Cambridge A Level Mathematics 9709 — 2017 Feb/March Paper 1 · Variant 2

1 Find the set of values of k for which the equation 2x2 + 3kx + k = 0 has distinct real roots. [4] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 2 (3 ) 4 2 k k − × × M1 Attempt 2 4 b ac − 2 9 8 k k − > 0 soi Allow 2 9 8 k k − . 0 A1 Must involve correct inequality. Can be implied by correct answers 0, 8/9 soi A1 k < 0, k > 8/9 (or 0.889) A1 Allow (‒∞, 0) , (8/9, ∞) Total: 4

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@ A5 1 2 In the expansion of + 2ax2 , the coefficient of x is 5. Find the value of the constant a. [4] ax ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 5C2 ( ) 3 2 2 1 2ax ax       soi B1 Seen or implied. Can be part of an expansion. 2 3 1 10 4 5 a a × × = soi M1A1 M1 for identifying relevant term and equating to 5, all correct. Ignore extra x 8 a = cao A1 Total: 4

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3 12h h The diagram shows a water container in the form of an inverted pyramid, which is such that when the height of the water level is h cm the surface of the water is a square of side 12h cm. (i) Express the volume of water in the container in terms of h. [1] [The volume of a pyramid having a base area A and vertical height h is 3Ah.]1 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ Water is steadily dripping into the container at a constant rate of 20 cm3 per minute. (ii) Find the rate, in cm per minute, at which the water level is rising when the height of the water level is 10 cm. 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Mark scheme: 3(i) 3 1 12 V h = oe B1 Total: 1 3(ii) 2 d 1 d 4 V h h = or ( ) 2/3 d 4 12 d h v V − = M1A1 Attempt differentiation. Allow incorrect notation for M. For A mark accept their letter for volume - but otherwise correct notation. Allow V ′ d d d d d d h h V t V t = × 2 4 20 = × h soi DM1 Use chain rule correctly with ( ) d 20. d V t = Any equivalent formulation. Accept non-explicit chain rule (or nothing at all) d d h t       = 2 4 20 10 × = 0.8 or equivalent fraction A1 Total: 4

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4 C 8 cm 702 rad D B 8 cm A In the diagram, AB = AC = 8 cm and angle CAB = 270 radians. The circular arc BC has centre A, the circular arc CD has centre B and ABD is a straight line. (i) Show that angle CBD = 1409 radians. 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(ii) Find the perimeter of the shaded region. 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Mark scheme: 4(i) Total: 1 4(ii) ½ sin 7 8 BC π = or 8 2 5 sin sin 7 14 BC π π = or ( )( ) 2 2 2 2 8 8 2 8 8 cos 7 BC π = + − M1 BC = 6.94(2) A1 arc CD = their 6.94 9 /14 π × M1 Expect 14.02(0) arc 8 2 / 7 CB π = × M1 Expect 7.18(1) perimeter = 6.94 + 14.02 + 7.18 = 28.1 A1 Total: 5

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5 y y = tan x A x 0 O B y = cos x The diagram shows the graphs of y = tan x and y = cos x for 0 ≤x ≤0. The graphs intersect at points A and B. (i) Find by calculation the x-coordinate of A. 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(ii) Find by calculation the coordinates of B. 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Mark scheme: 5(i) 2 tan cos sin cos x x x x = → = 2 sin 1 sin x x = − M1 Use 2 2 cos 1 sin x x = − sin 0.6180 x = . Allow (‒1 + √5)/2 M1 Attempt soln of quadratic in sin x . Ignore solution ‒1.618. Allow x = 0.618 x-coord of A = 1 sin 0.618 0.666 − = cao A1 Must be radians. Accept 0.212π Total: 4 5(ii) EITHER x-coord of B is 0.666 their π − (M1 Expect 2.475(3). Must be radians throughout y-coord of B is tan( 2.475) or cos( 2.475) their their M1 x = 2.48, y = ‒0.786 or ‒0.787 cao A1) Accept x = 0.788π OR y-coord of B is – (cos or tan (their 0.666)) (M1 x-coord of B is 1 cos−(their y) or π + 1 tan−(their y) M1 x = 2.48, y = ‒0.786 or ‒0.787 A1) Accept x = 0.788π Total: 3

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6 Relative to an origin O, the position vectors of the points A and B are given by −−→ −−→ OA = 2i + 3j + 5k and OB = 7i + 4j + 3k. (i) Use a scalar product to find angle OAB. 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(ii) Find the area of triangle OAB. 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Mark scheme: 6(i) BA = OA ‒ OB = ‒5i ‒ j + 2k B1 Allow vector reversed. Ignore label BA or AB OA.BA = ‒10 ‒ 3 + 10 = ‒3 M1 soi by ±3 |OA|×|BA| = 2 2 2 2 2 2 2 3 5 5 1 2 + + × + + M1 Prod. of mods for at least 1 correct vector or reverse. / 3 cos 38 30 OAB + − = × M1 OAB = 95.1º (or c 1.66 ) A1 Total: 5 6(ii) ∆ OAB = 1 38 30 sin95.1 2 × . Allow ½ 38 74sin39.4 × M1 Allow their moduli product from (i) = 16.8 A1 cao but NOT from sin 84.9 (1.482c) Total: 2

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9 The point A 2, 2 lies on the curve y = x2 −2x + 2. (i) Find the equation of the tangent to the curve at A. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The normal to the curve at A intersects the curve again at B. (ii) Find the coordinates of B. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The tangents at A and B intersect each other at C. (iii) Find the coordinates of C. 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Mark scheme: 9(i) d 2 2 d y x x = − . At x = 2, m = 2 B1B1 Numerical m Equation of tangent is ( ) 2 2 2 y x −= − B1 Expect y = 2x ‒ 2 Total: 3 9(ii) Equation of normal ( ) 2 ½ 2 y x −= − − M1 Through (2, 2) with gradient = ‒1/m . Expect ½ 3 y x = − + 2 2 2 2 ½ 3 2 3 2 0 x x x x x − + = − + → − − = M1 Equate and simplify to 3-term quadratic ½, 3¼ x y = − = A1A1 Ignore answer of (2, 2) Total: 4 Question Answer Marks Guidance 9(iii) At ( ) ½, grad 2 ½ 2 3 x = − = − − =− B1 Ft their ‒½. Equation of tangent is ( ) 3¼ 3 ½ y x − = − + *M1 Through their B with grad their ‒3 (not m1 or m2). Expect 3 7 / 4 y x = − + 2 2 3 7 / 4 x x − = − + DM1 Equate their tangents or attempt to solve simultaneous equations 3 / 4, ½ x y = = − A1 Both required. Total: 4

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