TJC 9758 2024 Promo Solutions

1 Solution 1 Note that ( ) 2 2 7425− − =x y ( ) 22 22 7 110 2 yx − − = 22 1xy−=  replace x by 10 x 2 2 2 110 x y−=  replace y by 2 y 22 22 110 2 xy −=  replace y by 7y− ( ) 22 22 7 110 2 yx −−= The sequence of transformation are 1. Scaling parallel to x-axis by scale factor 10. 2. Scaling parallel to y-axis by scale factor 2. 3. Translation in the positive y-direction by 7 units Alternatively, 22 1xy−=  replace x by 10 x 2 2 2 110 x y−=  replace y by 7 2y− 22 2 7 110 2 x y− − =  replace y by 2 y ( ) 2222 2 2 2 77 1 110 2 2 10 2 yx y x −− − =  − = The sequence of transformation are 1. Scaling parallel to x-axis by scale factor 10. 2. Translation in the positive y-direction by 7 2 units. 3. Scaling parallel to y-axis by scale factor 2.

2 2 Let x, y and z be the number of plates of chicken rice, duck rice and pork rice sold. Total profit = $167.40 50 80 110 16740 5 8 11 1674 (1) x y z x y z  + + =  + + = Mr Chan earned $28.80 more in profit from the sale of pork rice than the chicken rice 110 50 2880 5 11 288 (2) zx xz  − =  − + = Total number of plates of rice sold being 3 times the plates of duck rice sold 3 2 0 (3) x y z y x y z  + + =  − + = Solving (1), (2), (3), we have 81, 72, 63x y z= = = Mr Chan sold 81 plates of chicken rice, 72 plates of duck rice and 63 plates of pork rice that day.

3 3a 3b 1 f ( )= −y x y =  x = −5 (−1,0) x y x y x = 4 ° (5,0) )

4 4a 2 22 22 4 2 1 12 5 2 5 2 2 11 2 5 2 5 xx x x x x x x x x x +− =+− + − + −= + + − + − + 2 2 4 d25 x xxx + −+ ( ) 22 2 2 2 11d 25 12 x xxx x −= + + −+ −+ 21 11ln 2 5 tan 22 xx x x C − −= + − + + +  4b ( )( )( ) 2 2 d sec 3 2 sec3 sec3 tan 3 3d 6 tan 3 sec 3 x x x xx xx = = 2tan3 sec 3 dx x x x ( ) 21 6tan3 sec 3 d6 x x x x=  ( ) 221 sec 3 1 sec 3 d6 x x x x=−   211 sec 3 tan 363 x x x C= − +

5 5(ai) ( ) ( )( ) ( ) ( ) ( ) ( )( ) 2 2 d 1 sinlnd 1 sin d d 1 sin 1 sin s 1 n co s ln 1 cos 1 sin cos 1 sin 1 sin 1 i 2cos 2cos 2se n sco o c s si i in c l 1 n s ns x xx x xx x x x x xx x x x x x x x xx +  − = =−+− − + += +− − − = − − + = = 5(aii) ( ) ( ) ( ) ( ) ( ) 1 2 3 2 1 d 1 1 1 1 1tand 1 1 1 1 1 1 1 2 1 2 2 1 d d 1 x x x x x xx x x x −     − = + − =− −+  − = −  − − − − 5(b) 3xyx= . Taking logarithm on both sides : 3ln ln xyx= , ln 3 lny x x= . Differentiating implicitly w.r.t. x on both sides : ( ) d11 d (3)ln 3y y x x xx=+ d1 d 3ln 3y yx x=+ , d d 3 (3ln 3) , 3 (ln 1). y x x yx xx =+ =+

6 6(a) 6(b) We need to add the graph of 5ln( 3) 1yx= + − The intersection points are ( 0.51585,3.5496)− and (9.5564,11.651) . From graph, 1 0.516x−  − (3 s.f.) or 9.56x (3 s.f.) 6(c) Substituting x with sinx− , we get 1 sin 0.516x− − − or sin 9.56x− 0.516 sin 1x or sin 9.56x− (N.A. since 1 sin 1x−   ) 0.542 2x 

7 7(ai) 11 22+ =  = −a b c a c b 1 2 1 2  = −  =  −  = a c c b c c c b c cb since =c c 0 Alternatively, Crossing both sides with c, we have 1() 2+  = a b c c c a c b c 0 +  = a c b c =−  a c c b =  7(aii) Dot both sides with c, we have 1() 2+  = a b c c c 2 21 1 1 (1)2 2 2 +  = = =a c b c c 10 cos 32bc += (since ac⊥ ) 11(1) 22  = b 1b = b is also a unit vector.

8 7(b) Since P lies on line segment DE, by Ratio Theorem, let 5 3 8 3 (1 ) 0 (1 ) 4 4 4 1 3 4 3 OP OE OD        −−           = + − = + − = − +           −−      Alternatively, 5 3 8 0 4 4 1 3 4 DE OE OD −           = − = − = −          −      Equation of line DE is 38 44 34  −       = + −      −    r ,  Since P is a point on line DE, 38 44 34 OP  −       = + −      −    for some  8 3 2 8 5 4 4 4 4 4 3 1 4 4 FP OP OF    −−           = − = − + − = −          −−      2 2 23 (8 5) ( 4 ) (4 4) 3FP   =  − + − + − = 2 2 264 80 25 16 16 32 16 9     − + + + − + = 296 112 32 0 − + = 26 7 2 0 − + = (3 2)(2 1) 0 − − = 12 23 or  = = Therefore, ( ) ( ) ( ) 1 2 1 2 1 2 8 3 1 4 4 2 4 3 1 OP  −  = − + =  −−  or ( ) ( ) ( ) 2 3 2 3 2 3 8 3 7 14 4 4 34 3 1  −  − + =  −− 

9 8 (ai) 1f( ) 1 cosxx −=− 2 1f ( ) 1 x x  = − [Refer to MF27] For 01 x , f ( ) 0x  meaning f is strictly increasing in the interval 01 x and hence is a 1-1 function. Let 1f1 2 k− −= which is equivalent to solving f ( ) 1 2k =− where 01 k 1 1 1 cos 1 2 cos 2 cos 02 k k k    − − − = − =  = = Alternative method to find 1f1 2 − − : Let 11 cosyx −=− 1cos 1 cos(1 ) xy xy − =− =− 1f ( ) cos(1 )xx− =− 1f 1 cos 1 122 cos 2 0   −    − = − −         = = 8(aii) x y (1, 1)

10 8(bi) 2h( ) ( 1)xx = − − Note that the maximum value of h is  when x = 1. y  hR ( , ] = − y = h(x) x