Cambridge A Level Mathematics 9709 — 2020 Oct/Nov Paper 1 · Variant 2

2 The first, second and third terms of a geometric progression are 2p 6, and p 2 respectively, + −2p + where p is positive. Find the sum to infinity of the progression. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 2 (−2p)² = (2p + 6) × (p + 2) or 2 2 2 6 2 p p p p − + = + − M1 OE. Using “a, b, c then b² = ac” or a =2p+6 , ar = −2p and ar² = p + 2 to form a correct relationship in terms of p only ( 2 ² 10 12 0 − − = p p ) p = 6 A1 a = 18 and r = −⅔ A1 (s∞ ) = their a ÷ (1 − their r) 5 18 3   = ÷     M1 Correct formula used with their values for a and r , r < 1 Both a & r from the same value of p. (s∞ = )10.8 A1 OE. A0 if an extra solution given SC B2 for s∞ = 2 6 2 6 2 2 1 1 2 6 2 + + − + − − + − p p or p p p p ignore any subsequent algebraic simplification. 5

More questions on Series

4 The sum, Sn, of the first n terms of an arithmetic progression is given by Sn n2 4n. = + The kth term in the progression is greater than 200. Find the smallest possible value of k. [5] ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 4 1 + x x S and S M1 Using two values of n in the given formula a = 5, d = 2 A1 A1 a + (n – 1) d > 200 → 5 + 2(k – 1) > 200 M1 Correct formula used with their a and d to form an equation or inequality with 200, condone use of n (k =) 99 A1 Condone ⩾ 99 Alternative method for question 4 ( 2 n 2a + (n – 1) d ) ≡ 2 4 + n n → 1 1 , 4 2 2 d a d   = − =     M1 Equating two correct expressions of n S and equating coefficients of n and 2 n d = 2, a = 5 A1 A1 a + (n – 1) d > 200 → 5 + 2(k – 1) > 200 M1 Correct formula used with their a and d to form an equation or inequality with 200, condone use of n (k =) 99 A1 Condone ⩾ 99 Alternative method for question 4 ( ) 2 2 1 4 ( 1) 4 1 − − → + − − − − k k sum sum k k k k M1 A1 Using given formula with consecutive expressions subtracted. Allow k+1 and k. 2 3 200 or 200 k + > = M1 A1 Simplifying to a linear equation or inequality (k =) 99 A1 Condone ⩾ 99 5

More questions on Series

@ A @ A 1 1 1 6 (a) Prove the identity x 1 [4] cosx −tan sin x +  tan x. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ @ A @ A 1 1 (b) Hence solve the equation x 1 2 tan2x for [2] cos x −tan sin x + = 0Å ≤x ≤180Å. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) 1 sin 1 1 cos cos sin    − +       x x x x B1 Uses “tanx = sinx ÷ cosx” throughout 1 sin 1 sin cos sin − +          x x x x or 1 sin² cos sin −       x x x M1 Correct algebra leading to two or four terms cos² cos sin x x x       A1 OE. A correct expression which can be cancelled directly to cos sin x x e.g. ( ) ( ) 2 2 cos 1 sin sin 1 sin x x x x − − cos² cos sin       x x x = cos sin       x x = 1 tanx A1 AG. Must show cancelling. If x is missing throughout their working withhold this mark. 4 6(b) Uses (a) → 1 2tan² tan = x x tan³x = 1 2 M1 Reducing to tan³x = k. (x =) 38.4° A1 AWRT. Ignore extra answers outside the range 0 to 180° but A0 if within. 2

