PhysicsAndMathsTutor.com

PhysicsAndMathsTutor.com

47 Mark Scheme June 00 (i) u =, u =, u = 8 The sequence is an Arithmetic Progression B B B For the correct value of u For both correct values of u and u For a correct statement (any mention of arithmetic) (ii) 00 ( + 99 ) 00 = 6 (i) r θ =, r θ = 6 B B (ii) r = 6 r = 6 B Hence θ = B (iii) Segment area is 6 6 sin 9. 6 cm * dep* 7 (i) (x + 7 x + ) dx 7 = x + x + x + c B 4 4 (i) (ii) [ ] 9 0 (ii) x = 6 cos BCA + 6 9 = 6 = So sin BCA = 0.948... Angles BCA and CAD are equal So sin ADC = sin CAD = 8 = 9 ADC = 8. o (i) f(-) = 0 - a + b = 0 f() = 6 7 + a + b = 6 Hence a = -, b = - 7 B B B 4 4 8 For correct interpretation of Sigma notation ie finding the sum of an AP or GP For use of correct n( a + ( n ) d ), or equiv, with n=00 and a & d not both = For correct value 00 For r θ = stated correctly at any point For r θ = 6 stated correctly at any point For showing given value correctly For correct value (or 0.67π) For use of = ab sin C, or equivalent For attempt at 6 For correct value (rounding to) 9.6 For expanding and integration attempt For at least one term correct For all three terms correct For addition of arbitrary constant, and no or dx kx For integral of the form For evaluating at least F(9), following attempt at integration For final answer of 6 only For relevant use of the correct cosine formula For attempt to rearrange correct formula For obtaining the given value correctly For correct answer for sin BCA in any form OR For substituting cos BCA = - / For attempt at evaluation For full verification For correct answer for sin BCA in any form For stating, using or implying the equal angles For correct use of the sine rule in ADC (sides must be numerical, angles may still be in letters) For a correct equation from their value in (i) For correct answer, from correct working For equating their attempt at f(-) to 0, or equiv For the correct (unsimplified) equation For equating their attempt at f() to 6, or equiv For the correct (unsimplified) equation For both correct values must follow two correct equations (ii) f() = 8 6 = 0 B For the correct verification (from correct a & 8

47 Mark Scheme June 00 6 (i) (ii) Hence f(x) = (x + ) (x ) 6 x x + + x + 8 4 7 4 7 x + 4 x + x x + c 4 8 7 (i) 0 ( ) = log (ii) log = Method A x y = 0 Hence x = log 0 ( y) i.e. x = 0 ( y) log 4 b) For recognition or use of two linear factors, or full division attempt by either (x + ) or (x ) For correct third factor (repeated) of (x + ), and full linear factorisation stated For 4 term binomial attempt or equiv For any one (unsimplified) term correct For any other (unsimplified) term correct For full, simplified, expansion correct + x For any correct use of n n+ For any two terms integrated correctly For any correct use of x n + using a negative index For all terms integrated correctly (must have at least 4 terms, including at least negative index) [No penalty for omission of +c in this part] For any relevant combination of log a ± log b For log must follow correct working only For correct answer For correct division of both sides by c For relevant use of a = b c = log b a For correct equation involving logs to base 0 For correct answer for x Method B x y = 0 x log y = log0 log y = x log0 e. = log y x 0 i. ( ) 4 For correct division of both sides by For taking logs of both sides For correct linear equation involving logs For correct answer for x Method C Method D y = x 0 log y = log x log y = log + log0 log y = log + x log0 i.e. = log y x 0 ( ) 0 x x = alog( b 0 ) x x = alog b + alog0 x = axlog0 a = a = x 4 4 a log b = 0 b = b = 7 8 (i) 00 000 x 0.9 = 7900 (ii) 00 000 x 0.9 x = 000 B Hence xlog 0.9 = log 0.0 So x = 8.4, 8 or 9; or n = 9.4, 9 or 0 i.e. 0 th year / 0 years / year is 00 4 For introducing logs throughout For correct RHS log + log 0 x For correct use of log a b = b log a For correct answer for x For substituting for y, and separating RHS into at least terms For attempting values for a and b For obtaining a = / For obtaining b = / For relevant use of ar or equiv For the correct answer 7900 For a correct equation or inequality For complete solution method by logs or trial For correct solution for their index allow integer values either side For correctly linking their index to date or 9

47 Mark Scheme June 00 (iii) Total is 00000 0.9 0 ( 0.9 ) = 97609 9 (a) (i) cos π = 6 (ii) tan π = 6 π Hence cos = = tan Other roots are π/ and π/6 (b) (i) 0.0(0.00 + (0.07 + 0.09) + 0.48) = 0.0774 (ii) Overestimate; tops of trapezia above the curve or equiv π 9 B B B B B B B 4 4 B number of years a n ( r ) For relevant use of r For correct (unsimplified) statement for their integer n (if no n stated then use their year 000) For answer 98000 or better, including decimal For any correct exact value For any correct exact value For correct verification (allow via decimals) For correct sketch of either y = tan x or y = cosx For second correct sketch, with both graphs in proportion (ie points of intersection) For one of π/ or π/6 (or equiv in degrees) For second correct value, and no others in range 0 x π State at least three of tan 0., tan 0., tan 0., tan 0.4 Substitute numerical values (must be attempt at y-coords, not x-coords) into correct trapezium rule, with h consistent with number of strips Obtain 0.0(tan 0. + (tan 0. + tan 0.)+ tan 0.4) or equiv in decimals (SC award if values are now decimals from using degrees gives final answer of 0.00) Obtain 0.077 or better For correct statement and justification 40

47 Mark Scheme January 006 (i) a+ 9d = 0, a+ 49d = 70 Attempt to find d from simultaneous a+ n d or equiv method Hence 0d = 60 d = Obtain d = equations involving ( ) a + (9 x ) = 0 or a + (49 x ) = 70 Attempt to find a from + ( ) a n d or equiv Hence a = 8 4 Obtain a = 8 (ii) For relevant use of S = ( 8 + ( ) ) = 0 ( ( ) ) n a+ n d For showing the given result correctly AG 6 (i) Δ= = For use of ca sin B or complete equiv. For correct value 4. (ii) b = 0 + 7 0 7 cos80 For attempted use of the correct cosine formula Hence length of CA is. cm For correct value. (iii) 0sin 80 sin C = = 0.889....66... Hence angle C is 6.9 o For correct value 6.9 6 x = 4x+ 64x B (i) ( ) + = (ii) ( 64) ( 4) 9 4 (i) ( ) ( ) (ii) perimeter =.8 + 0.8 + + = 7cm area = 0.8.8 = 7.cm 6 6 For use of the sine rule to find C, or equivalent Obtain and 4x Attempt x term, including attempt at binomial coeff. Obtain 64x Attempt coefficient of x from two pairs of terms Obtain correct unsimplified expression Obtain 9 Use rθ at least once Obtain at least one of 7cm or 6cm Obtain 7 Attempt area of sector using kr θ Find difference between attempts at two sectors Obtain 7. / 8

47 Mark Scheme January 006 (i) 4.8 B* For correct value of r used r = = 0.96 S = = 0.04 a B For correct use of to show r dep* given answer AG (ii) n = S n ( 0.96 ) B For correct, unsimplified, S n 0.96 n n Hence 0.96 > 0.99 0.96 < 0.008 For linking S n to 4 (> or =) and multiplying through by 0.04, or equiv. For showing the given result correctly, with correct inequality throughout AG n log0.96<log 0.008 B For correct log statement seen or implied (ignore sign) log 0.008 For dividing both sides by log 0.96 Hence n > 8. log 0.96 Least value of n is 9 6 For correct (integer) value 9 8 6 (a) 4 x x+ c For kx For correct first term x, or equiv B B 4 For correct second term 4x For +c (b)(i) a a Obtain integral of the form kx - 4x dx= 4x 4 = 4 a Use limits x = a and x = Obtain 4 = 4, or equivalent a (ii) 4 B State 4, or legitimate conclusion from their (b)(i) 8 7 (i)(a) log 0 x log 0 y B For the correct answer (ii) (b) + log 0 x + log 0 y log 0 x log 0 y = + log 0 x + log 0 y Hence log 0 y = Sum of three log terms involving 0, x, y For correct term log 0 x For both correct terms and log 0 y For relevant use of results from (i) For a correct, unsimplified, equation in log 0 y only So y = 0 0. For correct use of a = log 0 c c = 0 4 For the correct value 0. 8 a 9

47 Mark Scheme January 006 8 (i) + k+ + 6= 0 k = For attempting f( ) For equating f(-) to 0 and deducing the correct value of k AG 0BOR Match coefficients and attempt k Show k = - 4BOR B Following division, state remainder is 0, hence (x + ) is a factor, hence k = - EITHER: ( x+ )( x 7x+ 6) B For correct leading term x For attempt at complete division by f(x) by (x + ) or equiv. For completely correct quadratic factor = x+ x x For all three factors correct ( )( )( ) (ii) (iii) OR: f() = 6 0 + 6 = 0 Hence (x ) is a factor Third factor is (x ) = x+ x x Hence f(x) ( )( )( ) 4 f( ) d = + 6 6 B B For further relevant use of the factor theorem For correct identification of factor (x ) For any method for the remaining factor For all three factors correct For any two terms integrated correctly For all four terms integrated correctly x x x x x x 40 8 6 For evaluation of F() F( ) = + + = 9 4 For correct value 9 8 6 y 4 x B B For sketch of positive cubic, with three distinct, non-zero, roots For correct explanation that some of the area is below the axis 9 (i) 4 8 y 6 4 4 / / 4 Ø Ù x B B B For correct sketch of one curve For correct shape and location of second curve, on same diagram For intercept 4 on y-axis 6 (ii) (See diagram above) 8 β = 80 α sin x = 4cos x= 4 sin x (iii) ( ) 0 B For correct identification of intersections in correct order For attempt to use symmetry of the graphs For the correct (explicit) answer for β sin x For use of tan x = cos x For use of cos x = sin x Hence 4sin x+ sin x 4 = 0 For showing the given equation correctly ± 6 sin x = 8 B For correct solution of quadratic o Hence β α = 8.0... 6.97... 6 Attempt value for x from their solutions 6 For the correct value 6

