RI C1 Basics Tut Sect A (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2025 Year 5 _______________ Tutorial 1: Basics Page 1 of 5 Tutorial 1: Basics Section A (Basic Questions) 1 Find the range of values of k for the line 2yx k to intersect the curve 2228yx at two distinct points. [  ] Solution To find the intersection between the line and the curve, we substitute 2yx k into 2228 ,yx  2 2 22 22 8 888 0 xk x xk x k     For line to intersect curve at 2 distinct points, the above equation must have 2 distinct roots  discriminant = 2264 32( 1) 0kk   2 10k  Since 2 0 for all real ,kk 2 1 1 0 for all real kk . Required range of values of k is  . 2 Sketch the graph of ln 2yx for 2x , showing clearly the equations of any asymptotes and the coordinates of any intersec tions with the axes. By drawing a suitable straight line graph on the same diagram, dete rmine the number of solutions to the equation 21e2x x  . [2] Solution 2x (0, ln2) (-0.5,0) (0, 1) y = 2x + 1 That is to say, any (real) values of k 2 10k will satisfy the inequality . Observations  2x is a vertical asymptote for ln 2yx and should be clearly drawn with a dotted line.  The question asked for the coordinates of the intercepts  1l n 2 . So, y-intercept for the straight line is above the y-intercept for the log-graph.

Raffles Institution H2 Mathematics 2025 Year 5 _______________________________________________________________________________________________ ______________ Tutorial 1: Basics Page 2 of 5  21 21 e2 e2 21 l n 2 Draw 2 1 x x x x xx yx        Since there are 2 points of intersection between the two graphs, there are 2 solutions to the equation 21e2x x  . 3 Express 22 s i n 22 c o s in the form sin( )R  , where 0R and 0. 2  [4sin ] 4   Solution  22 2 2 2 sin 2 2 cos sin( ) sin cos cos sin Comparing, we have cos 2 2 ---------(1) sin 2 2 --------- 2 (1) (2) 8 8 RRR R R R                  4 (2) tan 1 (1) 4 Thus 2 2 sin 2 2 cos 4sin . 4 R          4 Given that cos Ap and 270 360A , express each of the following in terms of p. (a) sin A , (b) sin 2A , (c) cos .2 A [(a) 21 p (b) 221pp (c) 1 2 p ] Solution (a) 2sin 1Ap  2 sin 2 2sin cos 21 A AA p p    (b) 2 cos 1 cos 2cos 1 cos 22 2 AA AA    (c) But 135 1802 A (i.e. 2 A lies in the 2nd quadrant) cos 1 1cos .22 2 AA p     Observation  A