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 lies in the 4th quadrant. So, cos 0A while sin 0A and tan 0A . Which also means that 0p . A 1 3 p

Raffles Institution H2 Mathematics 2025 Year 5 _______________________________________________________________________________________________ ______________ Tutorial 1: Basics Page 3 of 5 5 ABCD is a parallelogram with 3 4AB   . Coordinates of points A and C are (1,1) and 8,8 respectively. (i) Find the coordinates of point D. (ii) AC and BD intersect at the point E. Find the coordinates of E. (iii) Find the length of the two si des of the parallelogram AB and BC. Deduce the geometrical relationship between A, B, C and D, and state the relationship between the two diagonals. [(i) (5,4) (ii) (4.5,4.5) (iii) 5] Solution (i) Since ABCD is a parallelogram, 3 4DC AB      . 83 5 84 4OD OC DC        Coordinates of D are (5,4). (ii) Diagonals of a parallelogram bisect each other, so point E is the midpoint of AC and BD. 81 3 . 511 81 3 . 522AE AC              13 . 5 4 . 5 13 . 5 4 . 5OE OA AE                      Alternatively,  18 4 . 511 18 4 . 522OE OA OC                     Coordinates of E are (4.5,4.5). (iii) 223 34 54AB DC        2251 4 43 541 3BC AD           Since AB=DC=BC=AD, ABCD is a rhombus, and the diagonals are perpendicular to each other. D C(8,8) A(1,1) B E

Raffles Institution H2 Mathematics 2025 Year 5 _______________________________________________________________________________________________ ______________ Tutorial 1: Basics Page 4 of 5 6 Solve the equation 23 5x using 2 different methods, namely: (i) numerically, (ii) graphically. [ 4 or 1x ] Solution (i) 23 5 23 5 o r 5 4 or 1 x x x     (ii) To solve the equation graphically, we draw the graphs of 23yx and 5y on the same diagram and find the points of intersection. From the graphs, we can see that they intersect at 2 points. 23 23 , i f 23 0 (2 3), if 2 3 0 323 , i f 2 3(2 3), if 2 yx xx xx xx xx            So, we consider  235 o r 23 5 4o r 1 xx x x       7 Given that 4x  , find the possible values of 13 x . [11, 13] Solution 44xx   When 4x , 13 13 ( 4 ) 1 1 1 1x    When 4x , 13 13 ( 4 ) 1 3 1 3x    The possible values of 13 x is 11 and 13.

Raffles Institution H2 Mathematics 2025 Year 5 _______________________________________________________________________________________________ ______________ Tutorial 1: Basics Page 5 of 5 8 Draw the graphs of 1y x and 12 2yx on the same diagram. Hence, solve the equation 112 2x x . Solution From the diagram, the 2 graphs intersect at 2 points - 1 point on the right of  1, 0 and 1 point on the left of  1, 0 . 1, if 1 01 (1 ) , i f 1 0 1, if 1 (1 ) , i f 1 xxx xx xx xx          For the point of intersection on the right of  1, 0 , 112 2 3 12 2 3 x x x x    For the point of intersection on the left of  1, 0 ,  112 2 1 32 6 x x x x     The solution to the equation 112 2x x is 2 or 63x .