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       1 , 1ab   M1 – Identify and simplify required sections by symmetry M1 – Correct integration by parts applied M1 – Split numerators and apply by parts. Must see [l n ( 5 ) ]x kx  . Condone w/o +c B1– Correct limits (or equivalents) A1 – Both a and b correct Total : 8 marks x4 -4 y O

6 Qn Suggested Solution Marking Scheme 6 (i)     1 1 2 2 22 22 2 2 2 2 2 cose ln cos 1d 1 d 1 d1 d dd d d12 2 2 dd d d dd1 (shown) dd xy yx y yx x yxy x yy y yxx y x xxx yyxx y xx                        Alternative 1 1 2 cos cos e d1 ed 1 x x y y x x            32 11 2 2 22 311 2 2 2 2 2 2 31122 2 2 1 2 1 cos cos cos cos cos cos cos cos d1 1 ee 1 2 2d1 1 e1 e 1 ddLHS 1 dd 11e 1 e 1 1 e 1 e RHS (shown) xx xx xx x x y x x xx xx x yyxx xx xx x x x x y                         B1 M1– Differentiate wrt x again. Two out of three terms correct. AG1 – All terms correct B1 M1– Differentiate wrt x again. Correct application of chain rule or product rule. AG1 – Show LHS = RHS (ii)   32 2 2 32 2 32 2 32 dd d d d12 dddd d dd d13 2 0 ddd yy y y yxx x x xxx x yy yxx xxx        23 22 2 2 23 dddW h e n 0 , e, e, e, 2 edd d yy yxy x xx          , 23 2 23 2 2e 1 ... 2! 3! e 1 ... 23 xxyx xxx           M1 – Any pair of terms in LHS correct (as evident of correct implicit differentiation) M1– First 3 terms correct and provide value for 3 3 d d y x A1

7 Qn Suggested Solution Marking Scheme (iii)  23 2 22 2 1 1 cos cos dde e 1 ...dd 2 3 e e 1 ... 1 x x xxxxx xx x                 22 2 0.5 0.5 2 1cosde =e 1 0 . 5 0 . 5d 1 3 eo r 3 . 6 1 ( 3 . s . f )4 x x xy x x             M1– Differentiate both sides of the equation in (ii) B1 A1 Total: 9 marks

8 Qn Suggested Solution Marking Scheme 7 (i) 32 5 pOD  ab  3 4OE  ab B1 B1 (ii) OD qOE     , where q is a constant 32 3 54 33 4 54 5 21 1 54 2 p q qq pq p        ab a b M1 – Collinear; form OD qOE     using answer in (i) M1 – Comparing coefficient A1 (iii) Shortest distance from the point E to OB    3 45 1 320 3 ()20 OBOE OB        ab b ab bb ab bb 0    3 20k  M1 – Formula M1 – Obtain the expression k ab using properties of cross product A1 – Correct k value, can be shown in the form .k ab (iv) It is the length of projection of a onto b. B1 Total: 9 marks a bˆab

9 Qn Suggested Solution Marking Scheme 8 (i)  2 1 11 xxyx x x     The equations of asymptotes: yx   and 1x B1 – yx   B1 – 1x (ii)  2 d 1d 1 y x x   At stationary point,   2 2 d 01 0d 1 1 1 where 0 y x x x x          For C to have 2 stationary points for 0x , 10 1   01   M1 – Correct differentiation based on expression in (i) B1 – Correct simplification of d 0d y x  to a quadratic equation at stationary point M1 – Use smaller root > 0 A1 – 01   (iii) G1 – Asymptotes + shape G1 – x-intercepts (condone if not written in coordinates form) (iv) G1 – Asymptotes + x-intercepts (equidistance from vert asymptote & condone if not written in coordinates form) G1– Shape (symmetrical about vert asymptote) Total: 10 marks yx  x y x y 1x 1y (1 , 0)(1 , 0) (1 , 0) O O

10 Qn Suggested Solution Marking Scheme 9(a)  4 2i4 3 121 i2434 1 i3 164 44 3 i 8e 8e , 0 , 1 , 2 8e , 0 , 1 , 2 k k z z zk zk               B1 – Correct argument for 4z M1 – Apply DM’s Thm correctly A1 – Correct answer with correct k values or listing of roots    *44 * 1 i1 3 164 i1 364 113 i 4 2 8e 2e , 0 ,1 , 2 22 k k wz zw zwk            M1 – Attempt to make use of  *41 4 z or equivalent B1 – Correct relationship between z and w A1 – Correct answer 9(b) 777 33 3 2 128 3437 pp q q   Consider   7 3 21 37 arg 3arg 7 3 33 3 13arg 4 33 pq p q                    7 3 128 64 64 3cos i sin i343 3 3 343 343 p q           M1 – Award once 7 3 2 7 is seen M1 – Award once   7 arg 3argp q is seen A1 – Correct answer 3  A1 Smallest integer value of n is 3. B1 Total: 11 marks