Cambridge A Level Mathematics 9709 — 2024 May/June Paper 3 · Variant 2

1 (a) Sketch the graph of y = x - 2a , where a is a positive constant. [1] (b) Solve the inequality 2x - 3a 1 x - 2a . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) B1 Correct shape, roughly symmetrical. Both sections should be solid straight lines. If not drawn with a ruler the intention must be clear. Allow construction lines if dashed or clearly fainter. 2a marked on each axis (must be 2a, not just 2). Needs to extend into negative x. If a is given a value, then B0. Ignore y = 2x – 3a if seen. 1 1(b) Solve linear equation or inequality to obtain critical value 5 3 x a  or exact equivalent. B1 Ignore x a  if seen. Obtain 5 3 x a  or exact equivalent B1 Accept 10 6 x a  or   5 3, . a  Must be strict inequality. Need a clear final solution: x a  or x a  must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. Alternative Method for Question 1(b) Solve quadratic equation     2 2 2 3 2 x a x a    to obtain critical value 5 3 x a  or exact equivalent (B1)   2 2 3 8 5 0 x ax a    Ignore x a  if seen. Obtain 5 3 x a  or exact equivalent (B1) Accept 10 6 x a  or   5 3, . a  Must be strict inequality. Need a clear final solution: x a  or x a  must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. 2 2a 2a y x O

More questions on Quadratics

2 Express 2 in partial fractions. [5] 2x - 5x - 12 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 State or imply the form 2 3 4 B C A x x     B1 2 3 4 Dx E F x x    and 2 3 4 P Qx R x x    are also valid. Use a correct method for finding a constant M1 SC: If score B0, they can score M1 A1 for one correct constant. B0 M1 A0 available if they substitute two values to form simultaneous equations but get an incorrect answer, or they substitute one value and make an arithmetic error. Obtain one of 3, 2 and 4 A B C    A1 SC: If the horizontal equation is correct apart from an incorrect value for A, the other A marks may be available. Obtain a second value A1 SC: If denominator factorised as    3 2 4 x x   can score a maximum of B0 M1 A1 A1 A0 for a split involving 3 terms. Obtain a third value A1 ISW Statement of the final split is not required. Question Answer Marks Guidance 2 Alternative method for Question 2 Divide numerator by denominator (M1) Obtain 2 3 2 5 12 Px Q x x           (A1) 6 20 3 (2 3)( 4) x x x           State or imply the form 2 2 5 12 2 3 4 Px Q x x D E x x        (B1) Must deal with the 3 separately or include it correctly on both sides in their split. Obtain one of 2 and 4 D E   (A1) SC: If denominator factorised as    3 2 4 , x x   then can score a maximum of B0 M1 A1 A1 A0 for a split involving three terms. Obtain a second value (A1) ISW Statement of the final split is not required. 5

More questions on Algebra

3 The variables x and y satisfy the equation a 2 y - 1 = b x - y , where a and b are constants. (a) Show that the graph of y against x is a straight line. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that a = b3 , state the equation of the straight line in the form y = px + q , where p and q are rational numbers in their simplest form. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) Use logarithms to obtain a correct expression without powers e.g.     2 1 ln ln y a x y b    2 1 log . a y x y b   Do not condone missing brackets unless recovered later. Separate terms and factorise to obtain   2ln ln ln ln y a b x b a    B1 Or equivalent, e.g. 2 2 ln ln ln ln b a y x a b a b   or   2 log log 1. a a y b x b    Clear explanation of linear form. From correct work only. B1 E.g. equation matches the linear form y mx c   or . py qx r   Condone if they compare with , y mx c   but do not actually state that it must therefore be a straight line. Stating “this is a linear equation” without comparing to a relevant standard form scores B0. B0 if they have m = … and c = … correct but never actually mention y = mx + c. 3 3(b) Use 3 a b  and log laws to simplify their equation M1 3 7 7 ln ln ln ln b b y x b b         Denominator reduced to a single log term. Obtain 1 3 7 7 y x   A1 Accept 3 7 7 x y   but not 3. 7 x y   Alternative method for Question 3(b) Use 3 a b  to obtain   3 2 1 y x y b b    or equivalent (M1) Or 1 3 log . a b  Obtain 1 3 7 7 y x   (A1) Accept 3 7 7 x y   but not 3. 7 x y   2

