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

1 6 1 Find the term independent of x in the expansion of 2x + x2 . [3]

Mark scheme: −1 1f ( x ) = 2 − ( x − 2 ) A1 [3] cao 2

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2 A curve has equation y = 3x3 −6x2 + 4x + 2. Show that the gradient of the curve is never negative. [3]

Mark scheme: 2 1 Allow +√ or √. Dep on final ans as (v) ( x − 2 ) = ( y − 2 ) M1 n 2 f of x 1 x = 2 ± ( y − 2 ) M1 2 1 ( ) ( )

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3 (i) Sketch, on a single diagram, the graphs of y cos 2θ and y 1 for 0 [3] 2 = = ≤θ ≤2π. (ii) Write down the number of roots of the equation 2 cos 2θ −1 = 0 in the interval 0 ≤θ ≤2π. [1] (iii) Deduce the number of roots of the equation 2 cos 2θ −1 = 0 in the interval 10π ≤θ ≤20π. [1]

Mark scheme: (3 x − 2 ) 2 ≥ 0 A1 [3] 3 (i) Correct cosine curve for at least 1 Range − 1 → 1 . Ignore labels on B1 oscillation θ axis Exactly 2 complete oscillations in[2,0π ] B1 1 Line y = correct B1 [3] 2 (ii) 4 Ft their graph. Accept 30 ° , 150 ° , B1√ [1] 210 ° , 330 ° (iii) 20 B1√ [1] Or 5 × their part (ii) 4 (i) 3 B1 [1] (ii) f ( x ) = x 2 − 6 x (+ c ) M1A1 Dependent on c present Subst (,3 −4 ) M1 cao c = 5 → f ( x ) = x 2 − 6 x + 5 A1 [4] 5 (i) Arc AB = r θ M1 θ OC = r sin θ or BC = r cos θ M1 oe eg BC = r sin etc tan θ r (1 + θ + cos θ + sin θ ) A1 [3] OC & BC reversed loses M1A1 correctly derived 1 2 π (ii) Sector OAB = × 10 × (= 31 .42 ) M1 oe ∆ in terms of π and 10 2 5 1 ∆OCB =  π  π  Allow OC & BC reversed 2 10 cos  10 sin  M1  5  5  (ie max 4/6) (= 23.78 ) Total area = 55 2. A1 [3] 6 (a) a + 5 d = 23 B1 Solution of 2 linear equations ( )

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5 C B q rad r q rad O A r The diagram represents a metal plate OABC, consisting of a sector OAB of a circle with centre O and radius r, together with a triangle OCB which is right-angled at C. Angle AOB = θ radians and OC is perpendicular to OA. (i) Find an expression in terms of r and θ for the perimeter of the plate. [3] (ii) For the case where r = 10 and θ = 15π, find the area of the plate. [3]

Mark scheme: 5(2 a + 9 d ) = 200 B1 Attempt solution, expect d = 6 a = −7 M1

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7 x 2y 3y 3x y 4x The diagram shows the dimensions in metres of an L-shaped garden. The perimeter of the garden is 48 m. (i) Find an expression for y in terms of x. [1] (ii) Given that the area of the garden is A m2, show that A = 48x −8x2. [2] (iii) Given that x can vary, find the maximum area of the garden, showing that this is a maximum value rather than a minimum value. [4]

Mark scheme: 1 7 (i) y = oe B1 [1] 6(48 − 8 x ) (ii) A = 4 xy + 2 xy or 3 xy + 3 xy = 6 xy M1 A = x (48 − 8 x ) = 48 x − 8 x 2 A1 [2] AG δA (iii) = 48 − 16 x B1 δx Attempt to solve derivative = 0 A = 72 cao M1A1 Expect x = 3 δ 2 A = − 16 (< 0 ) ⇒ Maximum B1 [4] www Accept other complete methods 2 δx x x + y y + z z

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8 Relative to an origin O, the point A has position vector 4i + 7j −pk and the point B has position vector 8i −j −pk, where p is a constant. (i) Find −−→OA.−−→OB. [2] (ii) Hence show that there are no real values of p for which OA and OB are perpendicular to each other. [1] (iii) Find the values of p for which angle AOB = 60◦. [4]

