GCE Mathematics Advanced GCE Unit 76: Further Pure Mathematics Mark Scheme for June
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76 Mark Scheme June A B C ( )( 9) 9 B For correct form seen anywhere with letters or values A 6 A( 9) ( BC)( ) B, C 6 9 6( ) 6( 9) (i) Asymptote = y asymptote y METHOD ( y6) (y) b acy6 (y) y y6 y : this is true y So y takes all values METHOD dy 7 Obtain OR d B B For correct A (cover up or otherwise) For equating coefficients at least once.(or substituting values) into correct identity. For correct B and C For correct final statement cao, oe For correct equation For dividing out (remainder not required) For correct equation of asymptote (ignore any etras) N.B. answer given For forming quadratic in For considering discriminant For correct simplified epression in y soi For completing square (or equivalent) and correct conclusion www dy For finding either by direct d differentiation or dividing out first For correct epression oe. dy, d so y takes all values. For drawing a conclusion For correct conclusion www Alternate scheme: Sketching graph Graph correct approaching asymptotes from both side Graph completely correct Eplanation about no turning values Correct conclusion B B B B A graph with no eplanation can only score
76 Mark Scheme June (i).., B.8 B e.7 F ( ).8 (.86) e.79 For correct For correct For dividing by For estimate of e e F( ) (iii) F ( ).78 (.78) B For true F( ) obtained from d ( ln ) d TMDP anywhere in (i) deduct once (but answers must round to given values or A) B For y and y F( ) drawn, crossing as shown B For lines drawn to illustrate iteration (Min horizontal and vertical seen) Staircase B For stating staircase
76 Mark Scheme June (i) rcos, y rsin a cos sin r cos sin for f sin sin a cos cos a sin cos sin cos For substituting for and y For correct equation oe (Must be r =..) For correct limits for θ (Condone <) N.B. answer given For replacing θ by in their f( ) For correct simplified form. (Must be convincing) f( ) f For correct reason for (iii) (iv) r a.. a B B B For correct value of r.oe Closed curve in st quadrant only, symmetrical about Diagram showing, tangential at O
76 Mark Scheme June (i) d sin y cos y dy dy d sin y For implicit diffn to d y d cosy oe For using sin ycos y to obtain N.B. Answer given + taken since sin has positive gradient B For justifying + sign f(), f () B For correct values f ( ) f ( ) f ''() =, f '''() Use of chain rule to differentiate f () Use of quotient or product rule to differentiate f (). For correct values www, soi (iii) 6 sin Alternative Method: f() =, f'() = f'( )... 8 f ''( )... 9 f '''( )... f '(),f ''(),f '''() sin 6 sin ln( ) 6 B For correct series (allow!) www For correct values Correct use of binomial Differentiate twice Correct values Correct series Bft For terms in both series to at least f.t. from their multiplied together For multiplying terms to at least For correct series up to www For correct term in www
76 Mark Scheme June 6(i) n I ( ) d n n n ( ) n ( ) d n In n ( ) d n n n n I n d n I ni ni For integrating by parts (correct way round) For correct first stage For splitting suitably For obtaining correct relation between I n and In I I n n I n n 6 For correct result (N.B. answer given) For evaluating I [OR I by parts] For using recurrence relation [OR ] times (may be combined together) 6 6 6 9 I 9 7 I I I For [OR ] correct fractions I For correct eact result
76 Mark Scheme June 7(i) y = tanh - B Both curves of the correct shape (ignore overlaps) and labelled y = tanh y = tanh B B gradient = at = stated For asymptotes y = ± and = ± (or on sketch) y = tanh - B Sketch all correct k k tanh d ln(cosh ) ln(cosh k) For substituting limits into ln cosh For correct answer (iii) Areas shown are equal: = tanhk y = k For consideration of areas For sufficient justification tanh k tanh d = rectangle ( k tanh k ) For subtraction from rectangle For correct answer N.B. answer given ktanh k ln(cosh k) Alternative: Otherwise by parts, as tanh OR ln PTO for alternative schemes 6
76 Mark Scheme June 7(iii) Alternative method By parts: tanh k I tanh d u tanh dvd du d v tanh k tanh k I tanh k k k k k ktanh k ln(sech k) ktanh kln(sech k) Alternative method By substitution Let y tanh tanh y tanh k tanh ln( ) tanh ln( tanh ) d sech d y y When, y When tanh k, y k tanh k d k tanh d sech d I y y y I ytanh y tanh y dy ktanh kln cosh k k k u y dvsech y dy du dy v tanh y For integrating by parts (correct way round) For getting this far Dealing with the resulting integral For substitution to obtain equivalent integral Correct so far For integration by parts (correct way round) Final answer 7
76 Mark Scheme June 8(i) cosh u du cosh usinh udu B For correct result cosh u d coshusinhudu sinhu cosh udu For substituting throughout for For correct simplified u integral cosh u du sinh ucosh uu For attempt to integrate For correct integration cosh u ( ) ln ( ) ( c) ln( ) B 7 For substituting for u For correct result oe as f( ) ln(g( )) (iii) V ( ) d ( ) ln For attempt to find d For correct integration (ignore π) V B For statement that volume is infinite (independent of M mark) 8
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