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 cos 1 0−= −− =kt t k t 2 22 1 2 22 10 122    − − =       + − = − = k kk k kk Since  = k , 1 or 1.=− = When 1 = =k , 1 2πcos 23 kt − = − = When 1 = =−k , 1 πcos 23 kt − = − =

7 © NJC 2025 [Turn_over (ii) ( ) s sin sin 2 cos cos 2 sin 2 0.5 cos 2 0.5 cos3 0.5 cos cos 2 1 0.5 in tt t t t tt t t t t pq        =      −    =−  − − = + − =− cos3 0.5 1 cos3 0.5 2 2 t t  −= =− pq p Greatest length 1 1 0.5 2 = − − 2 3 2 = or 1.06 (3 s.f.)

8 © NJC 2025 4 Suggested Solution (i) d d PP khT =−  293 6.5Th=− d d 293 6.5 P kP hh −= − , where 0k  or d d 293 6.5 P kP hh= − , where 0k  (ii) 1d d 293 6.5 1 dd 293 6.5 Pk P h h kPhPh =− − =− −   Since 0P , ln 293 6.56.5 ln 293 6.5 6.5 6.5 ln ln 293 6.56.5 e e , where e 293 6.5 293 6.5 k hC k h C k b kP h C P AA Ah Ah −+ − = − + = ==  =− =− Since 0,k 06.5 kb= . (iii) When 0h= , 101300P= , ( ) ( )101300 293 ... 1bA= When 2h= , 80000P= , ( ) ( ) ( ) 80000 293 6.5 2 80000 280 ... 2 b b A A = −  = (1) / (2), 101300 80000 293 280 101300 293 80000 280 ln ln 5.2015 (5 s.f.) b b = = =

9 © NJC 2025 [Turn_over 5.2015 8 101300 293 1.4935 10 A − = = 5.201581.4935 10 293 6.5Ph −=  − When 8.848h= , ( ) 5.201581.4935 10 293 6.5 8.848 32500 (to 3 s.f.) P −=  − = The atmospheric pressure is 32 500 Pascals. (iv) Either The model suggests that when altitude is greater than 45.1 km, atompheric pressure increases (to infinity) as altitude increases, which is not possible. Or As the temperature T measured in Kelvin can only be positive, the model is invalid if 293 6.5 0Th= −  . 101300 O 45.1

10 © NJC 2025 5 Solutions ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 12 12 11 P P1 12 1 12 11 12! ! 12 ! 12! 11 ! 13 ! 1 1 ! 12 ! 1 11 ! 13 12 ! 13 Shown1 kk kk Yk Yk ppk ppk pkk pkk pk k k pk k k kp kp − −+− = =−  −= − − −= −−− −−= −− − − −= − Given mode is 4, ( ) ( ) ( ) ( ) ( ) P 4 P 3 P4 1P3 9 >141 9 4 4 4 13 YY Y Y p p pp p =  = = = − −  and ( ) ( ) ( ) ( ) ( ) P 4 P 5 P5 1P4 8 <151 8 5 5 5 13 YY Y Y p p pp p =  = = = − −  Hence 45 .13 13p