47 Mark Scheme January 006 ( ) 4 4 x = 8x 6x + 6x 96x + 6 4 4 (i) u =, u =, u4 = B B (ii) Sum is ( ( ) ) ( ( ))... ( ( ) i.e. 0 ( + ( ) ) = 0 y = 4x + c + + + + + + ) Hence = 4 4 + c c = So equation of the curve is y = 4x 6 6 4 (i) Intersect where x + x = 0 x =, (ii) Area under curve is 4x x 8 i.e. ( 4 ) ( 8 ) + =9 Area of triangle is 4½ Hence shaded area is 9 4½ = 4½ 6 Attempt binomial expansion, including attempt at coeffs. Obtain one correct, simplified, term Obtain a further two, simplified, terms Obtain a completely correct expansion For correct value for u For correct values for both u and u4 For correct interpretation of Σ notation For pairing, or 0 0 For correct answer 0 For attempt to integrate For integral of the form kx For 4x, with or without +c For relevant use of (4, ) to evaluate c For correct value (or follow through on integral of form kx ) For correct statement of the equation in full (aef) For finding x at both intersections For both values correct For integration attempt with any one term correct For use of limits subtraction and correct order For correct area of 9 Attempt area of triangle (½bh or integration) Obtain area of triangle as 4 ½ Obtain correct final area of 4 ½ OR Area under curve is ( ) = [ ] x x + x = 8 ( + ) ( 4) = 4 x x dx 8 Attempt subtraction either order For integration attempt with any one term correct x x + x Obtain ±[ ] For use of limits subtraction and correct order Obtain ± 4 ½ - consistent with their order of subtraction Obtain 4 ½ only, following correct method only 8

47 Mark Scheme January 006 (i) x = x x+ x = sin cos cos cos 0 ( x )( x ) Hence cos cos + = 0 cos x = x = 60 cos x = x = 80 (ii) tan x = x = or o Hence x = 67. or 7. o o o 4 4 For transforming to a quadratic in cos x For solution of a quadratic in cos x For correct answer 60 o For correct answer 80 o [Max out of 4 if any extra answers present in range, or in radians] SR answer only is B, B justification ie graph or substitution is B, B For transforming to an equation of form tanx = k For correct solution method, i.e. inverse tan followed by division by For correct value 67. For correct value 7. OR sin x = cos x sin x = cos x = sin x = ± cos x = ± o o Hence x = 67. or 7. 8 6 (i) (a) 00 + 9 = 9 (b) 40 ( 00 + 9 ) = 67400 (ii) 9 00r = 00 r =.09... B 40 00(.09 ) Hence total is = 49.09 9 Obtain linear equation in cos x or sin x Use correct solution method For correct value 67. For correct value 7. [Max out of 4 if any extra answers present in range, or in radians] SR answer only is B, B justification ie graph or substitution is B, B For relevant use of a + (n )d For correct value 9 For relevant use of n ( a + l ) or equivalent For correct value 67400 9 For correct statement of 00r = 00 Attempt to find r For correct value.0 For relevant use of GP sum formula For correct value 49 ( s.f. or better) 9

47 Mark Scheme January 006 7 (i) AC = + 8 8 cos 0.8 = 6.796... Hence AC=7.90 cm (ii) Area of sector = 7.90.7 =. 0 Area of triangle = 7.90 sin.7 = 0. 9 Hence shaded area =. cm (iii) (arc) DC = 7.90.7 =. 4 (line) DC = 7.90 + 7.90 7.90 7.90 cos.7 DC =.9 Hence perimeter =.cm 8 (i) ( ) f = 4a+ b = 6 ( ) f = 0 a b= Hence a =, b = 7 (ii) Quotient is Remainder is 8 x + x 9 4 0 6 B Attempt to use the cosine formula Correct unsimplified expression Show the given answer correctly Attempt area of sector using ( ) r θ Attempt area of Δ ACD, using () sinθ Obtain. Use rθ to attempt arc length Obtain.4 r, or equiv Attempt length of line DC using cosine rule or equiv. Obtain. For equating f() to For correct equation 4a + b = 6 For equating f( ) to 0 For correct equation a b = For attempt to find a and b For both values correct For correct lead term of x For complete division attempt or equiv For completely correct quotient For attempt at remainder either division or f( ) For correct remainder 0

47 Mark Scheme January 006 9 (i) (ii) { ( ) (iii) ( ) A 0. + 0. + 0. + 0. + 0..09 x = x log 6 0 = log 0 6 0 6 0 x = = 0 0 Hence log log 6 log log log0 + log0 = log0 log = + 0 log 0 OR () x x = = 6 6 x log0 = log0 6 log0 6 x = log0 log0 + log0 = log0 log0 = + log0 OR () x x = = 6 6 Hence OR x = x = log ( x ) log 0 = 0 log x = + log log log0 log0 = log0 6 log0 x log = log 6 log x 0 0 = 6 x = 0 + x = log 6 0 0 6 0 0 } B B 4 4 Attempt sketch of any exponential graph, in at least first quadrant Correct graph must be in both quadrants For identification of (0, ) State, or imply, at least three correct y-values For correct use of trapezium rule, inc correct h For correct unsimplified expression For the correct value.09, or better x and attempt at logs For equation ( ) = 6 Obtain x log ( ) = log ( ) 6, or equivalent For use of log 6 = log + log For showing the given answer correctly For equation x = 6 and attempt at logs Obtain x log = log 6, or equivalent For use of log 6 = log + log For showing the given answer correctly n Attempt to rearrange equation to = x Obtain = For attempt at logs For showing the given answer correctly Use log + log = log 6 Obtain xlog = log 6 Attempt to remove logarithms x = Show ( ) 6 correctly

47 Mark Scheme Jan 007 + 9d = 7 Attempt to find d, from a + (n )d or a + nd Hence d = Obtain d = S n = 00 / {( ) + ( 99 ) } Use correct formula for sum of n terms = 60 4 Obtain 60 π (i) 46 80 = 0.80 / 0.80 Attempt to convert to radians using π and 80 (or π & 60) 4 Obtain 0.80 / 0.80, or better (ii) 8 x 0.80 = 6.4 cm B State 6.4, or better (iii) ½ x 8 x 0.80 =.6 /.7 cm Attempt area of sector using ½r θ or r θ, with θ in radians Obtain.6 /.7, or better (i) ( 4x ) dx = x x + c Obtain at least one correct term Obtain at least x x (ii) y = x - x + c B State or imply y = their integral from (i) 7 = + c c = 4 Use (,7) to evaluate c So equation is y = x x + 4 Correct final equation o 4 (i) area = 8 sin 60 B State or imply that = 8 Use ac sin B sin 60 = o or exact equiv = 0 6 Obtain 0 6 only, from working in surds o (ii) ( ) AC = + 8 8 cos 60 Attempt to use the correct cosine formula Correct unsimplified expression for AC AC = 7.8 cm Obtain AC = 7.8, or better (a) (i) 4x+ 7 log x B Correct single logarithm, as final answer, from correct working only (ii) log x+ 7 x = 4 x+ 7 x = 9 B State or imply = log 9 4 x + 7 = 9x Attempt to solve equation of form f(x) = 8 or 9 x =.4 Obtain x =.4, or exact equiv 9 (b) log0 dx ( log0 + log0 6 + log0 9) x B State, or imply, the correct y-values only 4.48 Attempt to use correct trapezium rule Obtain correct unsimplified expression 4 Obtain 4.48, or better 6 8

47 Mark Scheme January 007 7 6 (i) ( ) + 4x = + 8x + 6x + 40x B Obtain + 8x Attempt binomial expansion of at least more term, with each term the product of binomial coeff and power of 4x Obtain 6x 4 Obtain 40x (ii) 8a + 008 = 00 Multiply together two relevant pairs of terms Hence a = - ¼ Obtain 8a + 008 = 00 Obtain a = - ¼ 7 (i) (a) B Correct shape of kcosx graph B (90, 0), (70, 0) and (0, ) stated or implied 7 (b) cos x = 0.4 Divide by, and attempt to solve for x x = 66.4 o, 94 o Correct answer of 66.4 o /.6 rads Second correct answer only, in degrees, following their x sin x (ii) tan x = Use of tan x = (or square and use sin x + cos x ) cos x x = 6.4 o, -7 o Correct answer of 6.4 o /.6 rads Second correct answer only, in degrees, following their x 8 (i) -8 6 4 + = - Substitute x = -, or attempt complete division by (x + ) Obtain, as final answer (ii) 7 8 + + = 0 A.G. B Confirm f() = 0, or equiv using division (iii) x = B State x = as a root at any point f(x) = (x )(x 6x ) Attempt complete division by (x ) or equiv Obtain x 6x + k Obtain completely correct quotient 6 ± 6 + 44 x = Attempt use of quadratic formula, or equiv, to find roots = ± or ± 0 6 Obtain ± or ± 0 4 9 (i) u =.. 0 Use.r 4, or find u, u, u 4 =.64 tonnes A.G. Obtain.64 or better N (ii).(.0 ) 8 9 9 Use correct formula for S N.0 Correct unsimplified expressions for S N (.0 N ) ( 9 0.0.) ( ) Link S to 9 and attempt to rearrange N N.0 0. Hence.0 N. 4 Obtain given inequality convincingly, with no sign errors (iii) log.0 N log. Introduce logarithms on both sides and use log a b = b log N log.0 log. Obtain N log.0 log. (ignore linking sign) N.44.. Attempt to solve for N N = trips 4 Obtain N = only 9 0

47 Mark Scheme January 007 0 (i) 0 = B Verification of (9, 0), with at least one step shown 9 a x dx = x 6 x 9 9 (ii) [ ] a Attempt integration increase in power for at least term For second term of form kx ½ For correct integral = ( a 6 a ) ( 9 6 9) Attempt F(a) F(9) = a 6 a + 9 Obtain a 6 a + 9 a 6 a + 9 = 4 Equate expression for area to 4 a 6 a + = 0 Attempt to solve disguised quadratic ( a )( a ) = 0 a =, a = Obtain at least a = a =, a = but a > 9, so a = 9 Obtain a = only 0 0

47 Mark Scheme June 007 (i) u = B State u = u = 9.6, u 4 = 7.68 (or any exact equivs) B Correct u and u 4 from their u (ii) 0 ( 0.8 S = ) Attempt use of 0 0.8 S n ( r ) a n = r, with n = 0 or 9 = 74. Obtain correct unsimplified expression Obtain 74. or better OR List all 0 terms of GP 4 x = x + 4x x + 6x x + 4x x x 4 ( ) () () () ( ) 4 A Obtain 74. x + + * Attempt expansion, using powers of x and / x (or the two terms in their bracket), to get at least 4 terms * Use binomial coefficients of, 4, 6, 4, dep* Obtain two correct, simplified, terms Obtain a further one correct, simplified, term = 4 + + + 6 (or equiv) Obtain a fully correct, simplified, expansion x 8x 4 + x 4 x OR * Attempt expansion using all four brackets * Obtain expansion containing the correct powers only (could be unsimplified powers eg x. x - ) dep* Obtain two correct, simplified, terms Obtain a further one correct, simplified, term Obtain a fully correct, simplified, expansion OR ( x+ ) 00 log = log Introduce logarithms throughout ( x + ) log = 00log Drop power on at least one side 00 log log Obtain correct linear equation (now containing no powers) x + = Attempt solution of linear equation x =46 Obtain x = 46, or better (x + ) = log 00 Intoduce log on right-hand side x + = 00log Drop power of 00 Obtain correct equation Attempt solution of linear equation Obtain x = 46, or better 4 (i) area { ( 7 9 ) } + + + + Attempt y-values for at least 4 of the x =,.,,., only Attempt to use correct trapezium rule Obtain { ( 7 9 ) } + + + +, or decimal equiv 0..766....94 4 Obtain.94 or better (answer only is 0/4) (ii) This is an underestimate *B State underestimate as the tops of the trapezia are below Bdep*B Correct statement or sketch the curve 6 8

