Cambridge A Level Mathematics 9709 — 2011 May/June Paper 1 · Variant 1

2 7 1 Find the coefficient of x in the expansion of x + x2 . [3]

Mark scheme:   1 7C2 x5  SOI and leading to final answer B2 B1 for 2/3 parts correct leading to ans.  x 2  84 or 84x as final answer B1 If no answer: 84x seen scores B2, else 2 [3]  2  7C2 x5  2  scores SCB1 only  x  dv 2 M1  

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2 The volume of a spherical balloon is increasing at a constant rate of 50 cm3 per second. Find the rate of increase of the radius when the radius is 10 cm. [Volume of a sphere = 43πr3.] [4]

Mark scheme:  dv  22  =  4πr M1 dr   2 A1 SOI at any point = 4π × 10 dv dr dt M1 Correct link between differentials with = OE used dt dv dr dr finally as subject dt 50 1 50 = or 0.0398 2 A1 Allow . 4π × 10 8π 400π [4] 0 Non-calculus methods 4

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3 (i) Sketch the curve y = (x −2)2. [1] (ii) The region enclosed by the curve, the x-axis and the y-axis is rotated through 360◦about the x-axis. Find the volume obtained, giving your answer in terms of π. [4]

Mark scheme: 3 (i) Correct shape – touching positive x-axis B1 Ignore intersections with axes [1] (ii) (π ) ∫ ( x − 2 ) 4 d x M1 Use (π ) ∫ y 2 d x & attempt integrate but expansion before integn needs 5 terms  ( x − 2) 5  (π )   A1 5   (π )[0 − ( −32/) 5]) M1 Use of limits 0, 2 on their (π ) y ∫ 2 dx 32π or 6.4π A1 cao Rotation about y-axis max 1/5 5 [4] B1

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4 Q B R S 5 cm P C 6 cm k 2 cm j A D i 6 cm The diagram shows a prism ABCDPQRS with a horizontal square base APSD with sides of length 6 cm. The cross-section ABCD is a trapezium and is such that the vertical edges AB and DC are of lengths 5 cm and 2 cm respectively. Unit vectors i, j and k are parallel to AD, AP and AB respectively. (i) Express each of the vectors −−→CP and −−→CQ in terms of i, j and k. [2] (ii) Use a scalar product to calculate angle PCQ. [4]

Mark scheme: B14 (i) CP = ‒6i + 6j ‒2k CQ = ‒6i + 6j + 3k B1 [2] (ii) Scalar product = 36 + 36 ‒ 6 M1 Use of x1 x 2 + y1 y 2 + z1 z 2 66 = |CP | |CQ | cos θ M1 Linking everything correctly |CP | = 76 , |CQ | = 81 M1 Correct magnitude for either Angle PCQ = 32.7° (or 0.571 rad) A1 cao 147.3° converted to 32.7° gets A0 [4] GCE AS/A LEVEL – May/June 2011 9709 11 2 sin 2 θ sin 2 θ M1 E ti f ti f i θ

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5 (i) Show that the equation 2 tan2θ sin2θ = 1 can be written in the form 2 sin4θ + sin2θ −1 = 0. [2] (ii) Hence solve the equation 2 tan2θ sin2θ = 1 for 0◦≤θ ≤360◦. [4]

Mark scheme: 2 sin θ sin θ M1 Equation as function of sin θ5 (i) = 1 1 − sin 2 θ 2 sin 4 θ + sin 2 θ − 1 = 0 AG A1 [2] (ii) ( 2 sin 2 θ − 1)(sin 2 θ + )1 = 0 M1 Or use formula on quadratic in sin2 θ ( ± 1) sin θ = A1 2 θ = 45°, 135° A1 θ = 225°, 315° A1 Provided no excess solutions in range [4] ( z −x3 ) B1 600 

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6 The variables x, y and ß can take only positive values and are such that ß = 3x + 2y and xy = 600. 1200 (i) Show that ß = 3x + x . [1] (ii) Find the stationary value of ß and determine its nature. [6]

Mark scheme:  600  ( z 3 x )6 (i) z = 3 x + 2  or x = 600 OE B1  x  2 [1] → AG d z 1200 dz 1800 (ii) = 3 − or = 2 − B1 d x x 2 dy y 2 = 0 → x = 20 or = 0 → y = 30 M1A1 Set to 0 & attempt to solve. Allow ±20 Ft from their x provided positive 120 z = 60 + = 120 A1√ Or other valid method 20 d 2 z 2400 d 2 z k = B1√ Dep. on (k > 0) or other = dx 2 x 3 dx 2 x 3 > 0 ⇒ minimum B1 valid method. [6] 31( + 2 x ) −1 B1

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dy 3 17 A curve is such that dx = and the point (1, 2) lies on the curve. (1 + 2x)2 (i) Find the equation of the curve. [4] (ii) Find the set of values of x for which the gradient of the curve is less than 3.1 [3]

Mark scheme: 31( + 2 x ) 7 (i) + ( c ) B1 − 1 31( + 2 x ) −1 y = + ( c ) B1(indep) Division by 2 y = necessary − 2 Sub (1, (1/2)) M1 Dependent on c present 1 3 = + c ⇒ c = 1 A1 Use of y = mx + c etc. gets 0/4 2 − 6 [4] (ii) (1 + 2x)2(>)9 or 4x2 + 4x – 8(>)0 OE M1 1, ‒2 A1 x > 1, x < –2 ISW A1 [3]