Mark scheme: x1 x 2 + y1 y 2 + z1 z 2 28 (i) (4i + 7 j − pk ). (8i − j − pk ) = 25 + p M1A1 [2] 2 (Not 25 + (− p ) ) 2 Ft provided equation has no real (ii) 25 + p = 0 ⇒ no real solutions B1√ [1] solutions OA.OB (iii) cos 60 = used M1 OA.OB must be scalar OA OB Not 65 − p 2 OA = 65 + p 2 or OB = 65 + p 2 M1 unless follows 65 + (− p ) 2 25 + p 2 1 his scalar (i ) 1 Scalar product = 25 + p 2 can score = or = A1√ 2 2 65 + p 2 65 + p 2 here if not scored in part (i) p = ± .387 or ± 15 A1 [4] 2 2 ( )

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9 A line has equation y = kx + 6 and a curve has equation y = x2 + 3x + 2k, where k is a constant. (i) For the case where k = 2, the line and the curve intersect at points A and B. Find the distance AB and the coordinates of the mid-point of AB. [5] (ii) Find the two values of k for which the line is a tangent to the curve. [4] [Questions 10 and 11 are printed on the next page.]

Mark scheme: 9 (i) x 2 + 3 x + 4 = 2 x + 6 ⇒ x 2 + x − 2 (= 0 ) M1 3-term simplification ( x − 1)( x + 2 ) = 0 → (8,1 ), (− 2,2 ) DM1A1 DM1 for attempted solution for x 2 2 cao ( 45 from wrong points scores AB = 3 + 6 = .6 71 or 45 or 3 5 B1 B0) − 1 5,  B1√ [5] Ft their coordinates  2  GCE AS/A LEVEL – October/November 2011 9709 11 (ii) x 2 + (3 − k )x + 2 k − 6(= 0 ) M1 Simplified to 3-term quadratic 2 (3 − k ) 2 − 4(2 k − 6 ) = 0 DM1 Apply b −ac4 = 0 as function of k only (3 − k )(11 − k ) = 0 DM1 Attempt factorisation or use formula Both correct NB Alternative methods for (ii) k = 3 or 11 A1 [4] possible If B0B0 then SCB1 for both y = 1 & ( ) ( )

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10 y y = Ö(1 + 2x ) C B x A O meeting the x-axis at A and the y-axis at B. The The diagram shows the curve y = √(1 + 2x) y-coordinate of the point C on the curve is 3. (i) Find the coordinates of B and C. [2] (ii) Find the equation of the normal to the curve at C. [4] (iii) Find the volume obtained when the shaded region is rotated through 360◦about the y-axis. [5]

Mark scheme: If B0B0 then SCB1 for both y 1 & 10 (i) B = ()1,0 C = (3,4) B1, B1 [2] x = 4 1 δy 1 − 1 − 2 required & at least one of 1 × 2 (ii) = × 2(1 + 2 x ) 2 M1A1 2 δx 2 for M1 Grad. of normal = −3 B1 y − 3 = −3( x − 4 ) or y = −3 x + 15 oe B1√ [4] Ft only from their C 2 1 2 2 1 x δy , square ( y − )1 & attempt ∫ (iii) y = 1 + 2 x ⇒ x = SOI B1 2 2 2 ( y − )1 n int 1 4 2 (π ) × × ( y − 2 y + 1)δy M1 ∫ 4 1  y 5 2 y 3  Apply limits 0 → their 1 (from their (π ) × − + y A1 B) 4  5 3  2 π 2 1  1  − + 1 (π ) × DM1 cao SCB1 for ∫ y δx →4 (scores 4  5 3  1/5) 2 π A1 [5] 15 ( ) 2 B 1 B1

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11 Functions f and g are defined by f : x →2x2 −8x + 10 for 0 ≤x ≤2, g : x →x for 0 ≤x ≤10. (i) Express f(x) in the form a(x + b)2 + c, where a, b and c are constants. [3] (ii) State the range of f. [1] (iii) State the domain of f −1. [1] (iv) Sketch on the same diagram the graphs of y = f(x), y = g(x) and y = f −1(x), making clear the relationship between the graphs. [4] (v) Find an expression for f −1(x). [3]

Mark scheme: 2 B 1, B1,11 (i) 2( x − 2 ) + 2 [3] For 2 , − 2 , 2 B1 (ii) 2 ≤ f ( x ) ≤ 10 oe B1 [1] Allow < etc. Ignore notation (iii) 2 ≤x ≤ 10 B1√ [1] Ft from part (ii). Ignore notation Or from int with y axis to int with (iv) f ( x ) ≈: half parabola from (,010 ) to (2,2 ) B1 their y = x g ( x ): line through 0 at ≈45 ° B1 f −1 ( x ): reflection of their f ( x ) in g ( x ) B1√ Everything totally correct B1 [4] GCE AS/A LEVEL – October/November 2011 9709 11 2 1 Allow +√or ‒√Dep on final ans as ( ) ( )

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