47 Mark Scheme June 007 (i) ( sin ) = sinθ + θ Use cos θ = sin θ sin θ = sin θ + sin θ + sin θ = 0 Show given equation correctly sinθ sinθ + = Attempt to solve quadratic equation in sin θ sin θ = or - Both values of sinθ correct θ = 4 o, 8 o, 70 o Correct answer of 70 o (ii) ( )( ) 0 Correct answer of 4 o For correct non-principal value answer, following their first value of θ in the required range (any extra values for θ in required range is max 4/) (radians is max 4/) SR: answer only (or no supporting method) is B for 4 o, B for 8 o, B for 70 o 7 6 (a) (i) 4 x 4 x = 4x x + c Expand and attempt integration Obtain ¼ x 4 x (A0 if or dx still present) B + c (mark can be given in (b) if not gained here) (b) (ii) 4 [ x ] 6 4 x Use limits correctly in integration attempt (ie F(6) F()) =(4 7) ( ¼ ) = ¾ Obtain ¾ (answer only is M0A0) 6 x dx = x + c B Use of = x x Obtain integral of the form kx - Obtain correct -x - (+ c) (A0 if or dx still present, but only penalise once in question) 8 70 { + 70 d} Attempt S 70 Obtain correct unsimplified expression (4 + 69d) = 9 Equate attempt at S 70 to 9, and attempt to find d d = 4 Obtain d = 7 (a) S 70 = ( ) ( ) OR { } 9 70 = + l Attempt to find d by first equating n / (a + l) to 9 l = 7 Obtain l = 7 + 69d = 7 Equate u 70 to l d = Obtain d = (b) ar = -4 B Correct statement for second term a r = 9 B Correct statement for sum to infinity 4 4 r = 9 9r or a = 9 ( 9 ) Attempt to eliminate either a or r a 9 r 9r 4 = 0 a 9a 6 = 0 Obtain correct equation (no algebraic denominators/brackets) ( r 4)( r + ) = 0 ( a + )( a ) = 0 Attempt solution of three term quadratic equation 4 r =, r = a = -, a = Obtain at least r = (from correct working only) Hence r = 7 Obtain r = only (from correct working only) SR: answer only / T&I is B only 9

47 Mark Scheme June 007 8 (i) 0.9 = 6. AB Use ( ) r θ = 6. AB = 6 AB = 6 Confirm AB = 6 cm (or verify ½ x 6 x 0.9 = 6.) (ii) 6 sin 0.9 = AC. 4 * Use Δ = bc sin A, or equiv dep* Equate attempt at area to.4 AC =.8 cm Obtain AC =.8 cm, or better (iii) BC = 6 +.8 6.8 cos 0. 9 Attempt use of correct cosine formula in ΔABC Correct unsimplified equation, from their AC Hence BC =. cm Obtain BC =. cm, or anything that rounds to this BD = 6 0.9 =.4 cm B State BD =.4 cm (seen anywhere in question) Hence perimeter =. +.4 + (.8 6) Attempt perimeter of region BCD = 4. cm 6 Obtain 4. cm, or anything that rounds to this 9 (i) (a) f (-) = + 6 4 = 0 B Confirm f(-) = 0, through any method (b) x = - B State x = - at any point f( x ) = ( x + )( x + x 4) Attempt complete division by (x + ), or equiv Obtain x + x + k Obtain completely correct quotient ± + 6 x = Attempt use of quadratic formula, or equiv, find roots = 4 4 x ( ) 6 Obtain ( ) (ii) (a) ( + ) + log x log ( 4x + ) log x = B State or imply that log (x + ) = log (x + ) Add or subtract two, or more, of their algebraic logs correctly ( x+ ) x ( ) = log 4 ( x+ ) x+ x 4x+ = x + 6x + 9 x = 8x + + 6x + x 4 = 0 ( ) 4 Obtain correct equation (or any equivalent, with single term on each side) B Use log a = a = at any point x Confirm given equation correctly (b) x > 0, otherwise log x is undefined B* State or imply that log x only defined for x > 0 x = ( 4) + B dep* State x = ( 4) + (or x = 0.7) only, following their single positive root in (i)(b) 4 0

47 Mark Scheme January 008 47 Core Mathematics Mark Total area of sector = ½ x x 0.7 Attempt sector area using (½) r θ = 4. Obtain 4., or unsimplified equiv, soi area of triangle = ½ x x sin0.7 = 8.98 Attempt triangle area using ½absinC or equiv, and hence area of segment = 4. 8.98 subtract from attempt at sector =.7 4 Obtain.7, or better area { ( 8) } + + + Attempt y-values at x =,,, 7 only Correct trapezium rule, any h, for their y values to find area between x = and x = 7 Correct h (soi) for their y values 6.7 4 Obtain 6.7 or better (correct working only) 4 (i) log a 6 B State log a 6 cwo 4 (ii) 4 (i) log0 x log y log0 y = log0 x 0 * Use b log a = log a b at least once x = log 0 dep* Use log a - log b = log a / b y x Obtain log 0 cwo y BD 6 = sin 6 sin 0 Attempt to use correct sine rule in BCD, or equiv. BD = 8.4 cm Obtain 8.4 cm 4 (ii) 8.4 = 0 + 0 x 0 x 0 x cos θ Attempt to use correct cosine rule in ABD cos θ = 0.998 Attempt to rearrange equation to find cos BAD (from a = b + c ± ( ) bc cos A ) θ = 66.4 0 Obtain 66.4 0 x d x = 8x Attempt to integrate y = 8 x + c 0 = 8 4 + c Use (4, 0) to find c Obtain correct, unsimplified, integral following their f(x) Obtain8x, with or without + c c = 4 Obtain c = -4, following kx only Hence y = x 4 6 State y = x 4 aef, as long as single power of x 8 8 6

47 Mark Scheme January 008 Mark Total 6 (i) u = 7 B Correct u u = 9, u = B Correct u and u (ii) Arithmetic Progression B Any mention of arithmetic (iii) ½ N (4 + (N ) x ) = 00 B Correct interpretation of sigma notation Attempt sum of AP, and equate to 00 N + 6N 00 = 0 Correct (unsimplified) equation (N 44)(N + 0) = 0 Attempt to solve term quadratic in N hence N = 44 Obtain N = 44 only (N = 44 www is full marks) 7 (i) Some of the area is below the x-axis B Refer to area / curve below x-axis or negative area (ii) Attempt integration with any one term correct Obtain / x / x [ ] 7 x = ( ) ( 0 0) x Use limits (and 0) correct order / subtraction 0 9 = 4 Obtain (-)4½ [ ] 7 7 x = ( ) ( 9 ) x Use limits and correct order / subtraction = 8 Obtain 8 / (allow 8.7 or better) Hence total area is / 6 7 Obtain total area as / 6, or exact equiv 8 SR: if no longer f(x)dx, then B for using [0, ] and [, ] 8 (i) u 4 = 0x0.8 Attempt u 4 using ar n- =. Obtain. aef 0 (ii) 0( 0.8 ) S 0 = Attempt use of correct sum formula for a GP 0.8 = 49.4 Obtain 49.4 (iii) N 0 0( 0.8 ) < 0. 0 Attempt S using a 0.8 ( 0.8) r Obtain S = 0, or unsimplified equiv 0 0( 0.8 N ) < 0.0 Link S S N to 0.0 and attempt to rearrange 0.8 N < 0.000 A.G. Show given inequality convincingly log 0.8 N < log 0.000 Introduce logarithms on both sides N log 0.8 < log 0.000 Use log a b = b log a, and attempt to find N N > 8.69, hence N = 9 7 Obtain N = 9 only 8 8

47 Mark Scheme January 008 9 (i) (90 o, ), (-90 o, -) B State at least correct values B State all 4 correct values (radians is B B0) (ii) (a) 80 - α B State 80 - α (b) -α or α 80 B State - α or α 80 (radians or unsimplified is BB0) Mark (iii) sinx = cos x sinx = ( sin x) Attempt use of cos x = sin x sin x sinx = 0 Obtain sin x sinx = 0 aef with no brackets (sinx +)(sinx ) = 0 Attempt to solve term quadratic in sinx sinx = - /, sinx = Obtain x = -9. o x = -9. o, -6 o, 90 o Obtain second correct answer in range, following their x 6 Obtain 90 o (radians or extra answers is max out of 6) Total SR: answer only (and no extras) is B B B 0 (i) (x + ) 4 =(x) 4 + 4(x) + 6(x) + 4(x) + 4 * Attempt expansion involving powers of x and (at least 4 terms) = 6x 4 + 60x + 600x + 000x + 6 * Attempt coefficients of, 4, 6, 4, dep* Obtain two correct terms 4 Obtain a fully correct expansion (ii) (x + ) 4 (x ) 4 = 0x + 000x Identify relevant terms (and no others) by sign change oe Obtain 0x + 000x cwo (iii) 9 4 (-) 4 = 660 and 760 800 = 660 A.G. B Confirm root, at any point 0x 680x + 800 = 0 Attempt complete division by (x ) or equiv 4x x + 0 = 0 Obtain quotient of ax + ax + k, where a is their coeff of x (x )(4x + 8x ) = 0 Obtain (4x + 8x ) (or multiple thereof) (x )(x )(x + ) = 0 Attempt to solve quadratic Hence x = ½, x = -½ 6 Obtain x = ½, x = -½ 0 SR: answer only is B B 9

