Cambridge A Level Mathematics 9709 — 2019 Feb/March Paper 1 · Variant 2

3 C 7 8 Y X A B D In the diagram, CXD is a semicircle of radius 7 cm with centre A and diameter CD. The straight line YABX is perpendicular to CD, and the arc CYD is part of a circle with centre B and radius 8 cm. Find the total area of the region enclosed by the two arcs. [6] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 3 Angle CBA = 1 7 sin 8 −      = 1.0654 or 1 17 cos 2.13 32 − −   = =     CBD B1 Sector BCYD = ( ) 2 ½ 8 2 1.0654 rad × × ×their soi or sector CBY ( ) 2 ½ 8 1.0654 rad = × ×their M1 Expect 68.1(9). Angle must be in radians (or their 61/360 × 2 × 82) Or sector DBY ( ) 2 2 2 7 8 7 or ½ 8 sin 2 1.0654 ∆ = × − × × × BCD their soi M1 Expect 27.1(1). Award M1 for ABC or ABD Semi-circle CXD = ( ) 2 ½ 7 76.9 7 π × = M1 M1M1 for segment area formula used correctly Total area = their68.19 ‒ their27.11 + their76.97 = 118.0–118.1 M1A1 Cannot gain M1 without attempt to find angle CBA or CBD 6

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4 A curve has equation y = 2x −1 −1 + 2x. dy d2y (i) Find and . 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(ii) Find the x-coordinates of the stationary points and, showing all necessary working, determine the nature of each stationary point. 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Mark scheme: 4(i) 2 d / d 2 2 1 2 − = − − + y x x B2,1,0 ‘( ) 2 2 1 − − x ’ and ‘2’) ( ) 3 2 2 d / d 8 2 1 − = − y x x B1 Unsimplified form ok 3 Question Answer Marks Guidance 4(ii) Set d / d y x to zero and attempt to solve – at least one correct step M1 x = 0, 1 A1 Expect ( ) 2 2 1 1 − = x When x = 0, 2 2 d / d 8 (or 0) = − < y x . Hence MAX B1 When x = 1, 2 2 d / d 8 (or 0) = > y x . Hence MIN B1 Both final marks dependent on correct x and correct 2 2 d / d y x and no errors May use change of sign of dy / dx but not at 1/ 2 = x 4

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5 Two vectors, u and v, are such that ` a ` a q 8 u = 2 and v = q −1 , 6 q2 −7 where q is a constant. (i) Find the values of q for which u is perpendicular to v. 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(ii) Find the angle between u and v when q = 0. 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Mark scheme: 5(i) u.v = 2 8 2 2 6 42 + − + − q q q B1 2 6 10 44 0 + − = q q oe M1 Simplify, set to zero and attempt to solve q = 2, ‒11/3 A1 Both required. Accept ‒3.67 3 Question Answer Marks Guidance 5(ii) u = 0 2 6      v = 8 1 7     −     −   u.v = ‒2 ‒ 42 M1 Correct method for scalar product |u| × |v| = 2 2 2 2 2 2 6 8 1 7 + × + + M1 Prod of mods. At least one methodically correct. 44 44 4 cos 11 40 114 4 285 θ − − − = = = √ × M1 All linked correctly and inverse cos used correctly 130.7 θ = ° or 2.28(05) rads A1 No other angles between 0° and 180° 4

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6 (i) The first and second terms of a geometric progression are p and 2p respectively, where p is a positive constant. The sum of the first n terms is greater than 1000p. Show that 2n > 1001. 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(ii) In another case, p and 2p are the first and second terms respectively of an arithmetic progression. The nth term is 336 and the sum of the first n terms is 7224. Write down two equations in n and p and hence find the values of n and p. 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Mark scheme: 6(i) 2 1 2 1 − = − n n p S soi M1 ( ) 2 1 1000 2 1001 − > → > n n p p AG A1 2 Question Answer Marks Guidance 6(ii) ( ) 1 336 + − = p n p B1 Expect np = 336 ( ) 2 1 7224 2  + − =   n p n p B1 Expect ( ) 7224 2 + = n p np Eliminate n or p to an equation in one variable M1 Expect e.g. 168(1 + n) = 7224 or 1 + 336/p =43 etc n = 42, p = 8 A1A1 5

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9 y y = x3 + x2 P x O 3 The diagram shows part of the curve with equation y = x3 + x2 . The shaded region is bounded by the curve, the x-axis and the line x = 3. (i) Find, showing all necessary working, the volume obtained when the shaded region is rotated through 360Å about the x-axis. 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(ii) P is the point on the curve with x-coordinate 3. Find the y-coordinate of the point where the normal to the curve at P crosses the y-axis. 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Mark scheme: 9(i) 3 2 d π = ∫ + V x x x M1 Attempt 2d ∫y x ( ) 3 4 3 0 4 3 x x π   +     A1 ( ) ( ) 81 9 0 4 π   + −     DM1 May be implied by a correct answer 117 4 π oe A1 Accept 91.9 If additional areas rotated about x-axis, maximum of M1A0DM1A0 4 Question Answer Marks Guidance 9(ii) ( ) ( ) 1/2 3 2 2 d 1 3 2 d 2 − = + × + y x x x x x B2,1,0 Omission of 2 3 2 + x x is one error (At x = 3,) y = 6 B1 At x = 3, 1 1 11 33 2 6 4 = × × = m soi DB1ft Ft on their dy / dx providing differentiation attempted Equation of normal is ( ) 4 6 3 11 − = − − y x DM1 Equation through (3, their 6) and with gradient ‒1/their m When x = 0, y = 7 1 11 oe A1 6

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10 y 1 2 y = 4x x O 1 The diagram shows the curve with equation y = 4x 2. (i) The straight line with equation y = x + 3 intersects the curve at points A and B. Find the length of AB. 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(ii) The tangent to the curve at a point T is parallel to AB. Find the coordinates of T. 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(iii) Find the coordinates of the point of intersection of the normal to the curve at T with the line AB. 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Mark scheme: 10(i) 1/2 4 3 = + → x x ( ) 1/2 2 1/2 ( ) 4 3 0 − + = x x OR 2 16 6 9 = + + x x x M1 Either treat as quad in 1/2 x OR square both sides and RHS is 3-term 1/2 1 or 3 = x ( ) 2 10 9 0 − + = x x A1 If in 1st method 1/2 x becomes x, allow only M1 unless subsequently squared x = 1 or 9 A1 4 or1 2 = y A1ft Ft from their x values If the 2 solutions are found by trial substitution B1 for the first coordinate and B3 for the second coordinate ( ) ( ) 2 2 2 9 1 12 4 = − + − AB M1 128 or 8 2 = AB oe or 11.3 A1 6 10(ii) dy/dx = 2 1/2 − x B1 2 1/2 − x = 1 M1 Set their derivative = their gradient of AB and attempt to solve (4, 8) A1 Alternative method without calculus: MAB = 1, tangent is y = mx + c where m = 1 and meets y = 4x1/2 when 4x1/2 = x + c. This is a quadratic with b2 = 4ac, so 16 – 4 × 1 × ܿ= 0 so c = 4 B1 Solving 4x1/2 = x + 4 gives x = 4 and y = 8 M1A1 3 Question Answer Marks Guidance 10(iii) Equation of normal is ( ) 8 1 4 − = − − y x M1 Equation through their T and with gradient ‒1/their gradient of AB. Expect 12 = −+ y x , Eliminate y (or x) → 12 3 or 3 12 −+ = + − = − x x y y M1 May use their equation of AB (4½, 7½) A1 3

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