Cambridge A Level Mathematics 9709 — 2019 Oct/Nov Paper 2 · Variant 2

2 (i) Solve the equation 4x 5 x . [3] + = −7 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Hence, using logarithms, solve the equation 5 2y , giving the answer correct to 3 significant figures. 2y+2 + = −7 [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 2(i) State or imply non-modular equation 2 2 (4 5) ( 7) x x + = − or pair of different linear equations Attempt solution of 3-term quadratic equation or pair of linear equations M1 Obtain 2 5 and 4 − A1 SC For 4 x = −only, from correct work, allow B1 3 2(ii) Apply logarithms and use power law for 2y k = where 0 k > from (i) M1 Obtain –1.32 only A1 AWRT 2

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4 The sequence x1, x2, x3, defined by à xn x1 1, ln 2xn = xn+1 = converges to the value !. (i) Use the iterative formula to find the value of correct to 4 significant figures. Give the result of each iteration to 6 significant figures. ! [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) State an equation satisfied by and hence determine the exact value of [2] ! !. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(i) Use iteration correctly at least once M1 Must see correct attempt at 3x Obtain final answer 1.359 A1 Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.3585, 1.3595] A1 Answer required to exactly 4 sf Must see to at least 5x 3 4(ii) Form correct equation in x (or α ) B1 ln 2 x x x = OE Obtain 1 2 e B1 2

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18 3 6 (a) Show dx ln 27. [4] 2x = thatÔ2 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ 1 60 3 (b) Find the exact value of 4 sin2 2x dx. Show all necessary working. [5] Ó 0 ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 6(a) Obtain 3 2 ln x or 3 2 ln(2 ) x or 3 2 ln kx Use subtraction law of logarithms correctly, showing sufficient detail M1 216 ln216 ln8 ln 8   − =     Use power law of logarithms correctly M1 ( ) ( ) ln ln n n kx kx = Confirm ln27 with sufficient working and no incorrect working A1 AG 4 6(b) Use appropriate identity to express integrand in form 1 2 cos3 k k x + *M1 1 0 k ≠ . Allow 3 2 2 x × for 3x Obtain correct 2 2cos3x − A1 Integrate to obtain form 3 4 sin3 k x k x + DM1 Obtain correct 2 3 2 sin3 x x − A1 Use limits to obtain 1 2 3 3 π − or exact equivalent A1 5

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7 The parametric equations of a curve are x 3 sin y 1 2 tan = 21, = + 21, for 0 1 ≤1 < 40. (i) Find the exact gradient of the curve at the point for which 1 [4] 1 = 60. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (ii) Find the value of at the point where the gradient of the curve is 2, giving the value correct to 3 significant figures.1 [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(i) Obtain d d 6cos2 x θ θ = Obtain d 2 d 4sec 2 y θ θ = B1 Divide d d y θ by d d x θ with θ equated to 1 6 π M1 Obtain 16 3 or exact equivalent A1 Allow FT on A1 if d d 3cos2 x θ θ = and d 2 d 2sec 2 y θ θ = 4 7(ii) Equate expression for d d y x to 2 with only one trigonometry ratio used *M1 Either cos2 or sec2 θ θ Obtain 3 1 3 cos 2θ = or 3 sec 3 = A1 Attempt correct steps to find a value of θ from 3 cos 2 m θ = , 0 1 m < < DM1 Obtain 0.402 θ = and no others within the range A1 AWRT SC: Allow FT if d d 3cos2 x θ θ = and d 2 d 2sec 2 y θ θ = 4

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