47 Mark Scheme June 008 47 Core Mathematics ( x) 6 = 6 + 6..(-x) +. 4.(-x) Attempt (at least) first two terms - product of binomial coefficient and powers of and (- )x = 64 76x + 60x Obtain 64 76x Attempt third term - binomial coefficient and powers of and (-)x Obtain 60x OR Attempt expansion involving all 6 brackets Obtain 64 Obtain 76x Obtain 60x SR if the expansion is attempted in descending order, and the required terms are never seen, then B B B for 4860x 4, -96x, 79x 6 4 (i) u = / B Obtain correct u u = - / B Obtain correct u from their u u 4 = B Obtain correct u 4 from their u (ii) sequence is periodic / cyclic / repeating B Any equivalent comment (i) ½ 8 θ = 48 State or imply (½) 8 θ = 48 Hence θ =. radians Obtain θ =. (or 0.477π), or equiv (ii) area = 48 ½ 8 sin. * Attempt area of Δ using (½) 8 sin θ = 48.9 d* Attempt 48 area of Δ = 6. Obtain 6. cm 4 (i) f() = 7a 6 a + = 0 * Attempt f() 6a = 4 d* Equate attempt at f() to 0 and attempt to solve a = 4 Obtain a = 4 OR * Attempt complete division / matching coeffs d* Equate remainder to 0 Obtain a = 4 (ii) f(-) = - 6 + 6 + Attempt f(-) = 0 Obtain 0 (or 6a 4, following their a) 6

47 Mark Scheme June 008 (i) xdy = (( y ) ) dy B Show x = y 6y + 7 convincingly term = ( y 6y + 7) dy A.G. B State or imply that required area = xdy + ( + ) =, + (4 + ) = 7 B Use x =, 4 to show new limits of y =, 7 (ii) [ ] y y + 7 y 7 Integration attempt, with at least one = ( 4 / 47 + 49) ( / 7 + ) correct All three terms correct = 6 / / Attempt F(7) F() = 4 / Obtain 4 /, or exact equiv 4 6 (i) ABC = 60 (0 + 0) = 00 o A.G. B Show convincingly that angle ABC is 00 o (ii) 0 CA = + 7 7 cos00 = 094.6 CA =. Attempt use of correct cosine rule Obtain. km (iii) sin C sin 00 = or sin A sin00 =. 7. Attempt use of sine rule to find angle C or A (or equiv using cosine rule) Correct unsimplified eqn, following their CA C = 6. o A =. o Obtain C = 6. o or A =. o (allow.4 o ) Hence bearing is 6 o Obtain 6 or 64 (or 90 o their angle C / 0 + their angle A) 4 7 (a) 4 ( x x + x ) dx Expand brackets and attempt integration, or other valid integration attempt 4 = 6 6 x x x + 4 (+ c) Obtain at least one correct term Obtain a fully correct expression B For + c, and no or dx (can be given in (b)(i) if not given here) 4 (b) (i) -6x - (+c) Obtain integral of the form kx - Obtain -6x - (+c) ] (ii) [ 6x B* State or imply that F( ) = 0 (for kx n, n -) = ¾ Bd* Obtain ¾ (or equiv) 7

47 Mark Scheme June 008 8 (i) Attempt sketch of exponential graph ( st quad) - if seen in nd quad must be approx correct Correct graph in both quadrants B State or imply (0, ) only (ii) 8 x = x x log 8 x = log ( x x ) Form equation in x and take logs (to any consistent base, or no base) could use log 8 xlog 8 = log + xlog Use log a b = b log a Use log ab = log a + log b,or equiv with log a / b x = + xlog Use log 8 = x ( log ) =, hence x = A.G. Show given answer correctly log OR 8 x = x x x = x x Use 8 x = x (x-) = x Attempt to rearrange equation to k = x log (x-) = log x Take logs (to any base) (x )log = x log Use log a b = b log a x ( log ) =, hence x = A.G. Show given answer correctly log 9 (a) (i) sinx.sinx = cosx cosx M Use tanx sinx cosx sin x cosx = cos x cos x cosx = cos x Use sin x cos x cos x + cosx = 0 Show given equation convincingly (ii) (cosx )(cosx + ) = 0 Attempt to solve quadratic in cosx cosx = / Attempt to find x from root(s) of quadratic x =. rad Obtain. rad or 70. o x =.0 rad Obtain.0 rad or 89 o (or π / 60 o - their solution) SR: B B for answer(s) only 4 (b) 0.x0.x{cos0+(cos0.+cos0.+cos0.7)+cos} Attempt y-coords for at least 4 of the correct x-coords Use correct trapezium rule, any h, for their y values to find area between x = 0 and x = Correct h (soi) for their y values 0.87 Obtain 0.87 4 8

47 Mark Scheme June 008 0 (i) u = + 4 x 0. Attempt use of a + (n )d = 9 km Obtain 9 km (ii) u 0 = x. 9 =. B State, or imply, r =. Attempt u 0, using ar n- u 9 = x. 8 =. Obtain u 0 =., and obtain u 9 =. OR B State, or imply, r =. Attempt to solve ar n- = Obtain n = 0 (allow n 0) n n (iii) (. ) B State or imply S > 00 N = (. ) (. ) (. ). n > Link (any sign) their attempt at S N (of a GP) to 00 and attempt to solve log n > log. Obtain 6, or. or better n >. ie Day 6 Conclude n = 6 only, or equiv eg Day 6 4 (iv) swum = x 0 = 60 km B Obtain 60 km, or x 0km run = ½ x 0 x (4 + 9 x 0.) Attempt sum of AP, d = 0., a =, n = 0 = 77. km 0 cycle = (. ) Attempt sum of GP, r =., a =, n = 0 (. ) = 9.0 km total = 666 km Obtain 666 or 667 km 4 9

47 Mark Scheme January 009 47 Core Mathematics 4 4 (i) ( x + 8x ) dx = x + 4x x + c (ii) Attempt integration increase in power for at least terms Obtain at least correct terms Obtain x + 4x x + c (and no integral sign or dx) x dx = 8x + c B State or imply x = x Obtain kx 4 Obtain 8 x + c (and no integral sign or dx) 6 (only penalise lack of + c, or integral sign or dx once) (i) 40 o π = 40 Attempt to convert 40 o to radians 80 (ii) arc AB = 7 7 = π Obtain π, or exact equiv 9 9 chord AB = 7 9 7 π Attempt arc length using rθ or equiv method 49 = 7. Obtain 7., π or unsimplified equiv 9 8 7 7 sin π =. Attempt chord using trig. or cosine or sine rules hence perimeter = 0. cm 4 Obtain 0., or answer that rounds to this (i) u = / B State u = / u = /, u = B State u = / and u = (ii) 4 k / = 0 Equate u k to 0 k = 6 Obtain 6 0 (iii) = ( + ) S Attempt sum of AP with n = 0 0 9 = 40 Correct unsimplified S 0 Obtain 40 6 4 4 ( + ) dx = [ x + x] 7 x Attempt integration increase of power for at least term + Obtain correct x x = ( + 6) ( 6) Use limits (any two of, 0, ), correct order/subtraction 4 = 4 4 Obtain 4 area of rectangle = 9 x 4 B State or imply correct area of rectangle 4 hence shaded area = 76 4 Attempt correct method for shaded area = 7 6 Obtain aef such as., OR Area = 9 (x 4 + ) Attempt subtraction, either order = 6 x 4 Obtain 6 x 4 (not from x 4 + = 9) 4 ( 6 ) dx = [ 6x x x ] Attempt integration Obtain ± ( 6x x )

47 Mark Scheme January 009 = ( - ) (- - ) Use limits correct order / subtraction = Obtain ± Obtain only, no wrong working 7 0 TA (i) = Attempt use of correct sine rule to find TA, or equiv sin 07 sin TA = 94 m Obtain 94, or better (ii) TC = 94 + 0 94 0 cos 70 Attempt use of correct cosine rule, or equiv, to find TC Correct unsimplified expression for TC, following their (i) = 874 m Obtain 874, or better (iii) dist from A = 94 x cos 70 = m Attempt to locate point of closest approach beyond C, hence 874 m is shortest dist Convincing argument that the point is beyond C, OR or obtain 89, or better perp dist = 94 sin 70 = 89 m SR B for 874 stated with no method shown 0 6 (i) S = 0. 9 Attempt use of S = r = 00 Obtain 00 0 0( 0.9 ) (ii) = 7 S 0 Attempt use of correct sum formula for a GP, with n = 0 0.9 = 9 Obtain 9, or better p p (iii) 0 0.9 < 0.4 B Correct 0 0.9 seen or implied 0.9 p- < 0.0 ( p ) log 0.9 < log 0.0 Link to 0.4, rearrange to 0.9 = c (or >, <), introduce log 0.0 p > log 0. 9 logarithms, and drop power, or equiv correct method p > 8. Correct method for solving their (in)equation hence p = 9 4 State 9 (not inequality), no wrong working seen 7 (i) 6k a = 4 * Obtain at least two of 6, k, a k a = 4 dep* Equate 6k m a n to 4 ak = A.G. Show ak = convincingly no errors allowed (ii) 4k a = 8 B State or imply coeff of x is 4k a 4k = Equate to 8 and attempt to eliminate a or k ( ) 8 k k = 6 Obtain k = 4 k = 4, a = 4 Obtain a = ½ SR B for k = ± 4, a = ± 8 a (iii) 4 4 ( ) = Attempt 4 k a, following their a and k (allow if still in terms of a, k) Obtain (allow x ) 9 6

47 Mark Scheme January 009 8 (a)(i) log xy = p q B State p + q cwo a + a x a y (ii) log ( ) = + p q Use log a b = bloga correctly at least once a Use log = log a log b correctly Obtain + p q b x 0 0 x (b)(i) log B State x 0 0 x log (with or without base 0) 0 x x (ii) log0 = log0 9 B State or imply that log = log x 0 0 x 0 = 9 Attempt correct method to remove logs x 9x 0 = 0 Obtain correct x 9x 0 = 0 aef, no fractions (x 0)(x + ) = 0 Attempt to solve three term quadratic x = 0 Obtain x = 0 only 9 (i) f() = + = 0 A.G. B Confirm f() = 0, or division with no remainder shown, or matching coeffs with R = 0 f(x) = (x )(x ) Attempt complete division by (x ), or equiv Obtain x + k Obtain completely correct quotient (allow x + 0x ) x = Attempt to solve x = x = ± 6 Obtain x = ± only 0 (ii) tan x =,, B State or imply tan x = or tan x = at least one of their roots from (i) tan x = x = π /, 4π / Attempt to solve tan x = k at least once tan x = x = π /, π / Obtain at least of π /, π /, 4π /, π / (allow degs/decimals) tan x = x = / 4, π / 4 Obtain all 4 of π /, π /, 4π /, π / (exact radians only) B Obtain π / 4 (allow degs / decimals) B 6 Obtain π / 4 (exact radians only) SR answer only is B per root, max of B4 if degs / decimals 7

