2024 RI H2 Math Prelim P2 (Soln) - new

RAFFLES INSTITUTION 2024 Year 6 H2 Mathematics Prelim Exam Paper 2 Questions and Solutions with comments ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments 1 The function f is defined by 2f: 2 xx x , for x  , 2x . (a) Sketch the graph of f and find its range. [3] Another function g is defined by g: 3 2xx  , for x  . (b) Show that the composite function fg exists. Find fg( )x and state the domain and range of fg. [5] (a) [3] Range of f = (, ) \ { 2 }  Note that the curve passes through origin. Other possible notations for range of f:  (, ) \ { 2 }   \{ 2 } (b) [5] gR[ 3 , ) fD(, 2 ) ( 2 , )    Since fRDg  , function fg exists. 2(3 2 ) 6 2 2fg( ) f (3 2 ) 32 2 12 xxxx xx        fgD   Note that 2(3)f( 3 ) 632  . fg gDD  gR [3, ) fgR (2,6] Hence, fgR( 2 , 6 ] . Give domain and range in set notations. Some students are confused about the domain and range of a composite function. x y 2x 2y O g f

2 ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments 2 The function f is defined by 432f( ) 4 5 ,zzA zB zC z   where A, B and C are real numbers. Given that 2i is a root of f( ) 0z  and 2()zk is a factor of f( ) ,z where k is a positive real number, find the values of A, B, C and k . [5] [5] Since all coefficients of f( )z are real and 2i is a root of f( ) 0z  , then 2i is also a root of the equation. The quadratic factor of the equation is       22i 2i 4 5zz z z  Then,    2432 2 22 45 ( 4 5) 2( 4 5 ) zA zB zC z z kz z zk z kz z          Comparing constants, 245 5 3kk   Since 0k  , therefore, 3k  . So,   432 2 43 2 45 6 9 ( 4 5) 10 38 66 45 zA zB zC z zz z z zzzz            10, 38, 66AB C   This question is generally well done. Some students tried to substitute the root in, then compare the real and imaginary parts to obtain 2 equations. However, 2 equations are not enough to solve for 3 unknowns, without using the condition that 2()zk is also a factor of f( )z .

3 ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments 3 (a) The points A , B and C on the plane  have position vectors a , b and c respectively. Show that a vector perpendicular to  is parallel to bccaab . [3] (b) p and q are non-zero vectors and  pp q q . (i) Find the relationship between p and q . [1] (ii) Find q . [1] (c) u is the position vector of a fixed point U relative to a fixed origin O. A variable point V has position vector v relative to O. Given that  0vvu , describe geometrically the set of all possible positions of the point V. [2] (a) [3]    , since AB BC           ba cb bcbbacab bc ca ab bb 0 bccaab     Since and AB BC     are parallel to  but not parallel to each other, so AB BC     is a vector perpendicular to both and ABB C    and hence perpendicular to  . Therefore a vector perpendicular to  is parallel to bccaab (Shown). Some students did not show working clearly for a “show” question. Some students started from the given expression which is to “verify” not to “show” Some students find the cross product of two position vectors which are not parallel to the plane. (b)(i) [1]   / / since is a scalar.pp q qp q p q Hence p is parallel to q. Note that parallel vectors may not be in the same direction, it can be opposite directions, since pq could be negative. (b)(ii) [1]     2 1 1       p p qq p p qq pp q q pqq q q   1q is not accepted since the magnitude of a vector is positive. (c) [2]    0 OV UV     vvu v vu     Since V is a variable point, so the set of all possible positions of the point V forms a sphere with OU as the diameter. It would be a circle if the vectors are 2-dimensional, so it is also accepted. U V O

4 ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments 4 (a) Given that 12e xy  , show that  2 2d12 d yx yx   and 2 2 dd(1 2 ) dd yyx yxx  . [3] (b) By further differentiation, obt ain the series expansion for y in terms of x up to and including the term in 3x . [3] (c) Verify that the same series expansion for y in part (b) is obtained if the standard series expansions for ex and (1 ) nx are used. [4] (a) [3] 12e xy  Method 1 12 1 2 e ln (1 2 ) xy yx   Differentiating wrt x, 11 22 1 2 2 2 1d 1 (1 2 ) (2) (1 2 )d2 d(1 2 ) d d(1 2 ) (shown)d y x xyx yxy x yxy x        Method 2 12 12 12 2 2 e d1 (2)ed 21 2 d12 e d d(1 2 ) (shown)d x x x y y x x yx x yxy x          Differentiating again wrt x, 22 2 2 2 dd d d(1 2 )(2) 2 2dd d d dd d(1 2 ) since 0 (shown)dd d yy y yxy xx x x yy yxy xx x         Most students have done well for the first part of the question. Students must take note that  212 12ee x x  . Recall that () mn m naa  Unfortunately, many students did not realise that they could attempt this part using implicit differentiation. They attempted by finding d d y x and 2 2 d d y x in terms of x first, resulting in a need for tedious simplification. (b) [3] Differentiating again wrt x, 32 2 32 2 32 32 dd d d(1 2 ) 2 ddd d dd d(1 2 ) 3 ddd yy y yx xxx x yy yx xxx      Generally, the students are able to complete part (b) well.

5 ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments When 0,x e,y  d e,d y x  2 2 d 0,d y x  3 3 d e.d y x  23 3 ee ( 0 ) e e 12! 3! 6 xx xyx x           (c) [4]  231 2 23 11 1 ( 2 ) 1 1 3 ( 2 )12 1 ( 2 ) . . . 22 2 2 2 2 2 3 ! 111. . . 22 xxxx xx x                   Method 1   23 23 111 ...12 22 11 ...1 22 23 2 32 23 2 3 3 3 ee ee 11e 1 ... 22 11 1 ... ... ...2! 2 3! 111 1e 1 ... 222 6 1e 1 ... (same results as 6 xx xx xx x xx x xx x xx x x x x xx                             in part ) (a) Method 2 23 23 111 ...12 22 11 1 22 23 2 3 22 333 3 ee eee e 11 1e1 1 2! 3! 2 11 2 1111 1e 1 ... 2! 2 3! 2 2 1e 1 ... (same results as in part )6 xx xx xxx xx x x x xx x x x x xx                                   (a)    Most of the students are able to expand  1 212 x correctly Many students have tried to expand 23111. . .12 22ee xx xx     as 23 2 23 3 23 1111 . . . 22 11 11. . .2! 2 2 11 11 ...3! 2 2 ... xx x xx x xx x          They did not realise that every subsequent term of this expansion will need to be taken in to account in order to obtain the correct constant term and the coefficients of 2,x x and 3x terms.

6 ______________________ 2024 Yr 6 H2 Math Prelim Exam Paper 2 Solution with Comments 5 The diagram shows the curve C with parametric equations  2 1xt ,  2 3yt . The curve C meets the axes at 16, 0 and 0, 16 . (a) Show that the line 16x meets C at the point P where 5t  . [1] The normal to C at P is denoted by l . (b) Find the cartesian equation of l . [3] (c) The line l meets C again at the point Q where xb . Show that the area of the region bounded by l , the lines 16x , xb and th