TJC_9758_2025_Prelim_P1_Solutions

1 9758/01/TJC/25 2025 TJC Prelim Exam H2 Math Paper 1 Solution 1(a) ( ) ( ) 23 2 1311 2 9 9 9 2 93 n n n n n n av v u u a − − −− =  =  = (constant) for all positive integers n. Therefore  nv is a geometric sequence. 1(b) 22 9 2 3 32v u a a     =  =          2 42aa =  = (rej 2− since a is positive) 2 Differentiate 31 ( )xy x y= + − w.r.t. x, we have 2dd 3( ) 1dd yyx x yxx = − −  At 0y= , 1x=− Gradient of tangent at A ( 1,0)− is d d d 331d d d 2 y y y x x x − = −  = 3(a) 2 2 22 ( ) aay bx x b x b== − − − (Completing Square)  replace x by xb+ 22 ay bx= − Translation of b units in the negative direction of the x-axis. 3(b) x y y = 0 x = 0 x = 4

2 9758/01/TJC/25 4(a) Using Sine Rule, 1 π πsin sin3 6 BC  =  + 1 π π πsin sin cos cos sin3 6 6 BC  = + 1 3 1 3 cos sin2 2 2 BC  = + 3 cos 3 sin BC  = + 4(b) 2 12 222 2 2 2 3 (since is small) 13 2 3 1 3 2 3 1 3 3 22 3 1 3 3 2 73 1 3 2 BC        −  −+  = + −       − − + −           − − +  = − + C A B 1

3 9758/01/TJC/25 5(a) 2 2 25 223 x xx − −− ( ) 2 2 2 25 2 2 3 023 x x x xx − − − − −− 2 2 02 6 19 23 x x x x − −− + − ( ) 23 29 22 2 2 023 x xx − − − −− --- (1) Since the numerator ( ) 23 29 2220x− − −  for all real values of x, 2 2 3 0xx− −  . ( )( )3 1 0xx− +  1x− or 3x 5(b) Replace x by 1x+ , from (a), 11x+ − or 13x+ 11x+ − has no solution 13x+ => 13x+ or 13x+ − 2x or 4x− −1 3

4 9758/01/TJC/25 6(a) Using ratio theorem, 3 4OX += ab 6(b) 23cos 62a b a b   = =  6(c) ( )OY OX= bb ( )1 34 = bba + b bb ( )2 1 34  =     b a b + b b b 2 2 1 3 cos46  =     b a b + b b 22 2 13 342 b +  =     2 3 3 8 += b (Shown) Alternative solution: 3 3 1 4 4 4OY XY   + =  = − =− + −  abb b a b 0XY OB•= 31 044  − + − • =  a b b 31 044 − • + − • = a b b b 231 cos 04 6 4  − + − =  a b b 223 3 1 04 2 4  − + − =   bb 1 3 3 2 3 3 4 8 8 +− =  = ( )2 3 3 , Shown8OY += b

5 9758/01/TJC/25 6(d) Area of 1 2OXY OX OY =  ( )2 3 313 2 4 8  ++ =    ab b ( ) ( ) 2 3 3 364 + = +  a b b ( )2 3 3 364 + =  +  a b b b ( )3 2 3 3 where 64 + =   = a b b b 0 ( )3 2 3 3 sin64 6 + =  ab ( ) 23 2 3 3 1 64 2 + =  b ( )radius of circle= = =ab ( ) 2 3 2 3 3 128  + = units2

6 9758/01/TJC/25 7(a) For Mabel: Amount of money Mabel saves in a year = $(101 + 102 + … +112) ( )12$ 101 112 $12782= + = Amount of money Mabe