More questions on Logarithmic and exponential functions

4 The equation of a curve is y e 2 x + y 2 e x = 6 . Find the gradient of the curve at the point where y = 1. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 Substitute 1 y  and obtain ex a  , where 0 M1 Must come from a quadratic in e .   2e e 6 0 x x    Ignore any negative solution. Obtain e 2 x  only A1 Or equivalent e.g. ln2. x  Condone x = 0.693….. State or imply   2 2 2 d d e 2 e e d d x x x y y y x x   B1 Accept y for d . d y x State or imply   2 2 d d e e 2 e d d x x x y y y y x x   B1 Accept y for d . d y x Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for x and y and obtain d ... d y x  M1 Independent. d d 2 4 4 2 4 0 d d y y x x            NB: Could also rearrange to obtain d d y x then substitute. Both steps are needed for M1. Obtain 5 4  or 1.25  A1 Correct answer from correct working only Accept 10 8 .  -1.2499 is A0. Question Answer Marks Guidance 4 Alternative method for Question 4: Dividing through by x e Substitute 1 y  and obtain ex a  , where 0 a  (M1) Must come from a quadratic in e . x   2e e 6 0 x x    Ignore any negative solution. Obtain e 2 x  only (A1) Or equivalent e.g. ln2. x  Condone x = 0.693….. State or imply   2 d d 2 d d y y y x x  (B1) Accept y for d . d y x State or imply   d d e e d d x x x y ye y x x   (B1) Accept y for d . d y x Differentiate RHS of given equation to obtain 6e x   , substitute for x and y and obtain d ..... d y x  (M1) Independent. d d 6 1 2 2 2 d d 2 y y x x           NB: Could also rearrange to obtain d d y x then substitute. Both steps are needed for M1. Obtain 5 4  or 1.25  (A1) Correct answer from correct working only Accept 10 8 .  -1.2499 is A0. 6

More questions on Differentiation

5 (a) It is given that the equation e 2 x = 5 + cos 3x has only one root. Show by calculation that this root lies in the interval 0.7 1 x 1 0.8 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that if a sequence of values in the interval 0.7 1 x 1 0.8 given by the iterative formula 1 x n + 1 = ln 5 + cos 3x n 2 ` j converges then it converges to the root of the equation in part (a). [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) Calculate the value of a relevant expression or values of a pair of expressions at 0.7 x  and 0.8 x  Need all relevant values but condone one error. Pairings must be clear for solutions involving four values (do not accept embedded values). M0 if working in degrees e.g -1.94…, 1.04… Complete the argument correctly with correct calculated values. (can be using the equation in the rubric or the equation in (b) or equivalent) A1 E.g. 4.95... 4.26...  and 4.06... 4.49...  -0.439…< 0, 0.690…> 0 1.4 < 1.5029…, 1.6 > 1.449… 1.1 > 1, 0.86 < 1 0.0515 > 0, -0.075 < 0. Allow values rounded or truncated to 2sf. 2 5(b) State   2 ln 5 cos3 x x   and take exponential of both sides to obtain 2e 5 cos3 x x   B1 Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts. 1 5(c) Use the iterative process correctly at least once M1 M0 if working in degrees (e.g. values heading for 0.89….). Obtain final answer 0.740 A1 Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or show that there is a sign change in the interval   0.7395, 0.7405 A1 E.g. 0.7,0.75150,0.73719,0.74105,0.74000,0.74028 0.75,0.73759,0.74094,0.74003,0.74028 0.8,0.72494,0.74443,0.73909,0.74053,0.74014,0.74025 Allow recovery. Allow truncation or rounding and condone small differences in the final decimal place. 3

More questions on Numerical solution of equations

6 y M O x The diagram shows the curve y = xe -ax , where a is a positive constant, and its maximum point M. (a) Find the exact coordinates of M. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 2 (b) Find the exact value of xe -ax d x . [5] y0a ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Use correct product rule *M1 Or equivalent. Condone incorrect chain rule. M0 if a value is used for a (not equivalent work). Obtain correct derivative A1 E.g. d e e d ax ax y ax x     Equate derivative to zero and solve for x DM1 Obtain 1 1 e , a a x y   A1 ISW Or exact equivalent. 4 6(b) Use integration by parts to obtain e e d ax ax px q x    *M1 Condone sign error in parts formula and omission of dx. M0 if a value is used for a (not equivalent work). Obtain 1 1 e e d ax ax a a x x     A1 OE Complete integration to obtain 2 1 1 e e ax ax a a x     A1 OE Correct use of limits 0 and 2 a in an expression of the form e e ax ax rx s    DM1   2 2 2 2 2 2 1 1 e e 0 a a a       Obtain   2 2 1 1 3e a   A1 ISW Or simplified 2-term equivalent, e.g. 2 2 2 e 3. e a  5

