Cambridge A Level Mathematics 9709 — 2024 Feb/March Paper 2 · Variant 2

1 Use logarithms to solve the equation 3 4 x + 3 = 5 2 x + 7 . Give your answer correct to 3 significant figures. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 Apply logarithms to both sides and apply power law at least once *M1 Obtain (4 x + 3)ln3 = (2 x + 7)ln5 A1 Or equivalent with x not in a power. Attempt solution of linear equation DM1 Obtain 6.78 A1 Or greater accuracy. 4

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2 (a) Sketch the graph of y = 3 x - 7 , stating the coordinates of the points where the graph meets the axes. [2] (b) Hence find the set of values of the constant k for which the equation 3 x - 7 = k ( x - 4) has exactly two real roots. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 State 3(7 , 0) and (0, 7) B1 Allow if only 73 and 7 shown on relevant axes. 2 2(b) State or imply that gradient of left-hand part of graph is –3 B1 State −3 k  0 B1 Using < and not ⩽. 2

More questions on Coordinate geometry

3 The polynomial p( )x is defined by p( x) = 6 x 3 + ax 2 + 3 x - 10 , where a is a constant. It is given that ( 2 x - 1) is a factor of p( )x . (a) Find the value of a and hence factorise p( )x completely. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Solve the equation p( cosec i) = 0 for - 90° 1 i 1 90° . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Substitute x = 1 , equate to zero and attempt solution M1 2 Obtain a = 31 A1 Divide by 2 x − 1 at least as far as 3x 2 + mx M1 Or equivalent (for example by inspection, …). Obtain 3 x 2 + 17 x + 10 A1 Obtain (2 x − 1)(3 x + 2)( x + 5) A1 5 3(b) Attempt solution of sin= k where k is valid constant from answer to part (a) M1 Obtain −11.5 A1 Or greater accuracy. 2

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4 y O 6 x The diagram shows the curve with equation y = 1 + e 0 .5 x . The shaded region is bounded by the curve and the straight lines x = 0, x = 6 and y = 0 . (a) Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures. 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(b) The shaded region is rotated completely about the x-axis. Find the exact volume of the solid produced. 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Mark scheme: 4(a) 2 3 B1 Or decimal equivalents Use y-values 2, 1 + e, 1 + e , 1 + e Use correct formula, or equivalent, with h = 2 M1 Obtain 15.7 A1 Or greater accuracy. 3 4(b) 0.5 x B1 Implied if π appears only at the end. State or imply that volume is π  (1 + e ) dx Integrate to obtain form k1 x + k 2e 0.5 x M1 Where k1k 2  0 with or without π . Obtain π( x + 2e 0.5 x ) or x + 2e 0.5 x A1 Obtain 4π + 2πe 3 A1 Or exact equivalent. 4

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5 y P O x x 3 The diagram shows part of the curve with equation y = . At the point P, the gradient of the curve x + 2 is 6. (a) Show that the x-coordinate of P satisfies the equation x = 3 12 x + 12 . 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(b) Show by calculation that the x-coordinate of P lies between 3.8 and 4.0 . 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(c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 3 significant figures. Show the result of each iteration to 5 significant figures. 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Mark scheme: 5(a) Differentiate using quotient rule *M1 OE 2 3 A1 OE ( x + 2)3 x − x Obtain ( x + 2) 2 Equate first derivative to 6 and simplify at least as far as x 3 = ... DM1 Confirm x = 3 12 x + 12 A1 Answer given – necessary detail needed. 4 5(b) Consider sign of x − 3 12 x + 12 , or equivalent, for 3.8 and 4.0 M1 Obtain −0.06... and 0.08..., or equivalents, and justify conclusion A1 Answer given – necessary detail needed. 2 5(c) Use iterative process correctly at least once M1 Obtain final answer 3.88 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show sign change in the A1 interval [3.875, 3.885] 3

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6 y A B x O M The diagram shows the curve with parametric equations x = 1 + t , y = ( ln t + 2)( ln t - 3), for 0 1 t 1 25 . The curve crosses the x-axis at the points A and B and has a minimum point M. dy 4 ln t - 2 (a) Show that = . 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(b) Find the exact gradient of the curve at B. 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(c) Find the exact coordinates of M. 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Mark scheme: 6(a) d y M1 Or chain rule if expression expanded before Use product rule to find differentiation. d t dy 2ln t 1 A1 Obtain = − dt t t d y d x M1 Divide attempt at by attempt at d t d t dy 4ln t − 2 A1 Answer given – necessary detail needed. Confirm = dx t 4 6(b) State or imply t = e 3 at point B B1 − 3 B1 Obtain gradient 10e 2 or exact equivalent 2 6(c) Equate first derivative to zero and obtain value for lnt or t M1 1 A1 Or exact equivalent. Obtain x-coordinate 1 + e 4 Obtain y-coordinate − 25 A1 Or equivalent. 4 3

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7 (a) Prove that sin 2i ( a cot i + b tan i) / a + b + ( a - b) cos 2i, where a and b are constants. 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[3] ry1 r 12 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ a = 11 for - r 1 a 1 r . [3](c) Solve the equation sin 23 a 2 cot 13 a + 7 tan 13 ` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 7(a) Use correct identity for sin2 or for cot B1 Express left-hand side in terms of sin and cos M1 Attempt to express k1 cos 2 + k 2 sin 2  in terms of cos2 only M1 Confirm a + b + ( a − b)cos2 A1 Answer given – necessary detail needed. 4 7(b) Obtain integrand 8 + 2cos2 B1 Integrate to obtain 8+ sin2 B1 Apply limits correctly to obtain 2 π + 1 3 − 1 B1 Or exact equivalent. 3 2 2 3 7(c) Use identity to obtain 5cos 2 = −2 B1 3 Obtain = 2.97 B1 Or greater accuracy. Obtain = −2.97 B1 Or greater accuracy; and no others between − π and π . 3

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