Cambridge A Level Mathematics 9709 — 2024 May/June Paper 2 · Variant 2

1 Solve the inequality 5x + 7 2 2x - 3 . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 Solve 5 7 2 3 x x to obtain 10 3  B1 Or inequality. Attempt solution of linear equation where 5x and 2x have different signs M1 Or inequality. Obtain 4 7  A1 State 10 4 3 7 , x x   A1 A0 if ‘… and …’ used. Alternative Method for Question 1 State or imply non-modulus equation 2 2 (5 7) (2 3)    x x (B1) Or inequality. Attempt solution of three-term quadratic equation (M1) Or inequality. Obtain 10 3  and 4 7  (A1) State 10 4 3 7 , x x   (A1) A0 if ‘… and …’ used. 4

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2 Use logarithms to solve the equation 6 2 x - 1 = 5e 3 x + 2 . Give your answer correct to 4 significant figures. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 1 ln6  x ln5 3 2   x *M1 Attempt solution of linear equation DM1 Obtain 9.256 A1 Or greater accuracy. 4

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3 y A B x O The diagram shows the curve with equation y = 8e -x - e 2 x . The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes. (a) Find the gradient of the curve at A. 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(b) Show that the x-coordinate of B is ln2 and hence find the area of the shaded region. 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Mark scheme: 3(a) Differentiate to obtain form 2 1 2 e e  x x k k M1 Where 1 2 0 k k , 1 8 k and 2 1 k . Obtain 2 8e 2e    x x A1 Substitute 0  x to obtain –10 A1 3 3(b) Attempt to find x-coordinate of B M1 2 8e e 0   x x . Obtain 3e 8  x and hence ln2  x A1 AG so necessary detail needed. A0 if decimals used. Integrate to obtain 2 1 2 8e e    x x B1 Use limits 0 and ln2 correctly to find area M1 For integral of form 2 3 4 e e  x x k k where 3 4 0  k k . 1 8  k and 2 1  k . Obtain 5 2 A1 OE 5

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4 A curve is defined by the parametric equations x = 4 cos 2 t , y = 3 sin 2t , for values of t such that 0 1 t 1 12 r. r . Give your answer in the Find the equation of the normal to the curve at the point for which t = 16 form ax + by + c = 0 where a, b and c are integers. 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Mark scheme: 4 Obtain forms 1 d cos sin d  x k t t t or 2 d cos2 d  y k t t *M1 Obtain correct 8cos sin  t t or 4sin2  t and 2 3cos2t A1 Attempt value of d d y x when 1 6 π  t *DM1 Need to see attempt at substitution. Obtain d 1 d 2 y x  A1 State or imply gradient of normal is 2 **M1FT Following their value of the first derivative. Attempt equation of normal **DM1 Not tangent and with attempt to find coordinates 3 3, 2       . Obtain 4 2 9 0    x y A1 Or equivalent of requested form. 7

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5 The polynomial p ( )x is defined by p ( )x = 9 x 3 + 18x 2 + 5x + 4 . (a) Find the quotient when p ( )x is divided by ( 3x + 2) , and show that the remainder is 6. 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............................................................................................................................................................ ............................................................................................................................................................ 2 p ( x) (b) Find the value of dx , giving your answer in the form a + ln b where a and b are integers. + 2 y0 3x [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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Mark scheme: 5(a) Carry out division at least as far as 2 1 3  x k x M1 Or equivalent (inspection, …). Obtain quotient 2 3 4 1   x x A1 Confirm remainder is 6 A1 Answer given – necessary detail needed. SC B1 for use of the factor theorem to show remainder is 6 if no other marks are awarded. Alternative Method for Question 5(a) Synthetic division –2/3 9 18 5 4 –6 8 –2 9 12 –3 6 (M1) Obtain quotient 2 3 4 1   x x (A1) Confirm remainder is 6 (A1) 3 Question Answer Marks Guidance 5(b) Identify integrand as 2 6 3 4 1 3 2    x x x B1FT Following their quotient. Integrate to obtain at least 3 x and 2 ln(3 2)  k x terms *M1 Obtain 3 2 2 2ln(3 2)     x x x x A1 Apply limits and appropriate logarithm properties DM1 Obtain 14 ln16  A1 5

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6 y M O x ln ( 2x + 1) The diagram shows the curve with equation y = . The curve has a maximum point M. x + 3 dy (a) Find an expression for . 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(c) Show by calculation that the x-coordinate of M lies between 2.5 and 3.0 . 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(d) Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. 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Mark scheme: 6(a) Attempt use of quotient rule M1 Or equivalent method. Obtain 2 2 1 2 ( 3) ln(2 1) ( 3)      x x x x A1 OE 2 6(b) Equate first derivative to zero and arrange as far as 2 1 ...  x M1 Confirm 3 0.5 ln(2 1)     x x x A1 Answer given – necessary detail needed. 2 Question Answer Marks Guidance 6(c) Consider sign of 3 0.5 ln(2 1)     x x x or equivalent for 2.5 and 3.0 M1 Obtain   0.07 0.0696...  and   0.4 0.4166... and justify conclusion A1 Answer given – necessary detail needed. 2 Alternative Method for Question 6(c) Consider the values of    3 f 0.5 ln 2 1     x x x and obtain   f 2.5 2.57 (2.5696…) and  f 3 2.58 (2.58339…) (M1) Conclude   f 2.5 3  and  f 3 2.5  so root lies in given interval (A1) Answer given – necessary detail needed. 2 6(d) Use iterative process correctly at least once M1 Obtain final answer 2.569 A1 Answer required to exactly 4sf. Show sufficient iterations to 6 sf to justify answer or show sign change in interval [2.5685, 2.5695] A1 3

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7 (a) Prove that 2 sin i cosec 2 i / sec i. 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(b) Solve the equation tan 2i + 7 sin i cosec 2 i = 8 for - r 1 i 1 r . 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(c) Find 8 sin 2 12 x cosec 2 x d x . 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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) 1 cosec2 sin 2    M1 Obtain 1 cos and confirm sec A1 Answer given – necessary detail needed. 2 7(b) Attempt to obtain quadratic equation in sec or cos only *M1 Obtain 2 7 2 sec 1 sec 8     involving one trigonometric ratio A1 Or equivalent, may be unsimplified, but reduce to 2 2sec 7sec 18 0      2 18cos 7cos 2 0      . Attempt to solve 3-term quadratic equation for sec, using a correct method, to find at least one value of  DM1 Or equivalent using cos. Obtain any two of the four correct solutions 0.952, 1.76   A1 Or greater accuracy. Obtain remaining two correct solutions A1 Or greater accuracy; and no others between π  and π . 5 7(c) Identify integrand as 2 1 2 2sec x B1 Integrate 2 1 2 sec k x to obtain 1 2 2 tan k x M1 Obtain correct 1 2 4tan x A1 Condone omission of ...c . 3

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