More questions on Integration

7 (a) Show that cos 4 i - sin 4 i / cos2i . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 8 r 4 4 2 2(b) Hence find the exact value of cos i - sin i + 4 sin i cos i d i . [6] 1 y- r ` j 8 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) Factorise LHS using difference of 2 squares *M1 2 2 2 2 cos sin cos sin       Simplify DM1 2 2 cos sin 1     must be seen or implied, e.g.     2 2 2 2 2 2 cos sin cos sin cos sin .           Obtain 4 4 cos sin cos2      from correct working A1 AG Alternative Method for Question 7(a) Use of correct rearrangements of double angle formulae (*M1) E.g. 2 2 1 cos2 1 cos2 2 2                  Only condone 2 2 1 cos2 cos2 1 2 2                  if correct expression for 2 sin  seen. Expand and simplify (DM1) Collect like terms. Condone recovery from missing brackets. Obtain 4 4 cos sin cos2      from correct working (A1) AG Alternative Method 2 for Question 7(a) Correct use of Pythagoras (*M1) E.g.   2 2 4 1 sin sin     or     2 2 2 2 cos 1 sin sin 1 cos        Expand and simplify (DM1) Condone recovery from missing brackets. Obtain 4 4 cos sin cos2      from correct working (A1) AG 3 Question Answer Marks Guidance 7(b) Use part (a) and correct double angle formula to obtain expression involving 2 sin 2 d   or 2 cos 2 d   M1 4 4 2 2 2 cos sin 4sin cos d cos2 sin 2 d             Allow BOD for 2 2sin 2if sin2 2sin cos     seen. 1 2 cos2 sin 2 d     B1 Seen or implied. Use of correct double angle formula on second part of the integral to obtain a form that can be integrated directly M1 e.g. 2 1 cos4 sin 2 d d 2       Obtain 1 1 2 8 sin4    A1 Condone a mixture of x and .  Correct use of limits π 8  in an expression of the form sin 2 sin 4 p q r      and evaluate the trig M1     1 1 1 2 16 8 2 2     Obtain 1 1 1 2 8 4 2    A1 ISW Or exact equivalent from exact working. 6

More questions on Integration

8 The points A, B and C have position vectors OA =- 2i + j + 4k , OB = 5i + 2j and OC = 8i + 5j - 3k , where O is the origin. The line l1 passes through B and C. (a) Find a vector equation for l1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ n ( 3i + j - 2k ) . The line l2 has equation r =- 2 i + j + 4k + (b) Find the coordinates of the point of intersection of l1and l2. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) The point D on l2 is such that AB = BD . Find the position vector of D. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 8(a) Correct direction vector seen or implied ( 3 3 3 BC    i j k ) B1 Condone 3 3 3 . BC    i j k Use a correct method to form a vector equation M1 Allow for the RHS with no LHS. Obtain   5 2      r i j i j k A1 ISW Must have r = … or ... x y z       , not l1 = … Or, equivalent vector form, e.g.   8 5 3       r i j k i j k or   5 2 3 3 3 .       r i j i j k Condone a column vector with i, j, k. 3 8(b) Use components to form two relevant equations in 2 unknowns For their l1 B0 if they use the same unknown for both lines. B1FT Two components of 5 2                 = 2 3 1 4 2                 seen or implied. Solve 2 relevant equations in 2 unknowns for  or  M1 For their l1. Obtain 2  or 3  A1 Or equivalent e.g. using BC  as direction vector gives 2 3 .  Obtain   7, 4, 2  No need to check the third equation – the question implies that the lines intersect. A1 Accept position vector. Condone a column vector with i, j, k. SC: B1 M1 A1 A1 if one component of their line is incorrect but they do not use that component. 4 Question Answer Marks Guidance 8(c) State   2 2 2 7 1 4 66 AB     B1 Or   2 66 AB  Condone a sign error in . AB  State BD  in component form B1 7 3 1 4 2 r r r              or equivalent.       2 2 2 3 7 1 2 4 66 AB BD r r r           2 14 60 0 r r   M1 Or equivalent equation in one unknown for their AB and their . BD OD    If you never see a correct form and they go direct to 2 2 9 49 1..... r r    then M0. 30 7 r   A1 Correct only. Ignore r = 0 if seen. 76 37 32 7 7 7 OD     i j k A1 Must be a vector. Condone if also have . OD OA    5

