NJC_9758_2025_Prelim_P2_Solutions

* © NJC 2025 [Turn_over 2025 NJC Preliminary Examination H2 Mathematics Paper 2 Solutions 1 Suggested Solution (i) ( ) 2 2 i3 31 2 w =− = − + = ( ) 1 1arg π tan 3 ππ 6 5π 6 w −=− =− = (ii) The point W represents w and the point B represents 2 w− . (iii) Since 2w = , the triangle OAB is an isosceles triangle. AB is parallel to the real-axis, so AOB ABO  = = . π2 6 π 12   = = ( ) πarg 2 12w− =− .

2 © NJC 2025 2 Suggested Solutions (i) ( ) ( ) 2 2 2 2 2 44 12 x a y xa y − + = − += (ii) Given ( ) 2 21 4 4xy− + = When 0x= , ( ) 2 20 1 4 4 3 2 y y − + = = Differentiate w.r.t. x, ( ) ( ) 2 21 4 4 d2 1 8 0 d d1 d4 xy yxy x yx xy − + = − + = −=− 4grad. of normal 1 y x= − Since gradient of 323 2yx=− is positive, from the diagram, the normal should have the y-intercept below the O i.e. 3 2− . For the point 30, 2  − , grad of normal 34 2 2301 − == − . Equation of line: 323 2yx=− (since y-intercept is 3 2− )

3 © NJC 2025 [Turn_over 2 Suggested Solutions (iii) Equation of normal: 323 2yx=− To find the intersection between normal and E, consider the positive square root of y. ( ) ( ) 2 2 2 1 4 4 41 4 xy xy − + = −−= By G.C., The coordinates of point of intersection are ( )0.530612,0.972069 . The y-intercept of E is 0.866025. O R

4 © NJC 2025 2 Suggested Solutions To find solid R, make x the subject of the formula, 21 4 4xy=  − Volume of solid R ( ) ( ) 2 0.972069 2 2 0.8660254 3 13π 0.530612 0.97206932 π 1 4 4 d 0.517 units (3 s.f.) yy =+  − − − = 

5 © NJC 2025 [Turn_over 3 Suggested Solutions (i) cos , 1 sin  =− t tp s cos n 2 i2   = −  t t q s cos c 2 os 0 sin sin 2 in sin cos cos 2 sin 2 c o 2 1 0 s 0 os c 2 sin tt tt t tt t t t t t pq           =      −     =  =  − − − −  −  − Method 1: Equate k -component to 0 ( ) sin 2 cos cos 2 sin 0 sin 2 cos cos 2 sin 0 sin 3 0 t t t t t t t t t − − = − + = −= For π 2π0, 33sin 3 =  =− tt . When π 3=t : ππcos cos 233 11 22 1 0 0    −    +      =     +−      =  = When 2π 3=t 2π 2πcos cos 233 11 22 1 0 0    −    +       =    − + −   =         =− Final answers: π ,13t == ; 2π ,13t = =−

6 © NJC 2025 Method 2: q is a scalar multiple of p 2 sin 2 s cos c 1 n os i         ==         − −  tt ttkkpq cos 2 c s os sin 2 in          =       −    − tk t k t t k From j-component: ( ) ( ) sin 2 sin 2cos sin sin 2cos sin sin 0 sin 2cos 0 sin 0 or 2cos 0 0 rejected 0 or cos 2 −= −= − − = − − = = − − = =   =− t k t t t k t t t k t t t k t t k kt t t From i-component: 2 2 cos 2 cos 2cos 1 cos 0 2cos cos 1 0 = − − = − − = t k t t k t t k t 2Sub cos into equation :2 2cos