ≤10. Using this formula, find dx (i) , [2] dt (ii) the rate of growth of a Green Anaconda snake which is 5 years old. [2]

Mark scheme: 3 (i) (k(2t – 1)–1/2 M1 k ≠ 1 0.7(2t – 1)–1/2 A1 oe [2] (ii) Sub t = 5 into their deriv M1 0.23(3) A1 Ignore units [2]

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4 C 3 cm 2.3 rad 2.3 rad O 3 cm 3 cm A B P The diagram shows points A, C, B, P on the circumference of a circle with centre O and radius 3 cm. Angle AOC = angle BOC = 2.3 radians. (i) Find angle AOB in radians, correct to 4 significant figures. [1] (ii) Find the area of the shaded region ACBP, correct to 3 significant figures. [4]

Mark scheme: 4 (i) 1.683(18…) B1 [1] (ii) (2) × ½ × 3²sin2.3 M1 Condone omission of factor 2 ½ × 3² × their 1.683 M1 NB M0 if using angle of 2.3 Triangle AOC + COB + sector M1 Two correct triangles + sector 14.3 A1 co [4]

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5 (a) The first and second terms of an arithmetic progression are 161 and 154 respectively. The sum of the first m terms is zero. Find the value of m. [3] (b) A geometric progression, in which all the terms are positive, has common ratio r. The sum of the first n terms is less than 90% of the sum to infinity. Show that rn > 0.1. [3]

Mark scheme: 5 (a) d = –7 used B1 co (m/2)[322 + (m – 1)(–7)] = 0 M1 Condone omission of (m/2). Statement 47 A1 co (condone m = 0) [3] a 1( − r n ) 9.0 a (b) < M1 Allow for =, <, >, ≤, ≥ 1 − r 1 − r 1 – rn < 0.9 M1 Needs inequality sign correct rn > 0.1 A1 co [3] GCE AS/A LEVEL – October/November 2010 9709 12 2

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6 A curve has equation y = kx2 + 1 and a line has equation y = kx, where k is a non-zero constant. (i) Find the set of values of k for which the curve and the line have no common points. [3] (ii) State the value of k for which the line is a tangent to the curve and, for this case, find the coordinates of the point where the line touches the curve. [4]

Mark scheme: 6 (i) kx2 – kx + 1 = 0 M1 y eliminated k2 – 4k < 0 M1 Applying b2 – 4ac < 0 or = or ≤ or ≥ 0 < k < 4 A1 co [3] (ii) k = 4 only B1√ ft from their k2 – 4k = 0. (Not k = 0) (2x – 1)2 = 0 M1 ft from their k x = ½ , y = 2 or (½, 2) A1, A1 [4] 2

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7 The function f is defined by f(x) = x2 −4x + 7 for x > 2. (i) Express f(x) in the form (x −a)2 + b and hence state the range of f. [3] (ii) Obtain an expression for f−1(x) and state the domain of f−1. [3] The function g is defined by g(x) = x −2 for x > 2. The function h is such that f = hg and the domain of h is x > 0. (iii) Obtain an expression for h(x). [1]

Mark scheme: 7 (i) (x – 2)2 M1 Must be “−2” ± k (x – 2)2 + 3 A1 co f(x) > 3 B1√ ft on their ‘3’ [3] (ii) x – 2 = (±) y − 3 M1 ± not required for M mark f –1 (x) = 2 + x − 3 A1 f(x) + removal of minus sign needed domain is x > 3 B1√ ft domain of f –1 = range of f or for f –1 [3] (iii) h(x) = x2 + 3 B1 co [1] 2

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8 y 2 y = 1 – x y = 3 x + 4 B A x O 2 The diagram shows part of the curve y = 1 and the line y = 3x + 4. The curve and the line meet at points A and B. −x (i) Find the coordinates of A and B. [4] (ii) Find the length of the line AB and the coordinates of the mid-point of AB. [3]

Mark scheme: 8 (i) 3x2 + x – 2 = 0 M1A1 Eliminates x or y. Sets quadratic to 0. (x + 1)(3x – 2) → x = –1 or ⅔ M1 Attempt to solve their equation (–1, 1), (⅔, 6) A1 co [4] (ii) AB2 = (5/3)2 + 52 M1 √ their coordinates from (i) AB = 5.27(0…) A1 Or (5√10)/3 oe mid-point = (–1/6, 7/2) B1√ ft from their (i) [3] 10 −a 6

