Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 1 · Variant 1

1 6 1 Find the term independent of x in the expansion of 2x + . [3] 4x2 ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 6C2 ( ) 4 2 2 1 2 (4 ) × × x x SC: Condone errors in 1 2 (4 ) − evaluation or interpretation for B1 only 4 2 1 15 2 4 × × B1 Identified as required term. 15 B1 3

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4 A runner who is training for a long-distance race plans to run increasing distances each day for 21 days. She will run x km on day 1, and on each subsequent day she will increase the distance by 10% of the previous day’s distance. On day 21 she will run 20 km. (i) Find the distance she must run on day 1 in order to achieve this. Give your answer in km correct to 1 decimal place. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the total distance she runs over the 21 days. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Identifies common ratio as 1.1 B1 Use of ( ) 20 1.1 20 = x M1 SOI ( ) 20 20 3.0 1.1     = =     x A1 Accept 2.97 3 4(ii) ( ) 21 1.1 1 3.0 1.1 1   −   × − their → 192 M1 A1 Correct formula used for M mark. Allow 2.97 used from (i) Accept 190 from x = 2.97… 2

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8 C B 6 cm 380 rad O A The diagram shows a sector OAC of a circle with centre O. Tangents AB and CB to the circle meet at B. The arc AC is of length 6 cm and angle AOC = 380 radians. (i) Find the length of OA correct to 4 significant figures. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the perimeter of the shaded region. 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(iii) Find the area of the shaded region. 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Mark scheme: 8(i) 3 π 6 8 OA× = ( ) 16 5.093 0 π OA = = A1 8(ii) 3 5.0930 tan π 16 AB their = × M1 Perimeter = 2 3.4030 6 12.8 × + = A1 8(iii) Area OABC = ( ) 2 ½ 5.0930 3.4030 × × × their their M1 Area sector = ( ) 2 3 ½ 5.0930 π 8 their × × M1 Shaded area = 17.331 15.279 − their their = 2.05 M1A1

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dy 1 9 A curve for which = 5x −1 2 −2 passes through the point 2, 3 . dx (i) Find the equation of the curve. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ d2y (ii) Find . [2] dx2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the coordinates of the stationary point on the curve and, showing all necessary working, determine the nature of this stationary point. 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Mark scheme: 9(i) 1/2 3 2 [ 5 1 5 2 ] x x − ÷ ÷ − B1 B1 ( ) 27 3 4 3 / 2 5 = − + × c M1 Substitute x = 2, y = 3 ( ) 3 2 2 5 1 18 17 17 7 2 5 5 15 5   −   = − = → = − +       x c y x A1 9(ii) ( ) [ ] 1/2 2 2 d / d ½ 5 1 5 −   = − ×   y x x B1 B1 9(iii) ( ) 1/2 5 1 2 0 5 1 4 1 x x x − − = → −= = M1A1 Set d 0 d y x = and attempt solution (M1) 16 17 37 2 25 5 15 y = − + = A1 Or 2.47 or 37 1, 15       2 d 5 1 5 2 2 4 d x y x = × = (> 0) hence minimum A1 OE

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10 D C A B Relative to an origin O, the position vectors of the points A, B, C and D, shown in the diagram, are given by ` a ` a ` a ` a −1 2 4 2 −−→ −−→ −−→ −−→ OA = 3 , OB = −3 , OC = −2 and OD = 2 . −4 5 5 −1 (i) Show that AB is perpendicular to BC. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Show that ABCD is a trapezium. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (iii) Find the area of ABCD, giving your answer correct to 2 decimal places. 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Mark scheme: 10(i) 2 1 3 4 2 2 3 3 6 , 2 3 1 5 4 9 5 5 0 −                       = − − = − = − −− =                       −            AB BC AB.BC = 6 ‒ 6 → = 0 (hence perpendicular) B1 AG 10(ii) 4 2 2 2 2 4 5 1 6             = − − = −             −       DC B1 Or: CD 2 4 6 −     =     −   AB = kDC M1 OE Expect k = 3 2 Or: DC.BC = 4 ‒ 4 = 0 hence BC is also perpendicular to DC Or: AB.DC = 1 or AB.CD = –1, angle between lines is 0 or 180 AB is parallel to DC, hence ABCD is a trapezium A1 10(iii) |AB| = 9 36 81 126 11.22 + + = = |DC| = 4 16 36 56 7.483 + + = = |BC| = 4 1 0 5 2.236 + + = = M1 Method for finding at least 2 magnitudes Area = 1 2 ( ) + × theirAB theirDC theirBC = 20.92 M1A1 OE

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