47 Mark Scheme June 009 47 Core Mathematics (i) cos θ = 6.4 7.0. 6.4 7.0 Attempt use of cosine rule (any angle) = 0.4 Obtain one of o, 4. o, 0.9 o,.0, 0.97, 0.9 θ = o or.0 rads Obtain o or.0 rads, or better (ii) area = 7 6.4 sin Attempt triangle area using (½)absinC, or equiv = 0. cm Obtain 0. (cao) (i) a + 9d = (a + d) * Attempt use of a + (n )d or a + nd at least once for u 4, u 0 or u 0 a = d Obtain a = d (or unsimplified equiv) and a + 9d = 44 a + 9d = 44 d = 44 dep* Attempt to eliminate one variable from two simultaneous equations in a and d, from u 4, u 0, u 0 and no others d =, a = 6 4 Obtain d =, a = 6 (ii) S 0 = 0 / (x6 + 49x) Attempt S 0 of AP, using correct formula, with n = 0, allow (a + 4d) = 70 Obtain 70 x x log 7 log Introduce logarithms throughout, or equiv with base 7 or x log 7 x log Drop power on at least one side Obtain correct linear equation (allow with no brackets) x log 7 log log Either expand bracket and attempt to gather x terms, or deal correctly with algebraic fraction x = 0. Obtain x = 0., or rounding to this, with no errors seen 4 (i)(x ) = x x x * Attempt expansion, with product of powers of x and +, at least terms 6 4 = x x 7x * Use at least of binomial coeffs of,,, dep* Obtain at least two correct terms, coeffs simplified 4 Obtain fully correct expansion, coeffs simplified OR (x ) = (x )(x 4 0x + ) M Attempt full expansion of all brackets = x 6 x 4 + 7x Obtain at least two correct terms Obtain full correct expansion 6 7 7 (ii) x dx x x x x c Attempt integration of terms of form kx n Obtain at least two correct terms, allow unsimplified coeffs Obtain 7 x x x x B 4 + c, and no dx or sign 7 8

47 Mark Scheme June 009 (i) x = 0 o, 0 o Attempt sin - 0., then divide or multiply by x = o, 7 o Obtain o (allow π / or 0.6) Obtain 7 o (not radians), and no extra solutions in range range (ii) ( cos x) = cosx Use sin x = cos x cos x cosx = 0 Obtain cos x cosx = 0 or equiv (no constant terms) cosx (cosx ) = 0 Attempt to solve quadratic in cosx cosx = 0, cosx = ½ Obtain 0 o (allow π / 6 or 04), and no extra solns in x = 90 o, x = 0 o B Obtain 90 o (allow π / or.7), from correct quadratic only SR answer only B one correct solution B second correct solution, and no others 6 x a dx x ax c Attempt to integrate Obtain at least one correct term, allow unsimplified Obtain x + ax (, ) a c M Substitute at least one of (, ) or (, 7) into integration attempt involving a and c (, 7) 8 a c 7 Obtain two correct equations, allow unsimplified M Attempt to eliminate one variable from two equations in a and c a =, c = Obtain a =, c =, from correct equations Hence y = x + x + 8 State y = x + x + 7 (i) f(-) = -6 + 6 8 Attempt f(-), or equiv = -0 Obtain -0 (ii) f(½) = ¼ + ¼ + ½ 8 = 0 AG Attempt f(½) (no other method allowed) Confirm f(½) = 0, extra line of working required 8 8 (ii i ) f(x ) = (x )(x + x + 8) Attempt complete division by (x ) or (x ½) or equiv Obtain x + x + c or x + 0x + c State (x )(x + x + 8) or (x ½)(x + 0x + 6) (iv) f(x) has one real root (x = ½) B State root, following their quotient, ignore reason because b 4ac = = -7 hence quadratic has no real roots as -7 < 0, B Correct calculation, eg discriminant or quadratic formula, following their quotient, or cubic has max at (-., -9.9) 9 6

47 Mark Scheme June 009 8 (i) ½ r. = 60 Attempt (½) r θ = 60 r = 0 Obtain r = 0 rθ = 0. = B State or imply arc length is.r, following their r perimeter = 0 + 0 + = cm 4 Obtain (ii)(a)u = 60 0.6 4 Attempt u using ar 4, or list terms = 7.78 Obtain 7.78, or better 0 (b) 60 0.6 S 0 0.6 Attempt use of correct sum formula for a GP, or sum terms = 49 Obtain 49, or better (allow 49.0 49. inclusive) (c) common ratio is less than, so series is B series is convergent or - < r < (allow r < ) or reference convergent and hence sum to infinity exists to areas getting smaller / adding on less each time S = 60 0.6 Attempt S using = 0 Obtain S = 0 a r SR B only for 0 with no method shown 9 (i) B Sketch graph showing exponential growth (both quadrants) B State or imply (0, 4) (ii) 4k x = 0k k x = k Equate 4k x to 0k and take logs (any, or no, base) x = log k k x = log k + log k k Use log ab = log a + log b x = log k k + log k Use log a b = b log a x = + log k AG 4 Show given answer correctly OR 4k x = 0k k x = k Attempt to rewrite as single index k x- = Obtain k x- = or equiv eg 4k x- = 0 x - = log k Take logs (to any base) x = + log k AG Show given answer correctly (iii) (a) area 0 4k 8k k Attempt y-values at x = 0, ½ and, and no others Attempt to use correct trapezium rule, y-values, h = ½ k k Obtain a correct expression, allow unsimplified (b) k k = 6 Equate attempt at area to 6 k 6 Attempt to solve disguised term quadratic k k = 9 Obtain k = 9 only 7

47 Mark Scheme January 00 47 Core Mathematics (i) ( cos x) = cos x Use sin x = cos x cos x + cos x = 0 A.G. Show given equation correctly (ii) (cos x )(cos x + ) = 0 Recognise equation as quadratic in cos x and attempt recognisable method to solve cos x = ½ Attempt to find x from root(s) of quadratic x = 60 o Obtain 60 o or π / or.0 rad x = 00 o 4 Obtain 00 o only (or 60 o their x) and no extra in range SR answer only is B B 6 + c * Attempt integration (inc. in power for at least one term) (i) ( 6x 4) dx = x 4x Obtain x 4x (or unsimplified equiv), with or without + c y = x 4x + c = 8 + c dep* Use (, ) to find c c = Hence y = x 4x + 4 Obtain y = x 4x + (ii) p 4p + = * Equate their y (from integration attempt) to p 4p 4 = 0 dep* Attempt to solve three term quadratic (p ) (p + ) = 0 p = - / Obtain p = - / (allow any variable) from correct working; condone p = still present, but A0 if extra incorrect solution 7 (i) ( x) 7 = 8 448x + 67x 60x Attempt (at least) two relevant terms product of binomial coeff, and x (or expansion attempt that considers all 7 brackets) Obtain 8 448x Obtain 67x 4 Obtain 60x (ii) 60 ( / 4 ) = - / 4 Attempt to use coeff of x from (i), with clear intention to cube / 4 Obtain - / 4 (w 6 ), (allow / 4 from +60x in (i)) 6

47 Mark Scheme January 00 4 (i) Attempt y-coords for at least 4 of the log ( + )d 0 x x (log + log. + correct x-coords only log 6 + log 6. + log 7) Use correct trapezium rule, any h, to find area between x = and x = Correct h (soi) for their y-values. 4 Obtain. (ii) B Divide by, or equiv, at any stage to log0 ( + x) dx = log ( x) dx 0 + obtain 0.78 or 0.77, ½. following their answer to (i) 0.78 B Explicitly use log a = ½ log a on a single term 6 {( ) ( )} [ ] 9x x + dx = 9x x + 0x Attempt subtraction (correct order) at any point = ( 9 + 0) (9 / +0) Attempt integration inc. in power for at least one term = 4 8 / Obtain ± ( / x + 0x) or x and / x + x = / Obtain remaining term of form kx - OR Obtain ± 9x - or any unsimplified equiv [ x + 9x ] [ ] x + x Use limits x =, correct order & subtraction = [( + ) ( + 9)] [(9 + ) ( / + )] 7 Obtain /, or exact equiv = 6 0 / = / 7 6 (i) f( ) = 0 4 + 9a b + = 0 Attempt f( ) and equate to 0, or equiv method a b = Obtain a b =, or unsimplified equiv f() = 6 + 4a + b + = Attempt f() and equate to, or equiv method a + b = Obtain a + b =, or unsimplified equiv Hence a =, b = 4 Attempt to solve simultaneous eqns 6 Obtain a =, b = 4 (ii) f(x) = (x + )(x x + ) Attempt complete division by (x + ), or equiv Obtain x x + c or x + bx +, from correct f(x) ie quotient is (x x + ) Obtain x x + (state or imply as quotient) 9 6

47 Mark Scheme January 00 7 (i) = 0 + 4 0 4 cos θ Attempt to use correct cosine rule in Δ ABC cos θ = 0.46 θ =.0 A.G. Obtain.0 radians (allow. radians) SR B only for verification of.0, unless complete method (ii) arc EF = 4.0 = 4.4 B State or imply EF = 4.4cm (allow 4.0) perimeter = 4.4 + 0 + + 6 Attempt perimeter of region - sum of arc and three sides with attempt to subtract 4 from at least one relevant side =.4 cm Obtain.4 cm (iii) area AEF = ½ x 4. Attempt use of (½) r θ, with r = 4 and θ =.0 = 8.8 Obtain 8.8 area ABC = ½ 0 4 sin. Attempt use of (½)absinθ, sides consistent with angle used = 6.4 Obtain 6.4 or better (allow 6.8 or 6.9) hence total area =.6 cm Obtain total area as.6 cm 0 8 (i) u = 8 + 4 Attempt a + (n )d or equiv inc list of terms = 0 A.G. Obtain 0 (ii) u n = n + ie p =, q = B Obtain correct expression, poss unsimplified, eg 8 + (n ) B Obtain correct n +, or p =, q = stated (iii) arithmetic progression B Any mention of arithmetic (iv) N ( + ( N ) ) ( 6 + ( N ) ) = 6 N 6 Attempt S N, using any correct formula (inc (n + ) ) Attempt S N, using any correct formula, 6N + N N N = with N consistent (inc (n + ) ) 9N + N = 0 * Attempt subtraction (correct order) and equate to 6 (9N + 7)(N 6) = 0 dep* Attempt to solve quadratic in N N = 6 Obtain N = 6 only, from correct working OR: alternative method is to use n / (a + l) = 6 Attempt given difference as single summation with N terms Attempt a = u N+ Attempt l = u N Equate to 6 and attempt to solve quadratic Obtain N = 6 only, from correct working 0 7

