NYJC_H2_Maths_P2_Solution

1 NYJC 2014 Preliminary Examination 9740/02 NANYANG JUNIOR COLLEGE JC2 PRELIMINARY EXAMINATION SOLUTIONS Higher 2 MATHEMATICS 9740/02 Paper 2 17th September 2014 3 Hours Additional Materials: Cover Sheet Answer Paper List of Formulae (MF15) READ THESE INSTRUCTIONS FIRST Write your name and class on all the work you hand in. Write in dark blue or black pen on both sides of the paper. You may use a soft pencil for any diagrams or graphs. Do not use staples, paper clips, highlighters, glue or correction fluid. Answer all the questions. Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place in the case of angles in degrees, unless a different level of accuracy is specified in the question. You are expected to use a graphic calculator. Unsupported answers from a graphic calculator are allowed unless a question specifically states otherwise. Where unsupported answers from a graphic calculator are not allowed in a question, you are required to present the mathematical steps using mathematical notations and not calculator commands. You are reminded of the need for clear presentation in your answers. At the end of the examination, fasten all your work securely together. The number of marks is given in brackets [ ] at the end of each question or part question.

1 By definition, nl m nl and nm . Therefore, .0ln . Let the foot of perpendicular from C to plane OAB be C’. //CC' ab    CC' t ab  for some scalar t () OC' OC CC' t    ab ca b      Therefore, there is a scalar such that ()t     ab c a b Multiplying both sides by a, () ( )t     aa b a c a b 0    ac aa ba (using result from first part) ()()   aa ab ac  C b O   B  A  C’ a c

2 Since hgR 0, , D  hgRD , gh exists.    ln( 2) gh g ln 2 e gh : 2, , 3 x xx xx x x      ghR1 , fg hx ax a    22xaaxa xa          22 0 21 0 xxa xa xa x       1 or 2x xa  Since ghD3 , , 3xa  3x a , hence range of values of x is  3, a  . Alternative solution:  fg hx x   22x ax x         22 0 21 0 xax x xx a        1 or 2xa x  Replacing by x xa , 1+a or 2xa xa  1 or 2x xa  Since ghD3 , , hence range of values of x is  3, . Range of values of x for (ii) is 3, a  . Maximum point  2 24 ,2 2 a a      1 fy x x O y

3 ln 1 1 xy x   1l n ( 1 ) 21 xy x   2(1 ) ln(1 )x yx dy 12(1 ) 2 dx 1xy x   2 22 dy d y 12(1 ) 4 dx dx (1 )x x    32 32 3 dy dy 22(1 ) 6 dx dx (1 )x x   (i) When 0x , 0y , 1 2 dy dx  , 2 2 3 2 dy dx  , 3 3 11 2 dy dx  23ln 1 1 3 11 0. . .12 2 2 6 2 xx x xx            