Cambridge A Level Mathematics 9709 — 2018 Oct/Nov Paper 1 · Variant 1

1 1 Showing all necessary working, solve the equation 4x −11x 2 + 6 = 0. [3] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 ½ ½ 4 3 2 − − x x oe soi Alt: 2 4 6 11 16 73 36 + = ⇒ − + x x x x M1 Attempt solution for ½ x or sub u = ½ x ½ 3/ 4 or 2 = x ( )( ) 16 9 4 − − x x A1 Reasonable solutions for ½ x implies M1 (ݔ= 2, 3/4, M1A0) 9 /16 oe or 4 = x A1 Little or no working shown scores SCB3, spotting one solution, B0 3

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2 A line has equation y = x + 1 and a curve has equation y = x2 + bx + 5. Find the set of values of the constant b for which the line meets the curve. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 2 2 5 1 1 4 0 + + = + → + − + = x bx x x x b M1 ( ) 2 2 ( 4 ) 1 16 − = − − b ac b M1 b associated with ‒3 & +5 or 1 − b associated with 4 ± A1 ( ) ( ) 2 2 2 0 2 0, 2, 1 4 − = + = = ± −= ± x or x x b (M1A1) Association can be an equality or an inequality b ⩾ 5, b ⩽ –3 A1 4

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3 Two points A and B have coordinates 3a, −a and −a, 2a respectively, where a is a positive constant. (i) Find the equation of the line through the origin parallel to AB. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) The length of the line AB is 31 units. Find the value of a. [3] 3 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(i) Gradient of AB = ‒3/4 B1 Accept ‒3a/4a 3 4 = − y x oe B1FT Answer must not include a. Ft on their numerical gradient 2 3(ii) ( ) ( ) ( ) 2 2 2 4 3 10 / 3 + = a a soi M1 May be unsimplified 2 25 100 / 9 = a oe A1 a = 2/3 A1 3

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4 The first term of a series is 6 and the second term is 2. (i) For the case where the series is an arithmetic progression, find the sum of the first 80 terms. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) For the case where the series is a geometric progression, find the sum to infinity. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) ( ) 80 80 12 79 4 2 =  + × −    S or [ ] 80 6 , 310 2 + = − l l M1A1 –12 160 A1 3 4(ii) 6 9 1 1 3 S∞= = − M1A1 Correct formula with 1 < r for M1 2

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5 (i) Show that the equation cos 1 −4 4 sin 1 − = 0 sin 1 5 cos 1 −2 may be expressed as 9 cos21 −22 cos 1 + 4 = 0. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence solve the equation cos 1 −4 4 sin 1 − = 0 sin 1 5 cos 1 −2 for 0Å ≤1 ≤360Å. 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Mark scheme: 5(i) ( ) 2 (cos 4)(5cos 2) 4sin 0 sin (5cos 2) θ θ θ θ θ − − − = − M1 ( ) ( ) 2 2 5cos 22cos 8 4 1 cos 0 sin (5cos 2) θ θ θ θ θ − + − − = − M1 Simplify numerator and use 2 2 1 = − s c . Accept numerator only 2 9cos 22cos 4 0 θ θ − + = www AG A1 3 5(ii) Attempt to solve for cosθ , (formula, completing square expected) M1 Expect cos 0.1978 θ = . Allow 2.247 in addition 78.6 , 281.4 θ = ° ° (only, second solution in the range) A1A1FT Ft for (360º ‒ 1st solution) 3

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dy 6 A curve has a stationary point at 3, 91 and has an equation for which = ax2 + a2x, where a is a 2 dx non-zero constant. (i) Find the value of a. 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(ii) Find the equation of the curve. 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(iii) Determine, showing all necessary working, the nature of the stationary point. 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Mark scheme: 6(i) 2 0 9 3 = + a a M1 Sub d 0 and 3 d y x x = = 3 = − a only A1 2 6(ii) ( ) 2 2 3 d 9 3 9 d 2 y x x x y x c x = − + → = − + + M1A1FT Attempt integration. 3 2 2 13 ½ + ax a x scores M1. Ft on their a. 9½ 27 40½ = − + + c DM1 Sub 3, 9½ = = x y . Dependent on c present 4 = − c A1 Expect 2 3 9 4 2 = − + − x y x 4 6(iii) 2 2 d 6 9 d y x x = − + M1 2 2 + ax a scores M1 At 2 2 d 3, 9 0 d y x x = = − < MAX www A1 Requires at least one of ‒9 or < 0. Other methods possible. 2

