2012 H2 Math Prelims Paper 1 Solution + Marking Scheme (final)

1 2012 Yr 6 H2 Math Preliminary Examination Paper 1 Suggested Mark Scheme Qn Suggested Solution Marking Scheme 1   2 2 2 2 3 1 1 3 1 1 31 i.e. 2 0 21 0 12 02 22 x x y y yy yy yy y x x               M1 – Cross-multiply or state that 2 10y   (in the case of combining into a single fraction) B1 – Correct answer in terms of y A1 Total: 3 marks

2 Qn Suggested Solution Marking Scheme 2     21 21 2 21 31 2 11 22 3 31 tan 2 d14 tan 211 2 d t a n 221 4 6 1 tan 264 n n n x xx x xxx n               1As , tan 2 2nn    21 33 21 2 3 tan 2 1 d14 6 2 4 7 384 x xx         M1 – Use f' ( ) f( ) d nkx xx and proceed to  1 1tan 2 nx  A1 – Evaluation with correct limits M1 – Can show implicitly A1 – Exact answer Total: 4 marks

3 Qn Suggested Solution Marking Scheme 3    1 2 2 2 22 2 2 9 121 39 13 11 222132 9 2 ! 9 1 213 18 216 1 23 9 108 18 21 0 7 3 27 324 x x xx xxx xxx xx x x xx                                        B1 – 1 2 1 9 xk   M1 – Correct use of binomial theorem A1 22 1 0 32 79 12 21 0 19 32 7 919 9 19 3 172 9 24345 19 3 243 15395 9 4 172 688 x x x            i.e. 1539, 688pq Alternatively, 19 3 172 9 24345 19 172 24312 5 19 5 172 60 243 34405 1539     i.e. 3440, 1539pq √M1 – Correct substitution (to award once 5 is seen) A1 Total: 5 marks

4 Qn Suggested Solution Marking Scheme 4 2 2 2 1f( 1 ) f( ) ( 2)! ( 1 )! (1 ) (1 ) ! 21 (1 ) ! 31 (1 ) ! rrrr rr rr r rr r r rr r            M1 – Simplify with (r1)! in the denominator. AG1 (i)  2 22 31 f( 1 ) f( )(1 ) ! f( 1 ) f( 2 ) n r n r rr rrr       f( 2 ) f( 3 ) f( 3 ) f( 4 ) f( 1 ) f( )nn      f( 1 ) f( ) 1 (1 ) ! n n n    M1 – List terms and show cancellation B1 – f( 1 ) f( )n A1 (ii) As , = 0. ( 1)! ( 1)( 2)...1 nnn nn n   2 2 31Thus 1 (1 ) !r rr r     B1 – Show 0( 1)( 2)...1 n nn  B1 Total: 7 marks

5 Qn Suggested Solution Marking Scheme 5(a) Volume of solid  ln 5 2 0 2ln 5 0 d 5d 38.44 (2 dp) by GC y xy ey        B1 – Correct formulation and limits. B1 – Answer to 2dp (accept 12.24) b(i)  , 0 4xx  B1 – Condone w/o set notation (ii) From the diagram    9 02 90 2 fd f | | d () 2 x xx x A BA B B     Consider:    ln(5 ) d ln(5 ) d 5 5ln(5 ) 1 d 5 ln(5 ) 5ln(5 ) 5l n ( 5 ) xx xxx x x x xx x x xx xc xx x c                     9 2 4 11 122 5 l n ( 5 ) 2l n 22 2 1 ln 2 Bx x x     