2024 RI H2 Math Prelim P1 (Soln) - new

RAFFLES INSTITUTION 2024 Year 6 H2 Mathematics Prelim Exam Paper 1 Questions and Solutions with comments       ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments 1 A function f is defined by 32f( ) .xa xb xc x d  The graph of f( )yx passes through the points (3 , 4 )  and (1, 8). Given that the graph of 1 f( )y x has a turning point at 1 4(2, ) find the values of ,, a n d abc d . [4] [4] 32f( ) f ( 3) 4 27 9 3 4 ----- (1) f (1) 8 8 ----- (2) xa xb xc x d abc d abcd        Since the graph of 1 f( )y x has a turning point at 1 4(2, ) , the graph of f( )yx has a turning point at (2,4) . f( 2 ) 4  and f( 2 ) 0  .  f (2) 4 8 4 2 4 ------ (3)abc d    2f( ) 3 2 f (2) 0 12 4 0 ------ (4) xa xb x c ab c          From the GC, 1, 1, 8 and 16ab c d    . Question was well done by most students. Some students had difficulty with understanding the statement: “ 1 f( )y x has a turning point at 1 4(2, ) ” with some working out  2 f()d 0d f( ) yx x x  .

2 ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments 2 [The volume of a sphere with radius r is given by 34 3 r and the surface area of a sphere with radius r is given by 24 r .] (a) The volume of an expanding sphere is increasing at a constant rate of 315 cm s .   Show that, at any instant, the rate of increase of the surface area is 21cm s ,k r  where r is the radius of the sphere and k is a constant to be determined. [3] (b) Find the exact rate of change of surface area of the expanding sphere when the surface area is 220 cm . [2] (a) [3] Volume, 34 3Vr         and Surface area,  24A r                 2d 4d V rr  d 8d A rr  2 dd d dd d dd d ddd 1 8 5 4 10 , where 10. (shown) AA r tr t Ar V rV t r r kr              Question was well done by most students. Common error: Some students related but then treated A or V as constants. So, by writing 3or3 Ar VVA r , one should actually get d1 d d3 d VA Arrr   or 2 1dd 3 dd A VV rr r r  and not d d3 VA r  or 2 3d d A V rr . (b) [2] When 20,A 242 0r                                       5r   (since  0r  ) d 10 2 5d5 A t    21cm s .

3 ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments 3 (a) Without using a calculator, solve exactly 2 1 1.1 x x x     [4] (b) Hence solve exactly 2 1 11    xx x . [4] (a) [4] 2 1 11 x x x     2 1 101 x x x      2 1 01 1 xxx x       2 22 01 x x x    Consider 2 22 0 :xx    2 22 4 ( 1 ) ( 2 ) 12(1) 3x     So  11 33 01 x x x       011 ) 3( 31xxx       1x or 13 13 x Most students are able to manipulate the inequality and solve for the roots. However, some students are unable to identify the regions correctly and not able to get the correct final answer. (b) [4] To solve 2 1 11    xx x      2 )1 1 ( () 1       xx x   Replace x with x  from part (a) Hence, 1 x  or 313 1   x i.e. 1x or 313 1   x 1 or 1 xx  or 13 13   x   Majority of the students are able to identify the replacement correctly. However, many are not able to open up the modulus and get the correct answer. Similar Question: Tut 2 Qn 8(c)                                                                                                                                                       1 –                                                                                        1 +                                                                                                                                                                                                                 -1

4 ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments 4 (a) Find cos d.cos3 cos x xxx    [3] (b) Find 12tan dx xx . Hence find the exact value of 1 12 1 tan dx xx   . [5] (a) [3] cos cos d dcos3 cos 2cos 2 cos 1 sec 2 d2 1 ln sec 2 tan 2 , 4 where is an arbitrary constant. xx xxxx x x xx x xc c         Students are able to use the factor formula to simplify the integration to 1 sec 2 d2 x x . However, quite a number of students are not able to integrate sec 2x . (b) [5]  12 22 12 4 23 12 4 2 12 4 tan d 2tan d22 1 tan d2 1 1tan ln 1 , where is an arbitrary constant.24 xx x xx x xx x xx xx x x xx c c                     Method 1  1 112 12 01 1 21 2 4 0 tan d 2 tan d ( of symmetry) 11tan ln 1 = ln 224 2 xx x xx x xx x              [Note that 12tanyx x  is an odd function since    211 2 1 2tan tan tanx xx xx x     ]           Method 2   1 12 1 01 12 12 10 01 21 2 4 21 2 4 10 tan d tan d tan d 11 11tan ln 1 tan ln 124 24 1 ln 242 xx x xx x xx x xx x xx x                     Most students are able to use the integration by parts to solve this problem. There was quite a number of students who are unable to open up the modulus function. Similar Question: Assignment 8B Qn 4, Mock Paper 1 Qn 5(b)     12 2 22 dtan d d1 2 d2 1 vux x x ux xvx x   

5 ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments 5 (a) Using the formulae for , prove that    sin 2 1 sin 2 1 2cos 2 sinrr r     . [1] (b) Hence find a formula for 1 cos 2 , where 0 , n r r     in terms of  sin 2 1n    and sin . [3] (c) Using the formula found in part (b), show that the sum of the series 22 2 2sin 10 sin 11 sin 12 ... sin 20 , for 0          is sin(41 ) sin(19 ) 4sink     , where k is a constant to be determined. [4] (a) [1]    sin 2 1 sin 2 1 sin 2 cos cos 2 sin sin 2 cos cos 2 sin 2cos 2 sin (proven) rr rrrr r               The question indicated the use of addition formulae, so it must be used instead of factor formulae Similar Question: Assignment 6B Qn 2 (b) [3]   11 2c o s 2s i n s i n 2 1 s i n 2 1 sin 3 sin sin 5 sin 3 sin 7 sin 5 ... nn rr rr r                    sin 2 1 sin 2 3 sin 2 1 sin 2 1 sin 2 1 sin nn nn n          Most did well on this part, but do remember that cancellations must be clearly shown, and that the final two lines are in terms of n, not r, because r is the running index. Similar Question: Assignment 6B Qn 2 (c) [4]  22 2 2 20 2 10 20 10 20 9 11 sin 10 sin 11 sin 12 ... sin 20 sin 1c o s ( 2 ) 2 1120 10 1 cos(2 ) cos(2 )22 11 1 sin(41 ) sin sin(19 ) sin 2 2 2sin 2sin 11 sin(41 ) sin(19 ) , where24 s i n r r rr r r rr                              11 (shown)2k  As this is a show question, the use of the answer in part (b) must be explicitly shown, especially the subtraction of the first 9 terms from the first 20 terms. Also, students should note that the number of terms from r = 10 to 20 is eleven . 6 (a) The diagram below shows the graph of f( )yx . sin A B 1 sin(2 1) sincos 2 2sin n r nr     

6 ______________________  2024 Yr 6 H2 Math Prelim Exam Paper 1 Solution with Comments