More questions on Differentiation

7 y y = x A 2, 2 y = k x3 −7x2 + 12x O x The diagram shows part of the curve with equation y = k x3 −7x2 + 12x for some constant k. The curve intersects the line y = x at the origin O and at the point A 2, 2 . (i) Find the value of k. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Verify that the curve meets the line y = x again when x = 5. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find, showing all necessary working, the area of the shaded region. [5] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) 2 = k(8 ‒ 28 + 24) → k = 1/2 B1 1 7(ii) When x = 5, y = [½](125 ‒ 175 + 60) = 5 M1 Or solve [ ]( ) [ ] 3 2 ½ 7 12 5 0, 2 − + = ⇒ = = x x x x x x Which lies on y = x, oe A1 2 7(iii) 3 2 1 [ ( 7 12 ) ] 2 x x x x dx − + − ∫ . M1 Expect 3 2 1 7 5 2 2 ∫ − + x x x 4 3 2 1 7 5 8 6 2 − + x x x B2,1,0FT Ft on their k 2 ‒ 28/3 +10 DM1 Apply limits 0 → 2 8/3 A1 OR 4 3 2 1 7 3 8 6 − + x x x B2,1,0FT Integrate to find area under curve, Ft on their k 2 ‒ 28/3 +12 M1 Apply limits 0 → 2. Dep on integration attempted Area ∆ = ½ × 2 × 2 or 2 2 0 d ½ 2   = =   ∫x x x M1 8/3 A1 5

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8 F E D 6 4 4 C B k j 3 O i 6 A The diagram shows a solid figure OABCDEF having a horizontal rectangular base OABC with OA = 6 units and AB = 3 units. The vertical edges OF, AD and BE have lengths 6 units, 4 units and 4 units respectively. Unit vectors i, j and k are parallel to OA, OC and OF respectively. −−→ (i) Find DF. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−→ (ii) Find the unit vector in the direction of EF. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Use a scalar product to find angle EFD. 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Mark scheme: 8(i) 6 2 = − + DF i k B1 1 8(ii) 6 3 2 = − − + JJJG EF i j k B1 | ( ) ( ) 2 2 2 | 6 3 2 = − + − + JJJG EF M1 Must use their JJJG EF Unit vector = ( ) 1 6 3 2 7 − − + i j k A1 3 8(iii) . JJJG JJJG DF EF = (‒6i + 2k).(‒6i ‒ 3j + 2k) = 36 + 4 = 40 M1 | JJJG DF | = √40, | JJJG EF | =7 M1 cosEFD = 40 7 40 oe M1 EFD = 25.4º A1 Special case: use of cosine rule M1(must evaluate lengths using correct method) A1 only 4

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9 A C 5 cm 150 rad O D B The diagram shows a triangle OAB in which angle ABO is a right angle, angle AOB = 150 radians and AB = 5 cm. The arc BC is part of a circle with centre A and meets OA at C. The arc CD is part of a circle with centre O and meets OB at D. Find the area of the shaded region. [8] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ 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Mark scheme: 9 Angle OAB / 2 / 5 3 /10 soi B1 Sector 2 1 3 5 2 10 π   = × ×     CAB their M1 Expect 11.78 5 8.507 sin 5 π = = OA M1A1 May be implied by 3.507 = OC Sector ( ) 1 3.507 ² 2 5 π = × × COD their M1 Expect 3.86 ( ) 1 3 5 8.507 sin 2 10 π ∆ = × × OAB their M1 Or 1 5 5 2 tan 5 π × × or 2.5 × ( ) 2 8.507 25 − their = 17.20 or 17.21 A1 Shaded area ( ) 17.20 1 7.21 11.78 3.86 1.56 or 1.57 − − = or A1 8

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10 A curve has equation y = 1 4x −3 −1. The point A on the curve has coordinates 1, 1 . 2 2 (i) (a) Find and simplify the equation of the normal through A. [5] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Find the x-coordinate of the point where this normal meets the curve again. [3] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (ii) A point is moving along the curve in such a way that as it passes through A its x-coordinate is decreasing at the rate of 0.3 units per second. Find the rate of change of its y-coordinate at A. 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Mark scheme: 10(i)(a) ( ) [ ] 2 d ½ 4 3 4 d y x x −   = − − ×   B1B1 When 1, 2 = = − x m B1FT Ft from their d d y x Normal is ( ) ½ ½ 1 − = − y x M1 Line with gradient ‒1/m and through A ½ = y x soi A1 Can score in part (b) 5 10(i)(b) ( ) ( ) ( )( ) ( ) 2 1 2 4 3 2 2 4 3 1 0 2 4 3 2 = → − = → − − = − x x x x x x M1A1 x/2 seen on RHS of equation can score previous A1 1/ 4 = − x A1 Ignore 1 = x seen in addition 3 10(ii) Use of chain rule: ( ) ( ) d 2 0.3 0.6 d y their t = − × ± = M1A1 Allow +0.3 or ‒0.3 for M1 2

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