47 Mark Scheme January 00 9 (i) Reasonable graph in both quadrants Correct graph in both quadrants B State or imply (0, 6) (ii) 9 x = 0 Introduce logarithms throughout, or equiv with log 9 x log 9 = log 0 Use log a b = b log a and attempt correct method to find x x =.8 Obtain x =.8 (iii) 6 x = 9 x Form eqn in x and take logs throughout (any base) log (6 x ) = log 9 x Use log a b = b log a correctly on log x or log 9 x or legitimate combination of these two log 6 + x log = x log 9 Use log ab = log a + log b correctly on log (6 x ) or log 6 log + log + x log = x Use log 9 = or equiv (need base throughout that line) x ( log ) = + log + log + log x = A.G. Obtain x = convincingly log log (inc base throughout) 8

47 Mark Scheme June 00 (i) f() = 8 + 4a a 4 a 6 = 0 a = (ii) f(-) = - + + 4 = -9 * d* ft Attempt f() or equiv, including inspection / long division / coefficient matching Equate attempt at f(), or attempt at remainder, to 0 and attempt to solve Obtain a = Attempt f(-) or equiv, including inspection / long division / coefficient matching Obtain -9 (or a, following their a) (i) area 8 4 7 0.8 B State or imply at least of the 4 correct y-coords, and no others Use correct trapezium rule, any h, to find area between x = and x = 0 Correct h (soi) for their y-values must be at equal intervals 4 (ii) use more strips / narrower strips B Obtain 0.8 (allow 0.7) Any mention of increasing n or decreasing h (i) ( + ½x) 0 = +x +.x + x B Obtain + x Attempt at least the third (or fourth) term of the binomial expansion, including coeffs Obtain.x Obtain x 4 (ii) coeff of x = ( x ) + (4 x.) + ( x ) = 00 ft Attempt at least one relevant term, with or without powers of x Obtain correct (unsimplified) terms (not necessarily summed) either coefficients or still with powers of x involved Obtain 00 7

47 Mark Scheme June 00 4 (i) u = 6, u =, u = 6 B State 6,, 6 (ii) S 40 = 40 / ( x 6 + 9 x ) = 440 Show intention to sum the first 40 terms of a sequence Attempt sum of their AP from (i), with n = 40, a = their u and d = their u u Obtain 440 (iii) w = 6 p + = 6 or 6 + (p ) x = 6 p = B 7 State or imply w = 6 Attempt to solve u p = k Obtain p = (i) sin 8 sin 6 Attempt use of correct sine rule θ = 4. o Obtain 4. o, or better (ii) a 80 ( x 6) = 0 o or 6 x π / 80 =.4 0 x π / 80 = 0.87 A.G. π ( x.4) = 0.87 Use conversion factor of π / 80 (ii) b area sector = ½ x 8 x 0.87 = 7.9 area triangle = ½ x 8 x sin 0.87 = 4. area segment = 7.9 4. =.4 4 8 Show 0.87 radians convincingly (AG) Attempt area of sector, using (½) r θ Attempt area of triangle using (½) r sin θ Subtract area of triangle from area of sector Obtain.4or.4

47 Mark Scheme June 00 6 a x 4x d x x x = ( / + 0) (9 + 8) Attempt integration Obtain / x +x = 64 / Use limits x =, correct order & subtraction b 6 y d y y 4y c 4 B Obtain 64 / or any exact equiv State y Obtain ky Obtain 4y (condone absence of + c) c 4 8x dx x = ( 0 ) ( -4 ) B State or imply x x Attempt integration of kx n = 4 Obtain correct -4x - (+c) ft 4 Obtain 4 (or -k following their kx - ) 7 (i) sin x cos sin x x sin x cos x cos x sin x cos x cos x cos x tan x Use either sin x + cos x =, or tan x = sin x / cos x Use other identity to obtain given answer convincingly. (ii) tan x = tan x tan x + tan x 6 = 0 (tan x )(tan x + ) = 0 tan x =, tan x = - x = 6.4 o, 4 o x = 08 o, 88 o B State correct equation Attempt to solve three term quadratic in tan x Obtain and - as roots of their quadratic Attempt to solve tan x = k (at least one root) ft Obtain at least correct roots 6 Obtain all 4 correct roots 8

47 Mark Scheme June 00 8 a log w = log 4 0 (w )log = 0 log 4 0log 4 w = log w = 7. * * d* Introduce logarithms throughout Use log a b = b log a at least once Obtain (w )log = 0 log 4 or equiv Attempt solution of linear equation y b log x 4 y 4 x y + = x 4 4 y x 4 9 Obtain 7., or better Use log a log b = log a / b or equiv Use f(y) = x 4 as inverse of log x f(y) = 4 Attempt to make y the subject of f(y) = x 4 4 Obtain y x, or equiv 9 (i) ar = a + d, ar = a + d ar ar = a ar ar + a = 0 r r + = 0 A.G. (ii) f (r) = (r )(r + r ) B Attempt to link terms of AP and GP, implicitly or explicitly. Attempt to eliminate d, implicitly or explicitly, to show given equation. Show r r + = 0 convincingly Identify (r ) as factor or r = as root r = Hence r = * d* Attempt to find quadratic factor Obtain r + r Attempt to solve quadratic Obtain r = only (iii) a r Equate S to a ( )( ) Obtain a a = 9 / / a = 4 Attempt to find a Obtain a = 4

47 Mark Scheme January 0 (i) ( + x) 7 = + 4x + 84x B Obtain + 4x Attempt third term Obtain 84x Needs to be simplified, so not 7 and 4x not 7 x x. B0 if other constant and/or x terms (from terms being sums not products). Must be linked by + sign, so, 4x is B0, but can still get for third term. Needs to be product of and an attempt at squaring x allow even if brackets never seen, so 4x gets. No need to see powers of explicitly. Coefficient needs to be simplified. Ignore any further terms, right or wrong. Can isw if they subsequently attempt simplification eg dividing by 4, but they won t then get the ft mark in part (ii). If manually expanding brackets they need to consider all 7, but may not necessarily show irrelevant terms. If the expansion is attempted in descending powers, only giving the first three will gain no credit in (i), unless they subsequently attempt the relevant terms in (ii) when we will then give appropriate credit for the marks in (i). This only applies if no attempt at the required terms is made in (i). A full expansion with the required terms at the end is marked as per original scheme. (ii) ( x)( + 4x + 84x ) coeff of x = 70 + 68 = 98 Attempt at least one relevant product Could be just a single term, or part of a fuller expansion considering terms other than x as well. Allow even if second x term isn t from a relevant product eg -70 + 84 gets A0. ft Obtain two correct unsimplified terms (not necessarily summed) either coefficients or still with powers of x involved Needs to come from two terms only, and can be awarded for unsimplified terms eg -x x 4x If fuller expansion then A0 if other x terms, but ignore any irrelevant terms. If expansion is incorrect in (i) and candidate only gives a single final answer in (ii) then examiners need to check and award either ft or M0. 6 Obtain 98 Allow 98x. Allow if part of a fuller expansion and not explicitly picked out. If clearly finding coefficient of x, allow as misread.

47 Mark Scheme January 0 (i) u =, u = 8, u = B B Obtain at least one correct term Obtain all three correct terms Just a list of numbers is fine, no need for labels. Ignore extra terms beyond u. (ii) arithmetic progression B Any mention of arithmetic Allow AP, but not description eg constant difference. Ignore extra description eg diverging as long as not wrong or contradictory. (iii) S = 00 / (0 + 60) or 00 / (x0 + 99x) = 4,0 (or S 00 S 00 = 60,700,0) Attempt relevant S n using correct formula Must use correct formula to sum an AP only exception is using (½ n )d rather than (n )d. Must use d = (or their d from (i) as long as constant difference). If (i) is incorrect they can still get full marks in (iii) as independent. They need to be finding the sum of 99, 00, 0 or 00 terms and make a reasonable attempt at a value of a consistent with their n if n = 99 then a = 0 / if n = 00 then a = or a = 0 / if n = 0 then a = / if n = 00 then a =. Allow slips on a = 0 as long as clearly intending to find u 0. If using ½ n (a + l) then there also needs to be a reasonable attempt at l. Attempting to sum from n = 0 to n = 00 gets both method marks together (assuming that the attempt satisfies above conditions). Attempt correct method to find required sum S 00 S 0 is M0. M0 is possible for correct method but with incorrect formula for S N (but must be recognisable as attempt at sum of AP). Need to show subtraction to gain, just calculating two relevant sums is not yet enough. Still need a = and d =. Obtain 4,0 Answer only gets full marks. 6 SR: if candidates attempt to manually add terms Attempt to sum all terms from u 0 to u 00 A Obtain 4,0

47 Mark Scheme January 0 (i) 0. 0. 0 0.. =.8 Attempt at least 4 correct y-coords, and no others If first term of 0 not explicit then other 4 terms need to be seen. Could be implied by eg (4 ), or implied by a table with correct x-coords in one column and attempts at y-coords in second column. Allow rounded or truncated decimals. Allow an error in rearrangement eg x -. Attempt correct trapezium rule, any h, to find area between x = and x =. Correct structure ie 0. x (any h) x (first + last + x middles) no omissions allowed. The first y-coord should correspond to attempt when x = (though may not be shown explicitly), and last to x =. It could be implied by using y 0 etc in rule, when these have already been attempted elsewhere and clearly labelled. It could use other than 4 strips, but these must be at equal widths. Using just one strip is M0. The big brackets must be seen, or implied by later working (omission of these can lead to.4 or.9 or 6. ). If ½ x k seen at start of rule then assume that ½ is part of a correct rule and the k is an incorrect strip width. Use correct h (soi) for their y-values must be at equal intervals Must be in attempt at the trapezium rule, not Simpson s rule. Allow if muddle over placing y-values. Allow if one y-value missing (including first or last) or extra. Allow if ½ missing. Using h = with only one strip is M0. 4 Obtain.8, or better More accurate solution is.89479 Answer only is 0/4. Using integration is 0/4. Using trapezium rule on the result of an integration attempt is 0/4. Using 4 separate trapezia can get full marks. If other than 4 trapezia, mark as above. (ii) Underestimate as tops of trapezia are below curve B* Bd* 6 State underestimate Convincing reason referring to trapezia being below curve Ignore any reasons given. Referring to gaps between curve and trapezia can get B. Could use sketch with brief explanation (but sketch alone is B0) must show more than one trapezium (but not nec 4) or imply this in the text. Trapezia must show clear intention to have top vertices on the curve. Sketching rectangles is B0. Triangle is B0. Explanation that refers to calculated area from integration is B0. Only referring to concave / convex is B0. Can get B for rate of change of gradient (or second derivative) is negative, but not for gradient is decreasing.

