Cambridge A Level Mathematics 9709 — 2024 Oct/Nov Paper 2 · Variant 3

1 The variables x and y satisfy the equation a 2y = e 3x + k , where a and k are constants. The graph of y against x is a straight line. 3 (a) Use logarithms to show that the gradient of the straight line is . [1] 2 lna ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that the straight line passes through the points ( 0 .4 , 0 .95) and ( 3 .3 , 3 .80) , find the values of a and k. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 3 B1 AG – necessary detail needed. State or imply 2 y ln a = 3 x + k and conclude that gradient is 2ln a 1 1(b) 3 M1 Equate to gradient of line 2ln a 3 2.85 1929 A1 Allow greater accuracy. Obtain = or equivalent and hence obtain a = 4.6 or a = e 2ln a 2.9 Substitute appropriate values to find value of k M1 Obtain k = 1.7 A1 Alternative Method for Question 1(b) Obtain 0.95 ( 2ln a ) = 3 ( 0.4 ) + k M1 OE or a1.9 = e1.2 + k Obtain 3.80 ( 2ln a ) = 3 ( 3.3 ) + k M1 OE or a 7.6 = e 9.9 + k 29 A1 Allow greater accuracy. 19 Obtain a = 4.6 or a = e Obtain k = 1.7 A1 4

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2 Solve the inequality x - 7 2 4 x + 3 . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 Attempt solution of equation or inequality, where signs of x and 4x are different M1 Obtain 54 … A1 OE … and finally no other value A1 Conclude x  54 A1  4  Allow  − ,  .  5  Alternative Method for Question 2 State or imply non-modulus equation ( x − 7) 2 = (4 x + 3) 2 or inequality B1 Attempt solution of three-term quadratic equation or inequality M1 Obtain finally 54 only A1 Conclude x  54 A1  4  Allow  − ,   5  4

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3 The function f is defined by f ( x) = tan 2 a 1 x k for 0 G x 1 r. 2 (a) Find the exact value of f la 2 r k. 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Mark scheme: 3(a) Differentiate to obtain form k tan 12 x sec 2 12 x M1 OE. May use identities before differentiation. Obtain correct tan 12 x sec 2 12 x A1 OE. Allow unsimplified. Substitute 23 π to obtain 4 3 A1 3 3(b) Express integrand as sec 2 12 x −+1 sin x B1 Integrate to obtain k1 tan 12 x − x + k 2 cos x M1 Where k1k 2  0. Obtain correct 2tan 12 x − x − cos x A1 Apply limits correctly to obtain 3 − 12 π or exact equivalent A1 4

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4 The polynomial p ( x) is defined by p ( x) = ax 3 - ax 2 - 15 x + 18 , where a is a constant. It is given that ( x + 2) is a factor of p ( x) . (a) Find the value of a. 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(b) Hence factorise p ( x) completely. 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(c) Solve the equation p ( cosec 2i) = 0 for - 90° 1 i 1 90° . 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Mark scheme: 4(a) Substitute x = −2, equate to zero and attempt solution M1 Obtain a = 4 A1 2 4(b) Divide by x + 2 at least as far as k1 x 2 + k 2 x M1 Obtain 4 x 2 − 12 x + 9 A1 Obtain ( x + 2)(2 x − 3) 2 or equivalent with integer coefficients only A1 3 4(c) Equate sin 2  to appropriate value from factorised form and attempt solution M1 2 Using their . 3 Obtain 54.7 A1 Or greater accuracy. Obtain –54.7 A1 Or greater accuracy. No others in −90 90. 3

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a 10 5 It is given that d x = 7 , where a is a constant greater than 1. y 2x + 1 a (a) Show that a = 3 0.5e 1 .4 ( 2 a + 1 ) - 0. 5 . 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(b) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures. 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Mark scheme: 5(a) Obtain integral of form k ln(2 x + 1) *M1 Obtain correct 5ln(2 x + 1) A1 Apply limits correctly and equate to 7 DM1 Apply appropriate logarithm property to reach at least a 3 = ... DM1 3 1.4 A1 AG – necessary detail needed. Confirm a = 0.5e (2a + 1) − 0.5 5 5(b) Use iterative process correctly at least once M1 Obtain final answer 2.18 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign change in the A1 interval [2.175, 2.185] 3

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6 A curve has parametric equations e 2 t - 2 3 t x = , y = e + 1. e 2 t + 1 d y (a) Find an expression for in terms of t. 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(b) Find the exact gradient of the curve at the point where the curve crosses the y-axis. 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Mark scheme: 6(a) Differentiate x using quotient rule or correct equivalent *M1 (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t A1 Obtain or equivalent (e 2 t + 1) 2 d y DM1 Attempt expression for in terms of t d x Obtain 1 2 et (e 2 t + 1) 2 or (unsimplified) equivalent A1 No fractions within fractions. Attempt to simplify (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t must be seen. 4 6(b) Identify t = 12 ln 2 at point where curve crosses y-axis B1 d y M1 Substitute non-zero value of t in their expression for and attempt simplification d x Obtain 9 2 2 or exact equivalent A1 3

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7 (a) Prove that cos ( i + 30°) cos ( i + 60°) / 1 3 - 1 sin 2 i . 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(b) Solve the equation 5 cos ( 2 a + 30°) cos ( 2 a + 60 °) = 1 for 0° 1 a 1 90° . 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(c) Show that the exact value of cos 20° cos 50° + cos 40° cos 70° is 1 3 . 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Mark scheme: 7(a) State (coscos30 − sinsin30)(coscos60 − sinsin60) B1 Expand and use correct exact values M1 Obtain 1 4 3(cos 2 + sin 2 ) − sincos or similarly simplified equivalent A1 Conclude 1 4 3 − 12 sin2 A1 AG – necessary detail needed. 4 7(b) Use identity to obtain value for sin4 *M1 Obtain sin4= 12 3 − 52 or 0.466… A1 Show correct process to obtain one value of  DM1 Obtain 6.9 and 38.1 A1 Or greater accuracy; and no others between 0 and 90. 4 7(c) Substitute = −10 to obtain cos20cos50 = 14 3 − 12 sin(−20) B1 3 1 B1 Substitute = 10 to obtain cos40cos70 = − sin20 4 2 Add and confirm 1 2 3 with clear indication that sin( −20) = − sin20 B1 AG – necessary detail needed. Alternative solution for Question 7(c) Rewrite as sin70cos50 + sin50cos70 or cos20sin40 + cos40sin20 or B1 sin70sin40 + cos70cos40 Obtain sin120 or sin60 or cos30 B1 Confirm 1 2 3 B1 AG – necessary detail needed. 3

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