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8 A television quiz show takes place every day. On day 1 the prize money is $1000. If this is not won the prize money is increased for day 2. The prize money is increased in a similar way every day until it is won. The television company considered the following two different models for increasing the prize money. Model 1: Increase the prize money by $1000 each day. Model 2: Increase the prize money by 10% each day. On each day that the prize money is not won the television company makes a donation to charity. The amount donated is 5% of the value of the prize on that day. After 40 days the prize money has still not been won. Calculate the total amount donated to charity (i) if Model 1 is used, [4] (ii) if Model 2 is used. [3]

Mark scheme: 8 (i) 1000, 2000, 3000... or 50, 100, 150... M1 Recognise series, correct a/d (or 3 terms ) 40 40 or M1 Correct use of formula 2(1000 + 40000 ) 2 ( 2000 + 39000) × 5% of attempt at valid sum M1 Can be awarded in either (i) or (ii) 41000 A1 cao [4] (ii) 1000, 1000 × 1.1, 1000 × 1.12 + ... or with a = 50 M1 Recognise series, correct a/r ( or 3 terms) 10001.1(40 − )1 M1 Correct use of formula. Allow e.g. r = 0.1 1.1 − 1 A1 Or answers rounding to this 22100 [3] GCE AS/A LEVEL – May/June 2011 9709 11

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9 S A B r P T 2q O In the diagram, OAB is an isosceles triangle with OA = OB and angle AOB = 2θ radians. Arc PST has centre O and radius r, and the line ASB is a tangent to the arc PST at S. (i) Find the total area of the shaded regions in terms of r and θ. [4] (ii) In the case where θ = 13π and r = 6, find the total perimeter of the shaded regions, leaving your answer in terms of √3 and π. [5] [Questions 10 and 11 are printed on the next page.]

Mark scheme: 9 (i) AS = r tan θ r M1 Or (AB) = 2r tan θ or ( AO ) = 2 1 2 cos θ Area OAB = r tan θ or (OAS ) = 2 r tan θ A1 2 1 r sin 2θ Or OAB = Area of sector = 12 r 2 × 2θ ( = r 2θ ) B1 A1 2 cos 2θ 1 Shaded area = r 2 (tan θ − θ ) OE [4] Or area sector (OPS ) = 2 r 2θ Allow e.g. r 2 tan θ − 1 2 r 2 2θ π 6 (ii) cos = ⇒ OA = 12 M1 3 OA AP = 6 A1 π AS = 6 tan (⇒ AB = 12 3 ) B1 3 π π Arc (PST) = 12π B1 Or arc ( PS ) = 6 or arc ( ST ) = 6 3 3 3 Perimeter = 12 + 12 3 + 4π A1 Allow unsimplified 4π [5] 2

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10 (i) Express 2x2 −4x + 1 in the form a(x + b)2 + c and hence state the coordinates of the minimum point, A, on the curve y = 2x2 −4x + 1. [4] The line x −y + 4 = 0 intersects the curve y = 2x2 −4x + 1 at points P and Q. It is given that the coordinates of P are (3, 7). (ii) Find the coordinates of Q. [3] (iii) Find the equation of the line joining Q to the mid-point of AP. [3]

Mark scheme: 10 (i) 2(x – 1)2 – 1 OR a = 2, b = –1, c = –1 B1, B1, B1 A = (1, –1) B1√ Allow alt. method for final mark [4] (ii) 2 x 2 − 5 x − 3 = 0 ⇒ ( 2 x + 1)( x − 3) = 0 OE in y M1, M1 Complete elim & simplify, attempt soln. x = − 1 2 , y = 3 1 2 A1 Additional (3, 7) not penalised [3] (iii) Mid-point of AP = (2, 3) B1√ Follow through on their A 1 − 1 Gradient of line = 2 = B1 − 5 5 2 − 1 1 Equation is y − 3 = ( x − 2) OE B1 Or y − 3 1 2 = − 1 5 5( x + 2 ) [3] 2 2

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11 Functions f and g are defined for x ∈> by f : x →2x + 1, g : x →x2 −2. (i) Find and simplify expressions for fg(x) and gf(x). [2] (ii) Hence find the value of a for which fg(a) = gf(a). [3] (iii) Find the value of b (b ≠a) for which g(b) = b. [2] (iv) Find and simplify an expression for f −1g(x). [2] The function h is defined by h : x →x2 −2, for x ≤0. (v) Find an expression for h−1(x). [2]

Mark scheme: 11 (i) fg(x) = 2x2 – 3, gf(x) = 4x2 + 4x – 1 B1, B1 fg & gf clearly transposed gets B0B0 [2] (ii) 2 a 2 − 3 = 4 a 2 + 4 a − 1 ⇒ 2 a 2 + 4 a + 2 = 0 M1 Dep. quadratic. Allow x for all 3 marks (a + 1)2 = 0 M1 Allow marks in (ii) if transposed in (i) a = –1 A1 [3] (iii) b2 – b – 2 = 0 → (b + 1)(b – 2) = 0 M1 Allow in terms of x for M1 only b = 2 Allow b = –1 in addition A1 Correct answer without working B2 [2] 1 (iv) f –1(x) = ( x − )1 B1 2 1 2 –1 f –1g(x) = ( x − 3) B1√ Must be simplified. Ft from their f 2 [2] (v) x = ( ± ) y + 2 M1 h–1(x) = − x + 2 A1 [2]

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