More questions on Trigonometry

2.7 The point 4, 7 lies on the curve y 2 = f x and it is given that f ′ x = 6x−1 −4x−3 (a) A point moves along the curve in such a way that the x-coordinate is increasing at a constant rate of 0.12 units per second. Find the rate of increase of the y-coordinate when x 4. [3] = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find the equation of the curve. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) ( ) f ' 4 5 2   =     *M1 Substituting 4 into ( ) f ' x d d d d d d y y x t x t   = ×     → d d y t       = 5 2 × 0.12 DM1 Multiplies their ( ) f ' 4 by 0.12 d d   =     y t 0.3 A1 OE 3 7(b) ( ) 1 1 2 2 6 4 1 1 2 2 x x c − − + − B1 B1 B1 for each unsimplified integral. Uses (4, 7) leading to c = (-21) M1 Uses (4, 7) to find a c value ( ) 1 1 2 2 8 or f 12 8 21 or 1 2 21 y x x x x x − = + − + − A1 Need to see y or f(x) = somewhere in their solution and 12 and 8 4

More questions on Integration

8 D A C 1 rad r cm r cm B In the diagram, ABC is an isosceles triangle with AB BC r cm and angle BAC radians. The = = = 1 point D lies on AC and ABD is a sector of a circle with centre A. (a) Express the area of the shaded region in terms of r and [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) In the case where r 10 and 0.6, find the perimeter of the shaded region. [4] = 1 = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 8(a) Use of correct formula for the area of triangle ABC M1 Use of 180–2θ scores M0. Condone 2π–2θ ( ) 2 1 2 sin π 2 r −θ or 2 1 2 sin2 r θ or 1 2 2 cos sin r r × × × θ θ or 1 2 2 cos sin r r × × θ θ A1 OE [Shaded area = triangle – sector] = their triangle area − 2 1 2 r θ B1 FT FT for their triangle area − 2 1 2 r θ (Condone use of 180 degrees for triangle area for B1) 3 8(b) Arc BD = rθ = 6 cm B1 SOI AC = 2rcosθ = ( 2×10cos0.6 = 20cos0.6 = 16.506) or ( ) ( ) 2 2 2 2 cos π 2 r r − −θ or ( ) sin π – 2 sin r × θ θ *M1 Finding AC or 1 2 AC (= 8.25) DC = 2rcosθ – r or ( ) ( ) 2 2 2 2 cos π 2 r r − −θ – r ( = 6.506) DM1 Subtracting r from their AC or r-rcosθ from their half AC (8.25-1.75) (Perimeter = 10 + 6 + 6.506 =) 22.5 A1 AWRT 4

More questions on Circular measure

10 y x O M 2 The diagram shows part of the curve y and its minimum point M, which lies on the 2 −x = 3 −2x x-axis. dy d2y (a) Find expressions for dx, dx2 and Ó y dx. [6] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Find, by calculation, the x-coordinate of M. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (c) Find the area of the shaded region bounded by the curve and the coordinate axes. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 10(a) d (d y x ) = [8] ( 3 [ 3 2 ) − × − +  x [ −1] ( ) 8 1 3 2 ³ x   = −     −   B2, 1, 0 B2 for all three elements correct, B1 for two elements correct, B0 for only one or no elements correct. d² d ² y x = ( ) 4 3 8 (3 2 ) 2 − −× × − × − x ( ) 4 48 3 2x     =   −   B1 FT FT providing their bracket is to a negative power ∫ydx = [(3− 2x)−1 ] [2 ÷ (−1 × −2)] [ − ½x²] (+c) 2 1 1 3 2 2 x c x   = − +   −   B1 B1 B1 Simplification not needed, B1 for each correct element 6 Question Answer Marks Guidance 10(b) d d y x = 0 → (3 – 2x)³ = 8 → 3 – 2x = k → x = M1 Setting their 2-term differential to 0 and attempts to solve as far as x = 1 2 A1 Alternative method for question 10(b) ( ) 2 2 2 0 0 2 (2 1) 0 (3 2 ) y x x x x x = → − = → − − = → = − M1 Setting y to 0 and attempts to solve a cubic as far as x = (3 factors needed) 1 2 A1 2 10(c) Area under curve = their 1 ² 1 1 2 0 1 2 3 2 0 3 2 2               − − −   − ×       − ×        M1 Using their integral, their positive x limit from part (b) and 0 correctly. 1 24 A1 2

More questions on Differentiation