47 Mark Scheme January 0 4 (a) log x - = log 0 (x )log = log 0 * Introduce logarithms throughout (or log 0 / log 0 ) & drop power Don t need to see base if taking logs on both sides, though if shown it must be the same base. If taking logs on one side only base must be explicit. x =.97 x =.97 d* Obtain (x )log = log 0, or equiv (eg x = log 0) Attempt to solve Condone lack of brackets ie x log = log 0, as long as clearly implied by later working. Attempt at correct process ie log0 / log ± or equiv (log0 + log) / log. Allow if log0 / log ± subsequently becomes log 4 ±, but M0 if log 4 appears before - is dealt with. Allow if processing slips when evaluating log0 / log eg. from incorrect brackets. 4 Obtain.97, or better Allow more accurate solution, such as.97 and then isw if rounded to.98. However,.98 without more accurate answer seen is A0. Answer only is 0/. Trial and improvement is 0/. (b) log x + log 9 = log (x + ) log (9x) = log (x + ) 9x = x + x = / 8 B State or imply log = log 9 or log Could be done at any stage. Must be correct statement when done, so LHS becoming log (x + 9) in one step is B0. Condone lack of base throughout question. Use log a + log b = log ab, or equiv Must be used to combine (or more) terms of log x + log k = log (x + ), with k most typically (but not exclusively) 6, 8 or 9. Could move log x and/or log 9 across to RHS and then use log a log b, but must still be log (x + ) as single term. Obtain correct equation with single log term on each side (or single log = 0) log (9x) = log (x + ), log x = log (x + ) / 9, log 9 = log (x + ) /x, log (x + ) / 9x = 0. Allow for correct equation with logs removed if several steps run together. 4 Obtain x = / 8 Allow 0.6 8 4

47 Mark Scheme January 0 (i) a 4 a r Equate S a r to 4a, or substitute r = 4 into S must be quoted correctly. Allow 4ar 0 for 4a. Initially using a numerical value for a is M0. a Once equation in a is seen ie 4 a assume that a has been r cancelled if this subsequently becomes 4. If initial r equation in a is never seen then assume that a = is being used and mark accordingly. r = 4 Attempt to find value for r or evaluate S Need to get as far as attempting r. Need to see at least one extra line of working between initial statement and given answer. Substituting numerical value for a is M0 (so M0 possible depending at what stage the substitution happens). r = 4 Obtain r = 4 (or show S = 4a) Allow r = 0.7. (ii) a = 9 * Attempt use of ar Must use r = ¾ not their incorrect value from (i). 4 Must be clearly intended as ar, so ( a / 4 ) = 9 is M0, unless correct expression previously seen. Can use equivalent method with ratio of 4 / ie 9 x ( 4 / ). a = 6 d* Equate to 9 and attempt to find a Must get as far as attempting value for a. Obtain a = 6 Answer only gets full credit. (iii) S 0 0 6 4 Attempt use of correct sum formula for a GP 4 = 6.8 Obtain 6.8, or better 8 Must be correct formula, with a = their (ii), r = ¾ and n = 0. More accurate answer is 6.79704 NB using n rather than n in the formula gives 6.79 (M0), and using n + gives 6.848 (M0). Must be decimal, rather than exact answer with power of ¾.

47 Mark Scheme January 0 6(a) x x x dx x x dx Simplify and attempt integration Need to attempt to divide both terms by x, or multiply entire numerator by x - allow if intention is clear even if errors when simplifying, or one term doesn t actually change. Need to simplify each of the two terms as far as x n before integrating. For integration attempt, need to increase power by for at least one term. = x 6x c Obtain at least one correct term Allow unsimplified terms. x Obtain x 6 Coefficients must now be simplified. Could be 6 x for second term. B 4 Obtain + c Not dependent on previous marks as long as no longer original function. B0 if integral sign or dx still present in answer. Ignore anything that appears on LHS of an equation eg y =, dx = or even =... (b)(i) a Any k, as long as numerical, including unsimplified. 4 6x dx x a Obtain integral of the form kx - Allow + c. Condone integral sign or dx still present. = 4 a - Attempt F(a) F() Must be correct order and subtraction. - a - / 8 is M0 unless clear evidence suggesting that there was an intention to subtract and that this is a sign error. Not dependent on first M mark, so substituting into their integration attempt (eg kx - ) can still get, but using kx -4 is M0. Obtain 4 a - Allow / 8 for ¼, but not (- / 8 ), but want not 6 /. A0 if + c, integral sign or dx still present. isw any subsequent work, usually equating to 0 or writing as inequality. (b)(ii) Bft 4 8 State 4, following their (i) Allow / 8. Do not allow 0 + ¼. Must appreciate that limit is required not inequality so <,, tends to ¼, ¼ etc are all B0. Picking large number for a and then concluding correctly is B. Condone denominator changing from to 0 (or even 0 being used as top limit) if final answer correct. For the ft mark their (i) must be of form a ± bx n, with n 4. If solution in (i) is incorrect but candidate restarts in (ii) and produces ¼ oe with no wrong working then allow B. 6

47 Mark Scheme January 0 7(i) tanx = ⅓ x = 8.4 o, 98.4 o x = 9. o, 99. o Attempt correct solution method Obtain one of 9. o or 99. o, or better Attempt tan - ( / ) and then halve answer. Allow radian equiv (0.6 or.7). ft Obtain second correct angle Maximum of marks if angles not in degrees. A0 if extra solutions in given range, but ignore extra outside range If A0 given, award ft for adding 90 o or π / to their angle. SR: if no working shown then allow B for each correct solution. Maximum of B if in radians, or extra solutions in given range. SR: if using tan x identity then Attempt to find x from solving quadratic equation in tan x, derived from correct tan x identity. Obtain at least one of 9. o or 99. o, or better (or radian equiv) Obtain second correct angle (ii) ( sin x) + sin x = 0 sin x sin x = 0 sin x (sinx ) = 0 sin x = 0, sin x = / x = 0 o, 80 o x = 4.8 o, 8 o Use cos x = sin x, aef Obtain sin x sin x = 0 Attempt to solve equation to find solutions for x Must be used not just stated. Must be used correctly, so sin x is M0. Allow aef, but must be simplified (ie no constant term; allow 0). Not dependent on first so could get M0 if cos x = sin x previously used. Must be quadratic in sinx (must have sinx term), but can still get if constant term in their quadratic as well. Candidates need to be solving for x, so need to sin - at least one of the solutions to their quadratic. Must be acceptable method if factorising then it must give correct lead term and one other on expansion (inc c = 0), if using formula then allow sign slips but no other errors. SR If solving the quadratic involves cancelling by sin x rather than factorising then M0, but give B if both 4.8 o and 8 o found (or radian equivs) Obtain two of 0 o, 80 o, 4.8 o, 8 o Must come from correct factorisation of correct quadratic equation ie sin x (sinx + ) = 0 leading to sin x = 0 and hence x = 0 o, 80 o is A0. Allow radian equivs 0, π (or.4), 0.7,.4. 8 Obtain all four angles Must now all be in degrees, with no extra in given range (ignore any outside range). SR If no working out seen, then allow B for each of 4.8 o and 8 o, and B for both 0 o and 80 o. Maximum of B if in radians or extra solutions in given range. 7

47 Mark Scheme January 0 8(i) ½ x x sin θ = 8 sin θ = 0.64 θ = π 0.694 =.4 * d* Attempt to solve (½)r sin θ = 8 to find a value for θ Attempt to find obtuse angle from their principal value. Allow if using r sin θ = 8. Need to get as far as attempting θ (acute or obtuse). ie π θ in radians, or 80 o θ in degrees (eg 40. o ). Obtain θ =.4, or better Allow answer rounding to.4 with no errors seen. Must be in radians, and clearly intended as only final solution (eg underlined if acute angle still present). A0 if angle then becomes.4π ie this is not isw. (ii) ½ x x.447 = 0.6 hence area = 0.6 8 =.6 cm * Attempt area of sector using (½)r θ Allow if using r θ. θ must be numerical and in radians, but allow if incorrect from their attempt at (i) eg.4π. Allow equivalent method using fraction of circle must be θ / π if using radians or θ / 60 if using degrees. Can get if using an acute angle from (i) (gives 8.68 from 0.694 or 8.6 from 0.69). Using an angle of 0.64 is M0 this is sin θ not θ. However, could still get if using other angle clearly associated with θ in (i). d* Attempt area of segment Subtract 8 from their sector area. Allow if new attempt made at area of triangle, even if their area isn t 8, eg could attempt ½ r (θ sin θ), with incorrect θ. Obtain area of segment as.6 Allow more sig fig as long as it rounds to.6 with no errors seen. Units not needed, and ignore if incorrect. (iii) arc = x.447 =. Bft State or imply arc length is θ θ must be numerical and in radians, or equiv method in degrees. ft on their angle in (i), including acute angle calculation may not be shown explicitly so examiners will need to check. chord = x sin. = 9.40 or AB = cos.447 or AB / sin.4 = / sin 0.47 or ½ x x AB x sin. = 8 Attempt length of chord AB Any reasonable method, and allow radian / degree muddle when evaluating. If using cosine rule, then must be correct formula even if slip when evaluating. Need to get as far as a = but not nec. If using right-angled trig then must use ½ θ to find relevant side, and double it. Could use sine rule or area of a triangle with angle of ½ (π θ). Obtain 9.40 (allow 9.4) Allow any answer in range 9.40 AB 9.4 (before rounding), including more sig fig, with no errors seen. perimeter =. + 9.40 =.6 cm 4 0 Obtain perimeter as.6 (allow.7) 8 Allow any answer in range.6 perimeter.7 (before rounding), inc more sig fig, with no errors seen.

