NJC_9758_2024_Promo_Solutions

2024 NJC SH1 H2 Maths Promotional Examinations Suggested Solutions Question 1 (Inequality) Suggested Solution 2 3 12 x xx + +− − ( ) ( )( ) 2 22 2 2 2 2 2 32 22 23 2 2 0 1 2 1 0 0 1 0 2 x x x x x x x x x x xx xx xx x xx + + −++ − + − + +  − + + +− + +− + −+    2x− or 1x or 1x=− Question 2 (Applications of Integration) Suggested Solution ( ) ( )   ( ) 2 2 2 2 2 2 2 2 d d cos s cos d cos c i os d c sin 2 2 2 sin s n cos 2 in cos 2 sin 2cos cos os d c 2 sin os 2 xx xx x x x x x x x x x x x x x x x x x c yx x x x x xx x x x x x x − + +− + = =− − =− =− =− = +− +− + + +=      V olume required ( ) ( )( ) ( ) 0 π 2 π2 2 0 3 =πd π2 π 2 π 2 ππ cos 2 sin 11 4 y xx x xx=−  + = − − =− − 

Question 3 (Applications of Integration, Transformation) Suggested Solution (i) y ( )0,0 ( ),0k x (ii) ( ) 2 2 0 d k x x k x− 2 0 32 d k x kx x=− 3 2 3 2 4 3 4 3 4 0 2 0 2 d d 4 3 4 3 3 2 kk k kk k x kx x x kx x x x x xkk k =− − + −    =− − + −       =  Question 4 (Transformation) Suggested Solution (a) x y O ( )1,0−

(b) 2 6 45 xy x=− + ( ) ( ) 2 2 6 45 6 45 x xy xx→ −=− =−− + − + ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 2 3 9 6 3 45 18 45 2 5 2 2 2 2 2 5 2 4 5 2 1 x xxy x x x x x xy x x x =− =− =−− + − + − +→ + + +=− =− =−+→ + − − + − ( ) ( ) 2 2h 21 xx x += − Question 5 (Vectors II) Suggested Solution (i) Equation of 1l : 7 5 12 6 6 2 8 2 4 5 5 10 5 −               − − = =              −        76 6 4 , 55          =+          r R . From line 2l : ( )10 85 2 5 5 1 10 ,10 28 82 y yzx k k z x k        = +   =+  −−− = = = =+        =+          r R Solving simultaneously: 1 7 0 5 5 8 2 65 64 k            =+      ++    ++ + 1, 4, 2. k =− =− = Sub 1=− into 7 6 6 554,,x zy  = =+ +=+ : The coordinates of P are ( )1, 2,0 .

(ii) Method 1 (Projection) Let the point on 2l closest to A be F. 7 1 6 6 2 4 5 0 5 PA OA OP ⎯⎯ → ⎯⎯ → ⎯⎯ →            = − = − =                PF ⎯⎯ → 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 11 22 22 11 11 2 2 26 224 115 1 8 3 8 2 16 33 2 16 3 2 2 2 PA ⎯⎯ →                   =                         =            ==        + + + +  +  + + + 1 8 3 11 3 2 16 3 22 3 0 16 3 16 3 OF OP PF ⎯⎯ → ⎯⎯ → ⎯⎯ →            = + = + =                Method 2 (Using perpendicular directions) Let the point on 2l closest to A be F. Since F lies on 2l , 5 10 2 82 OF    ⎯⎯ → + =+ + for some . R 5 7 2 10 2 6 4 2 8 2 5 3 2 AF OF OA    + − +           + − +