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9 P G F 10 cm 6 cm 6 cm D C E B a cm k j 10 cm O i 10 cm A The diagram shows a pyramid OABCP in which the horizontal base OABC is a square of side 10 cm and the vertex P is 10 cm vertically above O. The points D, E, F, G lie on OP, AP, BP, CP respectively and DEFG is a horizontal square of side 6 cm. The height of DEFG above the base is a cm. Unit vectors i, j and k are parallel to OA, OC and OD respectively. (i) Show that a = 4. [2] (ii) Express the vector −−→BG in terms of i, j and k. [2] (iii) Use a scalar product to find angle GBA. [4]

Mark scheme: 10 a 6 9 (i) = oe M1 or PDE is isos hence PD = 6 (M1) 10 10 a = 4 A1 AG [2] (ii) BG = –10j – 10i + 4k + 6j B2,1 Any acceptable notation. Loses 1 for = –10i – 4j + 4k each error. [2] (iii) BG.BA = 40 M1 Use of x1 x 2 + y1 y 2 + z1 z 2 40 cos GBA = M1 Modulus worked correctly for either 132 100 DM1 All ok – must be using ± BG . ± AB . GBA = 69.6° A1 Must be the acute angle [4] GCE AS/A LEVEL – October/November 2010 9709 12 8

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10 5 x 4 4 x 5 h 1 x 2 x The diagram shows an open rectangular tank of height h metres covered with a lid. The base of the tank has sides of length x metres and 2x1 metres and the lid is a rectangle with sides of length 4x5 metres and 5x4 metres. When full the tank holds 4 m3 of water. The material from which the tank is made is of negligible thickness. The external surface area of the tank together with the area of the top of the lid is A m2. 3 24 (i) Express h in terms of x and hence show that A = 2x2 + x . [5] (ii) Given that x can vary, find the value of x for which A is a minimum, showing clearly that A is a minimum and not a maximum. [5]

Mark scheme: 8 10 (i) h = 2 M1 Uses lbh = 4 x A1 co 1 2 1 5 4 A = x + 2 × xh + 2 xh + x × x M1 Allow 1 error but needs the lid 2 2 4 5 A = (3 / 2 ) x 2 + 3 xh 3 2 8 A = x + 3 x × 2 M1 For substitution of h as f(x) 2 x 3 2 24 A = x + A1 AG 2 x [5] dA 24 (ii) = 3 x − = 0 B1 Correct derivative. 2 dx x M1 Sets to 0 and attempts to solve. x = 2 A1 co d 2 A 48 = 3 + M1 Reasonable attempt – allow 1 error dx 2 x 3 > 0 when x = 2 hence minimum A1 co [5] AG (Result consistent with their f'')

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11 y x = 5 A 1 y = 1 B 4 (3x + 1) x O 1 1 The diagram shows part of the curve y = . The curve cuts the y-axis at A and the line x = 5 4 (3x + 1) at B. (i) Show that the equation of the line AB is y = −110x + 1. [4] (ii) Find the volume obtained when the shaded region is rotated through 360◦about the x-axis. [9]

Mark scheme: 11 (i) A = (0, 1) B1 B = (5, ½) B1 1 y − 1 = − ( x − 0) M1 ft their A,B 10 1 y = − x + 1 A1 AG 10 [4] 5 −1 / 2 5 2 (ii) Curve: (π)∫ 0 (3 x + )1 dx M1 Attempt ∫ 0 y dx (π not vital) 2π 1 2 5 [(3 x + )1 ] 0 A1A1 (π not vital). 2nd A mark is for ÷ 3. 3 2π [ 4 − ]1 DM1 Application of limits to their integral 3 (in either integral). Limits 0 to 5 only. [2π] 5 1 2 1 5 2 M1 Attempt ∫ 0 y dx (π not vital) Line: (π ) ∫ 0 ( 100 x − 5 x + )1 dx 1 3 1 2 5 10 1 (π )[ x − x + x ]0 A2,1 Also directly − ( − x + 3)1 300 10 3 10 125 25 10  1 3 3  (π )[ − + 5] or − ( − + )1 − 1 (π not vital) 300 10 3  2  35π – applying limits to their integral [ ] 12 35π 11π Volume = – 2π = DM1 Subtraction of their volumes 12 12 A1 co [9]

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