Cambridge A Level Mathematics 9709 — 2017 Oct/Nov Paper 1 · Variant 2

1 Find the term independent of x in the expansion of 2x −1 9 . [4] 4x2 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 EITHER: Term is 9 3 C × 26 × (−¼)3 OR1: ( ) ( ) 9 9 9 3 9 9 3 3 2 2 2 8 1 1 1 8 1 1 8 4 4 4   −     = − − −             x x or x x x x Term is 9 6 3 9 1 8 4 − × × C (B1, B1, B1) OE OR2: ( ) 9 9 3 1 2 1 8   −     x x Term is 3 9 9 3 1 2 8   × × −     C (B1, B1, B1) OE Selected term, which must be independent of x = −84 B1 4

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4 C P Q A B D O 6 cm The diagram shows a semicircle with centre O and radius 6 cm. The radius OC is perpendicular to the diameter AB. The point D lies on AB, and DC is an arc of a circle with centre B. (i) Calculate the length of the arc DC. 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(ii) Find the value of area of region P area of region Q, giving your answer correct to 3 significant figures. 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Mark scheme: 4(i) Pythagoras → r = 72 OE or 6 6 cos45 6 2 cos45 = → = = r r Arc DC = 72 × ¼π = 3 2 2 π , 2.12π, 6.66 M1 A1 Use of s=rθ with their r (NOT 6) and ¼π 3 4(ii) Area of sector BDC is ½ × 72 × ¼π (= 9π or 28.274…) *M1 Use of ½r²θ with their r (NOT 6) and ¼π Area Q = 9π – 18 (10.274…) DM1 Subtracts their ½ × 6 × 6 from their ½r²θ Area P is (¼π6² – area Q) = 18 M1 Uses {¼ π6² – (their area Q using 72 )} Ratio is 18 9 18 π − 18 10.274       → 1.75 A1 4

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7 Points A and B lie on the curve y = x2 −4x + 7. Point A has coordinates 4, 7 and B is the stationary point of the curve. The equation of a line L is y = mx −2, where m is a constant. (i) In the case where L passes through the mid-point of AB, find the value of m. 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(ii) Find the set of values of m for which L does not meet the curve. 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Mark scheme: 7(i) d 2 4 0 d = − = y x x 2, y → = = x 3 B1 B1 Midpoint of AB is (3, 5) B1 FT FT on (their 2, their 3) with (4,7) 7 3 m → = (or 2.33) B1 4 7(ii) Simultaneous equations → ( ) 2 4 9 0 − − + = x x mx *M1 Equates and sets to 0 must contain m Use of b²−4ac → (m + 4)² − 36 DM1 Any use of b²−4ac on equation set to 0 must contain m Solves = 0 → −10 or 2 A1 Correct end-points. −10 < m < 2 A1 Don’t condone ⩽ at either or both end(s). Accept −10 < m, m < 2. 4

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9 A O B C The diagram shows a trapezium OABC in which OA is parallel to CB. The position vectors of A and ` a ` a 2 6 −−→ −−→ B relative to the origin O are given by OA = −2 and OB = 1 . −1 1 (i) Show that angle OAB is 90Å. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ −−→ −−→ The magnitude of CB is three times the magnitude of OA. (ii) Find the position vector of C. 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(iii) Find the exact area of the trapezium OABC, giving your answer in the form a b, where a and b are integers. 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Mark scheme: 9(i) JJJG AB = 4 3 2      or 4 3 2 −     = −     −   JJJG BA e.g. JJJG AO . JJJG AB =− 8 + 6 + 2 = 0 → ˆ OAB = 90° AG OR 3, 38 , 29 = = = JJJG JJJG JJJG OA OB AB OA2 + AB2 = OB2 → ˆ OAB = 90° AG M1 A1 Use of dot product with either & AOorOA either AB or BA JJJG JJJG JJJJG JJJG . Must see 3 component products OR Correct use of Pythagoras. In both methods must state angle or Ө 90 ° = or similar for A1 3 9(ii) JJJG CB = 6 6 3     −     −   or 6 6 3 −     =       JJJG BC B1 Must correctly identify the vector. JJJG OC = JJJG OB + JJJG BC (or − JJJG CB ) = 0 7 4      M1 A1 Correct link leading to JJJG OC 3 Question Answer Marks Guidance 9(iii) 3, 9, 29 = = = JJJG JJJG JJJG OA BC AB (5.39 ) B1 For any one of these Area = ½(3 + 9) 29 or 3 29 + 3 29 M1 Correct formula(e) used for trapezium or (rectangle + triangle) or two triangles using their lengths. = 6 29 (1 1044,2 261 3 116) or A1 Exact answer in correct form. 3

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10 y y = 5x −1 P 2, 3 Q x O The diagram shows part of the curve y = 5x −1 and the normal to the curve at the point P 2, 3 . This normal meets the x-axis at Q. (i) Find the equation of the normal at P. 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(ii) Find, showing all necessary working, the area of the shaded region. 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Mark scheme: 10(i) ( ) 1 2 d 1 5 1 d 2 − = × − y x x × 5 ( = 5 6 ) of normal = m 6 5 − M1 Uses m1m2 = −1 with their numeric value from their dy/dx Equation of normal ( ) 6 3 2 5 − = − − y x OE or 5y + 6x = 27 or 6 27 5 5 − = + y x A1 Unsimplified. Can use = + y mx c to get 5.4 = c ISW Question Answer Marks Guidance 10(ii) EITHER: For the curve ( ) 5 1d ∫ − x x = ( ) 3 2 5 1 3 2 − x ÷ 5 (B1 Correct expression without ÷5 B1 For dividing an attempt at integration of y by 5 Limits from 1 5 to 2 used → 3.6 or 18 5 OE M1 A1 Using 1 5 and 2 to evaluate an integrand ( ) 2 may be ∫y Normal crosses x-axis when y = 0, → x= (4½) M1 Uses their equation of normal, NOT tangent Area of triangle = 3.75 or 15 4 OE A1 This can be obtained by integration Total area=3.6 + 3.75 = 7.35, 147 20 OE A1) OR: For the curve: ( ) ( ) 2 1 1 d 5 ∫ + y y = 3 1 5 3   +     y y (B2, 1, 0 –1 each error or omission. Limits from 0 to 3 used → 2.4 or 12 5 OE M1 A1 Using 0 and 3 to evaluate an integrand Uses their equation of normal, NOT tangent. M1 Either to find side length for trapezium or attempt at integrating between 0 and 3 Area of trapezium = ( ) 1 39 3 2 4½ 3 9 2 4 4 + × = or A1 This can be obtained by integration Shaded area = 39 12 147 7.35, 4 5 20 − = OE A1) Question Answer Marks Guidance 7

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