47 Mark Scheme January 0 9(i) f() = 08 + 8 + 0 = 0 hence (x ) is a factor B Show that f() = 0, detail required Substitute x = and confirm f() = 0 must show detail of substitution rather than just state f() = 0. Allow f() = -4 x + 9 x + 0 x = 0 for B. B State (x ) as factor (allow ( x) as the factor) Not dependent on first B. Must be seen in (i) so no back credit from (ii). Allow if not explicitly stated as factor (and allow f(x) = x ). Ignore other factors if also given at this stage. (ii) f(x) = (x )( 4x x + ) or f(x) = ( x)(4x + x ) or f(x) = (x + )( 4x + x ) or f(x) = ( x )(4x x + ) or Attempt complete division by (x ), or equiv (allow division by ( x) ) Obtain 4x x + c or 4x + bx + (or the negative of these if dividing by ( x) ) Must be a full attempt to find three term quadratic. Can use inspection, but must be a reasonable attempt at middle term, with first and last correct. Can use coefficient matching, but must be full method with reasonable attempts at all coefficients. Allow if actually factorising f(x). c, b non-zero constants. First option is likely to come from division, second option from inspection. Coefficient matching could lead to either. Allow for negative of either of these from factorising f(x). f(x) = ( 4x)(x x ) or f(x) = (4x )( x + x + ) Obtain (x )( 4x x + ) (or ( x)(4x + x ) ) Needs to be written as a product as per request in question paper. Allow (x )(4x + x ), but (x )(4x + x ) is A0. A0 if now linear factors and product of linear and quadratic never seen. If using one of the other two correct factors then all three marks are available, and apply mark scheme as above ie for full attempt at division or equiv, for lead term plus one other correct and for product of linear and quadratic. SR: If candidates initially state three linear factors and then expand to get the product of a linear and quadratic as requested award B if fully correct and simplified otherwise B0. (iii) 4x x + = 0 ( 4x) (x + ) = 0 x = ¼, x = Attempt to solve quadratic If factorising, needs to give two correct terms when brackets expanded. If using formula allow sign slips only need to substitute and attempt one further step. If completing the square must get to (x + p) = ± q, with reasonable attempts at p and q. Obtain (¼, 0), (, 0) Condone only x values given rather than coordinates. Allow if x = is still present as well. 9

47 Mark Scheme January 0 (iv) f(x)d x = 4 x x x x B Obtain 4 x x x x Allow unsimplified coefficients. Condone + c. F() F(¼) = (6) ( 0 / 6 ) = 6 0 / 6 F(¼) F( ) = ( 0 / 6 ) (4) = -4 0 / 6 * Attempt F() F( ¼ ) or F( ¼ ) F(-) Allow use of incorrect limits from their (iii). Limits need to be in correct order, and subtraction. Allow slips when evaluating but clear subtraction attempt must be seen or implied at least once. If minimal method shown then it must appear to be a plausible attempt eg F() = 98 or even F() F( ¼ ) = 98.4. Obtain at least one correct area, including decimal equivs Obtain 6 0 / 6 or 97 / 6 or 6.4 or 4 0 / 6 or / 6 or 4.4 Can get if both areas attempted and one is correct but the other isn t. d* Attempt full method to find total area including dealing correctly with negative area Need to see modulus of negative integral from attempt at F( ¼ ) F(-) (just changing sign from ve to +ve is sufficient). If values incorrect in (iii) then can only get this mark if their integral gives negative value. Need to have positive integral from F() F( ¼ ). Hence area = 6 0 / 6 + 4 0 / 6 = 40 0 / 8 Obtain 40 0 / 8 or / 8 or 40.8 Allow exact fraction (including unsimplified ie 044 / 6 ), or decimal answer to dp or better (rounding to 40.8 with no errors seen) SR: If candidate attempts F() F( ¼ ) and F(-) F( ¼ ) as an alternative method for dealing with negative area then mark as B correct integral M complete method obtain one correct area obtain correct total area Any attempts using this method must be fully supported by evidence of intention, especially - as top limit and ¼ as bottom limit used consistently throughout integration attempt. It should not be awarded if candidate appears to have simply confused their order of subtraction. 0

47 Mark Scheme June 0 (i) BC = 9 + 7 x 9 x 7 x cos40 o BC =.6 cm Attempt use of correct cosine rule Must be correct formula seen or implied, but allow a slip when evaluating eg omission of, or incorrect use of an additional big bracket. Allow even if subsequently evaluated in radian mode (.96). Allow if expression is not square rooted, as long as it is clear that correct formula was used ie either BC = or even just a = if the power disappears from BC. Obtain.6, or better Actual answer is.6449 so allow more accurate answer as long as it rounds to.64 (ii) area = 9 7 sin 40 = 49. cm Attempt triangle area using (½)absinC, or equiv Condone omission of ½ from this formula, but no other errors allowed. If using right-angled triangle, must use ½bh with reasonable attempt at perpendicular sides. Allow if subsequently evaluated in radian mode (7.00). If using 40 o, must be using sides of 9 and 7, not.6 from (i). If using another angle, can still get as long as sides used are consistent with this angle. Obtain 49., or better Actual answer is 49.7 so allow more accurate answer as long as it rounds to 49.7 Must come from correct working only. (iii) BD = sin 40 9 sin 6 BD = 6.49 cm Attempt use of correct sine rule, or equiv, to find length BD No further rearrangement required. Could have both fractions the other way up. Must be angles of 40 o and 6 o if finding BD directly. Must be attempting BD, so using 77 o to find AD is M0 unless attempt is then made to find BD by any valid method. Placing D on BC is M0. Obtain correct unsimplified expression involving BD as the only unknown Can still get even if evaluated in radians (40.07). If using a multi-step method (eg use 77 0 to find AD and then use cosine rule to find BD) then this A mark is only given when a correct (unsimplified) expression involving BD as the only unknown is obtained. 7 Obtain 6.49, or better Actual answer is 6.4976 so allow more accurate answer as long as it rounds to 6.49 Must come from correct working only not eg sin 7.

47 Mark Scheme June 0 (i) x 6 dx = 4x x + c B Obtain kx Obtain 4x Obtain x (don t penalise lack of + c) Any k, as long as numerical. Allow both and for equiv eg x x Allow for unsimplified coefficient as well (ie 6 /. ). Allow -x. Maximum of marks if or dx still present in final answer. Maximum of marks if not given as one expression eg the two terms are integrated separately and never combined. (ii) y = 4x x + c 7 = 4 + c c = hence y = x x 4 * State or imply y = their integral from (i) Must have come from integration attempt ie increase in power by for at least one term, but allow if - disappeared in part (i) ie at least one of the and the B must have been awarded in part (i). Can still get this if no + c. The y does not have to be explicit it could be implied by eg 7 = F(4). M0 if they start with y = their integral from (i), but then attempt to use y 7 = m(x 4). This is a re-start and gains no credit. d* Attempt to find c using (4, 7) M0 if no + c. M0 if using x = 7, y = 4. Obtain y = x x 4 Coefficients now need to be simplified, so -x is A0. Allow for equiv eg x x Must be an equation ie y =, so A0 for equation = or f(x) = 6

47 Mark Scheme June 0 (i) perimeter = r + rθ 6 + 8θ =. 8θ = 7. θ = 0.9 rads B* d* State or imply that arc length 8θ, or equiv in degrees ie θ / 60 x 6π Equate attempt at perimeter to. and attempt to solve for θ Allow B by implication for. / 8 or equivalent in degrees. Need to get as far as attempting θ. Must include radii and correct expression for arc length, either in radians or degrees. M0 if using chord length. Obtain θ = 0.9 rads Obtaining 0.9 and then giving final answer as 0.9π is A0 do not isw as this shows lack of understanding. Finding θ in degrees (.6 o ) and then converting to radians can get as long as final answer is 0.9 (and not eg 0.9006 from premature approximation). (ii) ½ x 8 x 0.9 = 8.8 Attempt area of sector using (½) r θ Obtain 8.8 Condone omission of ½, but no other error. Allow if incorrect angle from part (i), as long as clearly intended to be in radians. Allow equivalent method using fractions of the area. Allow working in degrees as long as it is a valid method. Allow if using 0.9π (even if 0.9 was answer to (i)), as long as clearly attempting (½) r θ with error on angle rather then (½) πr θ. Or any exact equiv. If 0.9 obtained incorrectly in part (i), full credit can still be gained in part (ii). Condone minor inaccuracies from working in degrees, as long as final answer is given as 8.8 exactly.

47 Mark Scheme June 0 4 (i) x + 4 = (y + ) x + 4 = y + y + x = y + y A.G. Attempt to make x the subject Allow for x = (y ± ) ± 4 only. Allow if (y + ) becomes y +, but only if clearly attempting to square the entire bracket squaring term by term is M0. Must be from correct algebra, so M0 if eg (x + 4) = x + 4 is used. Verify x = y + y Need to see an extra step from (y + ) 4 to given answer ie explicit expansion of bracket. No errors seen. (ii) ( y + y ) dy = [ ] y + y y = (9 + 9 9) (⅓ + ) B State or imply that the required area is given by ( y + y ) dy SR B for verification, using y = + ( y + y + 4), and confirming relationship convincingly, or for rearranging x = f(y) to obtain given y = f(x). No further work required beyond stating this. Allow if x appears in integral. Any further consideration of other areas is B0. = (9) ( ⅔) = 0⅔ Attempt integration Increase in power of y by for at least two of the three terms. Can still get if the - disappears, or becomes x. Allow for integrating a function of y that is no longer the given one, eg subtracted from, or using their incorrect rearrangement from part (i). ft Obtain at least two correct terms Allow for unsimplified coefficients. Allow follow-through on any function of y as long as at least terms and related to the area required. Condone, dy or + c present. Attempt F() F() for their integral Must be correct order and subtraction. This is independent of first so can be given for substituting into any expression other than y + y, including y +. If last term is x allow for using and throughout integral, but M0 if x value is used instead. 7 Obtain 0⅔ aef Must be an exact equiv so 0.6 is fine (but 9 / is A0). 0.7, 0.66 or 0 / + c are A0. Must come from correct integral, so A0 if from x. Must be given as final answer, so further work eg subtracting another area is A0 rather than ISW. 4 Answer only is 0/, as no evidence is provided of integration. SR Finding the shaded area by direct integration with respect to x (ie a C technique) can have if done correctly, 4 if non-exact decimal given as final answer but no other partial credit.

47 Mark Scheme June 0 Throughout this question, candidates may do valid work in the incorrect answer space. This can be marked and given credit wherever it occurs, as long as it does not contradict the working and final answer given in the designated space. (i) 4 B State 4, or B0 if other terms still present eg C 0 or 0. Could be part of a longer expansion, in which case ignore all other terms unless also solely numerical. (ii) nd term = x 4 x (kx) = 40kx rd term = 0 x x (kx) = 70k x 40k = 70k k =. B Obtain 40k as coeff of x Either stated, or written as 40kx. Allow unsimplified expression ie x 4 x k or x 4 x (kx), even if subsequently incorrectly evaluated. B0 if still C unless later clearly used as. Attempt coeff of x Needs to be an attempt at a product involving the relevant binomial coefficient (not just C unless later seen as 0), and an intention to square the final term (but allow for kx ). 67.k is M0 (from / x ). Obtain 70k Allow unsimplified ie 0 x x k or 0 x x (kx) even if subsequently incorrectly evaluated. Allow 70k following 0 x x kx ie an invisible bracket was used. Equate coefficients and attempt to solve for k Must be one linear and one quadratic term in k, and must be appropriate method to solve this two term quadratic eg factorise or cancel common factor of k. Condone powers of x still present when equated, as long as not actually used in solution method. Could still gain if incorrect, or no, binomial coefficients used each term must be product of powers of (poss incorrect), correct powers of k and any binomial coefficient used. Obtain k =. (ignore any mention of k = 0) Any exact equivalent, including unsimplified fraction. Could be implied by writing ( +.x). NB If expansion is given as 40kx + 70kx, and candidate then concludes that k = 40 / 70 this is B only as k never seen.