NJC 9758 2025 Prelim P1 Solutions

* © NJC 2025 2025 NJC Preliminary Examination H2 Mathematics Paper 1 Solutions 1 Suggested Solution Let the exchange rate for US Dollar, Japanese Yen and Chinese Yuan be 1 to U, 1 to J and 1 to C Singapore Dollars respectively. 150 5500 1000 419.30 250 9500 2200 797.20 425 1000 2000 913.10 U J C U J C U J C + + = + + = + + = By GC, 1.2856, 0.00862, 0.17905U J C= = = For Maybelline, ( )1.2856 1200 0.17905 568.40 275 a a += =

2 © NJC 2025 2 Suggested Solutions (i) ( )1 i 2 2 4izw+ + =− + 6 2 8 4i 3 4 2izw zw− −= + + = Solving simultaneously by elimination: ( ) ( ) i 6 1 i 2 4i 8 4i 7 i 6 8i 6 8i 7 i 7i 42 6i 56i 8 5 7 0 50 i 50i 50 1 z z z  − + + =− + + + += = + +−= +− ++ += =+ ( ) 3 4 2i =3 1 i 4 2i = 1 i w z= − − + − − −+ (ii) 22 1i 1i 1 i 1 i 1 i i 1 1 i 1i 1i 1 w z − −+= −+= + +− − + + + += = iwz= Alternative ( ) 2 i 1i ii 1i 1i i i 1 1i w z −+ + = + + += + = = iwz=

3 © NJC 2025 Rotating line segment OZ by π 2 anti-clockwise about the origin will give OW. Or line segment OW is a anti-clockwise rotation about the origin by π 2 from line segment OZ.

4 © NJC 2025 3 Suggested Solutions (i) x y =−y x a =− − by xa ( )0,a 0, b a   ( ),0a (ii) ( ) ( ) 2 − =− −− −= b xaxa x a b =−x a b or =+x a b (rejected since xa ) −x a b or xa

5 © NJC 2025 4 Suggested Solutions 221y q x= − − , 1q ( ) ( ) 22 2 2 22 1 1 y q x x y q − = − + − = ( )fyx= is a semi-circle. (i) (ii) 221y q x= − − Replace y with 1y+ : 2 2 2 211y q x y q x + = − − =− − Replace x with qx : ( ) 2 2 2 1 y q qx y q x =− − =− − Replace y with qy : 2 2 1 1 qy q x yx =− − =− − Translate 1 unit in the negative y direction. Scale by a factor of 1 q parallel to the x-axis. Scale by a factor of 1 q parallel to the y-axis. ( ),1q− 2 1xq=− − 2 1xq=− ( ),1q

6 © NJC 2025 5 Suggested Solutions (i) Given vertical asymptote at 2x= , 2p=− 228 2 x kxy x ++= − ( )( ) ( ) ( ) 2 2 2 4 2 8d d 2 x x k x kxy x x − + − + + = − Given 4 d 0d x y x =− = , ( )( ) ( )( ) ( ) ( )( ) 2 4 2 4 4 2 4 4 8 0kk− − − + − − + − + = ( )( ) ( )6 16 32 4 8 0kk− − + − − + = 28k= By long division or otherwise, 22 28 8 72 2 3222 xxyx xx ++ = = + +−− Equation of oblique asymptote of C is 2 32yx = + (ii) y x (8, 60) (-4, 12)

7 © NJC 2025 6 Suggested Solutions (a) (i) 1 1S r  = − and ( ) 3 3 1 1 r S r − = − , 1r  Method 1 ( ) ( ) 23 23 23 36 63 52 11 11 1 11 , since 011 11 1 1 2 20 2 1 0, since 0 r rr r rr rr r r r r r r r r r  −=−−  − = −− − = − − = − + − + = − + =  ( )( ) 4321 1 0r r r r r− + + − − = Since 1r , 432 10r r r r+ + − − = Method 2 ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 54 4 5 2 2 2 242 4 2 2 2 2 3 2 234 432 1 0 11 1 1 1 1 1 1 0 1 2 1 1 1 0 1 1 1 2 1 1 0 1 2 2 1 0 0 1 0, since rrr r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r r = + +− = + + − + + − − =  + + + + − − = − + + − + − − = − + − + − + − − = + − − − = + + − = − (a) (ii) By G.C., since 1r  0.661 (3 s.f.)r=− or 0.848 (3 s.f.)r=

8 © NJC 2025 (b) Sum of first 4n terms ( ) ( )( ) 2 4 2 7 4 12 28 8 2 n nd n dn dn = + − = + − Terms to be removed, 7 3 ,7 7 ,7 11d d d+ + + , … , (First term = 73 d+ ; common difference = 4d ; No. of terms removed = n terms. Sum of removed terms ( ) ( )( ) ( ) ( ) 2 2 7 3 1 42 14 6 4 42 14 2 42 72 n d n d n d dn d n d dn n nd dn = + + − = + + − = + + = + + Sum of remaining terms 22 2 28 8 2 7 2 6 3 21 n dn dn n nd dn dn dn n = + − − − − = − + Method 2 Sum every three consecutive terms as a single term i.e. ( ) ( )7 7 7 2 dd+ + + + , ( ) ( ) ( )7 4 7 5 7 6d d d+ + + + + , … 21 3 d + , 21 15 d+ , 21 27 d+ In the progression above, first term is 21 3 d+ , common difference is 12d , and there are n terms. Sum ( ) ( )( ) ( ) 2 2 21 3 1 122 21 3 1 6 6 3 21 n d n d n dn n n d dn dn n = + + − = + + − = − +

9 © NJC 2025 7 Suggested Solutions (i) 4e= tx , 2=yt , 0.t  Sub ln 4 xt = into 2=yt : ( ) 2 2 ln 4 1 ln16 xy yx = =  (ii) ( )d1 lnd8 y xxx= Steepest gradient: 2 2 d d y x ( ) ( ) ( ) 1 2 2 dd dd d1 lnd8 1d ln8d 11 ln8 y xx xxx xxx xx x − − = = =  = − +  2 2 d 0d y x = ( ) ( ) 2 2 2 1 1 1 ln 08 1 ln 1 0 xxx xx  − + =   − + =  ( )ln 1 0x− + = ex= o1s rince 0 f xx     R 1e,16P 

10 © NJC 2025 (iii) Point 1e,16   is the point with the greatest gradient. When 0,=t ( )40 e1==x , 0=y Point ( )1,0 is the initial point. For 0t , 1,x ( )d1 ln 0d8 y xxx= . 1e,16P  ( )1,0 y x