More questions on Vectors

9 The complex numbers z and ~ are defined by z = 1 - i and ~ =- 3 + 3 3i . (a) Express z~ in the form a + bi , where a and b are real and in exact surd form. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Express z and ~ in the form r e ii, where r 2 0 and - r 1 i G r . Give the exact values of r and i in each case. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) On an Argand diagram, the points representing ~ and z~ are A and B respectively. Prove that OAB is an isosceles right-angled triangle, where O is the origin. 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[3](d) Using your answers to part (b), prove that tan 12 3 - 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 9(a) State 3 3 3 3 3 3 i z   B1 Or exact equivalent with real and imaginary parts collected. Need brackets around the coefficient of i. Allow for a =, b = stated correctly. 1 Question Answer Marks Guidance 9(b) Obtain 2 z  B1 Obtain π 4 arg z  final answer B1 Obtain 6  B1 Obtain 2π 3 arg final answer B1 4 9(c) Note: The question does not require the diagram. If they use 5π 12 they need to demonstrate where it comes from. Complex number equivalent to AB is 3 3 3i.  Show 6, OA AB   hence isosceles B1 One mark for ‘isosceles’ and one mark for ‘right angle’. There will be alternatives e.g. use of Pythagoras (ratio of lengths is 1:1: 2 ), expressing each number in “vector” form and using scalar product or explaining the effect of multiplying by 1 – i. π 4 arg arg arg AOB z z        hence third angle is a right angle B1 2 B A O Im Re (-3+3 3)+(3+3 3)i -3+3 3i Question Answer Marks Guidance 9(d)   2 5 3 4 12 arg arg arg z z           M1 For showing correct use of their angles from part (b). Must demonstrate where 5π 12 comes from. 1 3 3 3 arg tan 3 3 3 z     M1 Correct method for their z from part (a). Must link to point B on diagram or to arg . z Need to see 1 3 3 3 tan 3 3 3    or 3 3 3 tan 3 3 3     and not just 1 1 3 tan . 1 3      5 12 3 1 tan 3 1      A1 Obtain given answer from full and correct working. 3

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10 (a) By writing y = sec 3i as cos3i, show that di = 3 sin i sec 4 i . 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(b) The variables x and i satisfy the differential equation 2 d i 4 x + 9 sin i = ( x + 3) cos i . ` j d x r . It is given that x = 3 when i = 13 Solve the differential equation to find the value of cosi when x = 0 . Give your answer correct to 3 significant figures. 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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: 10(a) Use of correct chain rule (and correct quotient rule) and 3 cos   4 cos sin k      or equivalent.   4 4 d 3 sin cos 3sin sec d y         Must be expressed in the given form A1 Obtain given answer from full and correct working (signs must be shown), but condone   3 d d sec ....   and '( ). y  2 Question Answer Marks Guidance 10(b) Separate variables:     4 2 3 sin d d cos 9 x x x            B1 Or 4 2 d d . 3sin 3 9 cos 9 x x x             Condone missing integral signs or missing d or d , x  but not both. Obtain   3 sec p A  B1 Correct form, p any constant but not 0. Use 2 2 2 d d 3 9 3 9 9 9 9 x x x x x x x                  and obtain   2 ln 9 q x  or   1 3 tan . x r C   *M1 Might have one third of both sides. Alt: substitute 3tan x   to obtain 1 tan d ; q    condone if have  in place of  in this method. Obtain   2 ln 9 q x  and   1 3 tan x r C   DM1 Obtain     ln cos q    OE. Obtain     3 2 1 3 2 3 sec ln 9 3tan x x C       or equivalent A1 Or might see a third of both sides. Must have 2 different variables. Use 1 3 , 3 x     in an equation including 3 sec , p    2 ln 9 q x  and 1 3 tan x r  to evaluate the constant of integration M1 Or as limits in a definite integral. Limits for  are 0 and 1 4 .  Obtain constant = 3 3 2 4 8 ln18    A1 OE, e.g. 1.308… to at least 3sf. Obtain cos 0.601  A1 Accept AWRT 0.601. 8

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