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 Mabel saves in 5 year $1278 5 $6390=  = For Janice: nth mth Amount at start of month ($) Amount at end of month ($) 1 Jan26 100 ( )100 1.003 2 ( ) ( ) 100 100 1.003 100 1 1.003 + =+ ( )( ) ( ) 2 100 1 1.003 1.003 100 1.003 1.003 + =+ n ( ) ( )( ) 2100 1.003 1.003 1.003 100 1.003 1.003 1 1.003 1 n n + + + − = − At Dec 2030, n =60. Therefore, Amount Janice saves in 5 years ( )( ) 60100 1.003 1.003 1 $6582.85 (2 d.p.)1.003 1 − == − [A1] Janice will have more money in her savings account than Mabel has in her piggy bank.

7 9758/01/TJC/25 7(b) At end of 4th year, Mabel will have $1278 x 4 = $5112 Janice will have ( )( ) 48100 1.003 1.003 1 $5169.97 > $51121.003 1 − =− At end of 3rd year, Mabel will have $1278 x 3 = $3834 Janice will have ( )( ) 36100 1.003 1.003 1 $3806.97 < $38341.003 1 − =− Therefore, the year when Janice’s saving is more than Mabel’s saving happen in 4th year (i.e. 2029). Let it be the kth month. Then set Janice’s saving > Mabel’s saving ( )( ) ( ) ( ) 36100 1.003 1.003 1 3834 2 100 1 (1)1.003 1 2 k k k + −   + + − − ( )   36100003 1.003 1 3834 199 032 k k k+ − − − +  Using GC, we have 4k  . Therefore, the month and year is Apr 2029.

8 9758/01/TJC/25 8(a) 2 2 4izw+ = − --- (1) i 2izw−= --- (2) From (1) : 2 4i 2 wz −−= Substitute into (2): 2 4ii 2i 2 2i 4 i 2 4i w w ww  − −  −=  + − − = Let w = a +ib, ,,ab  22 22 2i 4 i 2( i ) 4i 4 2 i (2 2 ) 4i a b a b a a b b + − + − + = − + − + − = Compare real and imaginary parts: ( ) 22 2 2 2 22 2 4 2 0 2 --- (3) 2 2 4 --- (4) Substitute (3) into (4): 2 2 4 --- (5) 2 2 4 4 8 4 4 (3 8) 0 0 (rejected since 4 2 from (5)) 8 or 3 aa a b b bb bb b b b bb bb b − =  = − + − = − − = +  − − = + + + = + +=  = +  − =− 8 2 i 3 2 4i ( 2 2 ) 8 2 2 2i 2i 2 3 3 w bz  = − − − − − = = − + − =− −  8(b) •A (2, -1) •C (-3, 2) Im(z) Re(z) O

9 9758/01/TJC/25 2 2 represents the complex number 2 i and represents the complex number 3 2i . BA z BC z −− − + − Since ,2ABC = A is obtained by rotating C through in anti-clockwise direction about .90 B 22i( 3 2i ) 2 izz− + − = − − ( ) 22 2 2 3i 2 i 2 i 1 i 4 2i 4 2i 1 i 2 6i 1 3i1 i 1 i 2 zz z z - - - = - - - = + + + += ´ = = +-+ 4 4 1 3i (2 i) 3 2i 3 2i+1 4i 2 2i AB DC z z = + − − =− + − =− + − =− − ( )3,2C - ( )2Bz Im Re ( )4Dz O ( )2, 1A -

10 9758/01/TJC/25 9(a) 9(b) 2 222 22 22 1 d 1 1 1 d1d d11 d 1 1 1d xtxt ytt t t t xtytyt t t t t −= +  = − =  + = −+ = −  = + =  Tangent parallel to y-axis  d d y x is undefined 2 1 0 1tt − =  = (c) At 2t = , d5 d3 y x = , 5 2x= , 3 2y= Equation of normal at 2t = is 3 3 5 2 5 2yx − =− −  or 3 35yx=− + (d) 11,x t y t tt= + = − ----- (1) 3 35yx=− + ----- (2) Sub (1) into (2): 22 2 1 3 1 35 3313 55 82 3055 1 or 2 (this is already used) 8 tt tt t t t tt tt − =− + +  − =− + −  − − =  =− = So coordinates of Q are 65 63,88 − (-2,0) (2,0)