Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 3 · Variant 1

1 Solve the equation 4 5x 5x, giving your answers correct to 3 decimal places. [4] −1 = ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................

Mark scheme: 1 State or imply non-modular equation ( ) ( ) 2 2 2 4 5 1 5 − = x x or pair of equations ( ) 4 5 1 5 − = ± x x M1 Obtain 4 5 3 = x and 4 5 5 = x (or 1 5 4 + = x ) A1 Use correct method for solving an equation of the form 5 = x a , or 1 5 + = x b where a > 0, or b > 0 M1 Obtain answers x = 0.179 and x = – 0.139 A1 Alternative method for question 1 Obtain 4 5 3 = x by solving an equation B1 Obtain 4 5 5 = x (or 1 5 4) + = x by solving an equation B1 Use correct method for solving an equation of the form 5 = x a , or 1 5 + = x b where a > 0, or b > 0 M1 Obtain answers x = 0.179 and x = – 0.139 A1 4

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3 The curve with equation y has one stationary point. = xe1−2x (a) Find the coordinates of this point. [4] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Determine whether the stationary point is a maximum or a minimum. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 3(a) Use correct product rule M1 Obtain correct derivative in any form A1 1 2 1 2 d e 2 e d x x y x x − − = − Equate derivative to zero and solve for x M1 Obtain x = 1 2 and y = 1 2 A1 4 3(b) Use a correct method for determining the nature of a stationary point M1 e.g. ( ) 2 1 2 1 2 2 d 2e 2 1 2 e d x x y x x − − = − − − Show that it is a maximum point A1 2

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5 (a) Show that the equation cot cot 2 21 + 1 = can be expressed as a quadratic equation in tan [3] 1. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Hence solve the equation cot cot 2, for 0 giving your answers correct to 3 decimal places. 21 + 1 = < 1 < π, [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 5(a) M1 Obtain a correct equation in tanθ in any form A1 e.g. 2 1 tan 1 2 2tan tan θ θ θ − + = Reduce to 2 tan 4tan 3 0 θ θ + −= , or 3-term equivalent A1 3 5(b) Solve a 3-term quadratic for tan θ and calculate θ M1 ( ) tan 2 7 θ = −± Obtain answer, e.g. 0.573 A1 Must be 3 d.p. Obtain second answer, e.g. 1.783 and no other A1 Ignore answers outside the given interval. Treat answers in degrees as a misread. ( ) 32.9 ,102.1 ° ° 3

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7 (a) Given that y ln ln x , show that = dy 1 [1] dx x ln x. = ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The variables x and t satisfy the differential equation dx x ln x t 0. dt + = It is given that x e when t 2. = = (b) Solve the differential equation obtaining an expression for x in terms of t, simplifying your answer. 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(c) Hence state what happens to the value of x as t tends to infinity. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 7(a) Show sufficient working to justify the given answer B1 1 7(b) Correct separation of variables B1 e.g. 1 1 ln d d t x x t x − =   Obtain term ( ) ln ln x B1 Obtain term ln − t B1 Evaluate a constant or use x = e and t = 2 as limits in an expression involving ( ) ln ln x M1 Obtain correct solution in any form, e.g. ln(ln ) ln ln 2 = − + x t A1 Use log laws to enable removal of logarithms M1 Obtain answer 2 e = t x , or simplified equivalent A1 7 7(c) State that x tends to 1 coming from e = k t x B1 1

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9 Two lines l and m have equations r 3i 2j 5k s 4i 3k and r i t 2j 2k respectively. = + + + −j + = −j −2k + −i + + (a) Show that l and m are perpendicular. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ (b) Show that l and m intersect and state the position vector of the point of intersection. 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(c) Show that the length of the perpendicular from the origin to the line m is 1 5. 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Mark scheme: 9(a) Use correct method to evaluate the scalar product of relevant vectors M1 ( ) 4 2 6 −− + Obtain answer zero and deduce the given statement A1 Need a conclusion or a statement in advance that the scalar product will be zero. 2 9(b) Express general point of l or m in component form, e.g. (3 + 4s, 2 – s, 5 + 3s) or (1 – t, – 1 +2t, – 2 + 2t) B1 Equate at least two pairs of components and solve for s or for t M1 Obtain correct answer s = – 1 and t = 2 A1 Verify that all three equations are satisfied A1 State position vector of the intersection – i + 3j +2k, or equivalent A1 Can come from 1 correct value and no contradictory statement. 5 Question Answer Marks Guidance 9(c) Taking a general point P on m, form an equation in t by either equating a relevant scalar product to zero, or equating the derivative of  OP to zero, or taking a specific point Q on m, e.g. (1, – 1, – 2), using Pythagoras in triangle OPQ *M1 e.g. 1 1 1 2 . 2 0 2 2 2 − − −+ = −+                t t t Obtain t = 7 9 A1 Carry out correct method to find OP DM1 Obtain 5 3 A1 Obtain the given answer from full and correct working. Alternative method for question 9(c) Take a specific point Q on m, e.g. (– 1, 3, 2) and use a scalar product to find QN, the projection of OQ on m *M1 Obtain QN = 11 3 , or equivalent A1 Use Pythagoras to obtain ON DM1 Obtain the given answer correctly A1 4

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