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    + − +           + − +          ++     = − =  = Since 2,AF l⊥ we have 1 2 0. 2 AF  =

21 4 2 2 0 3 2 2 2 8 4 6 4 0 9 12 4 3         −+       +  =      +    − + + + + + = =− =− 4 115 3 3 4 2210 2 33 16482 33 OF ⎯⎯ →  −    = + − =     +−    (iii) Let point A’ be the point which is the reflection of A in 2l . 2 PA PAPF ⎯⎯ → ⎯⎯ → ⎯⎯ → += 8 3 6 16 3 4 2 16 3 5 PA ⎯⎯ →         = +             8 3 6 2 3 2 12 16 3 4 20 3 20 316 3 5 17 3 17 PA ⎯⎯ → −−               =  − = =                      Equation of reflected line: 12 2 20 , 0 17  −       = +           r R .

Question 6 (Curve Sketching, Applications of Differentiation) Suggested Solution (i) 2 2 2 3 2 2 0x y xy y x+ + − − = 22 d d d2 2 3 3 4 1 0d d d y y yxy y x y x yx x x+ + + + − = 2 2 d 2 1 3 d 2 3 4 y xy y x x y x y − + −= ++ (ii) When 0, 1xy== or 1.y=− At ( )0,1 : d1 d2 y x =− At ( )0, 1− : d 1d y x =− (iii) Let .xk= ( ) ( ) ( ) ( ) ( )( ) 22 22 2 32 3 2 2 2 3 2 0 9 4 2 2 0 9 4 8 8 16 0 4 17 8 16 0 4 4 4 0 k y ky k k k k k k k k k k k k k k + + − + = − + − +  + + + +  + + +  + + +  22 22 2 2 1 1 14 0 4 4 4 4 8 8 1144 8 16 1 634 8 16 k k k k k k     + + = + + − +        = + − + =+ + for all real x. or Discriminant of 24 4 0kk+ + = is ( ) 21 4 4 4 63 0− =−  Since coefficient of 2 40k = , 24 4 0kk+ +  for all real x. Since 24 4 0kk+ +  for all real x, we need 4 0.k+ We have 4.k− The curve does not intersect the vertical line xk= when 4k− , thus has no parts where 4x− .

Question 7 (Vectors I and II) Suggested Solution (a) (i) p is perpendicular to q or p is a zero vector or q is a zero vector. (a) (ii) The locus of R is a circle with diameter OG. (b) 1 2 2 2 3 6 2 2 7 −           −=            −     1 2 2 QR OR OQ ⎯⎯ → ⎯⎯ → ⎯⎯ →  = − =  Projection of QR ⎯⎯ → onto p ( ) 2 22 2 6 7 2 6 7 1 2 2 −   − + +  =   2 11 10 89  −= 15 89 =

Question 8 (Applications of Differentiation, SLE) Suggested Solution (i) 1x= The stationary point on g( )yx= with x-coordinate 1 is a minimum point since g (1) 0  from the graph or x 1− 1 1+ g ( )x 0 0 0 Shape of tangent \ __ / (ii) g ( 1) 0 −= 243( 1) ( 1) ( 1) ( 1) 0 0 a b c d e a b c d e − + − + − + − + = + − + −= ( )g (1) 0 0 1a b c d e = + + + + = − −−−− We have ( ) 32g 4 3 2x ax bx cx d = + + + . ( ) ( ) ( ) 32 1g0 2 1 1 14 3 2 02 2 2 g 1 0 4 3 2 0 3 13 0224 a b c d a b c a b c d d   + + + = −   =      + + + =            −= + −−− − − + = − − −−−− We have ( ) 43 2g. x ax bx cx dx e = + + + + ( ) 4 3 2 1 27g 28 1 1 1 1 27 2 2 2 2 4 8 1 1 1 1 27 16 2 88 4 8 16 4 42 5 a b e a a b cd b c e c d e d  =−        + + + + =−                  +    + +  +    + + + =−           + =− −    −−−− From (ii), ( )05a b c d e− + − + = −−−−− By G.C., 2, 4, 0, 4, 2.a b c d e= = = =− =−

Question 9 (Vectors I) Suggested Solution (i) Method 1 Area of triangle ABO 1 2 ab= 2 2 a b c 0 abc   + + = +=− ( ) 2 12 ac aba ab CA    =− +=+ ++= and ( ) 2 2 bc abb ab CB    =− +=+ ++= ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( )( ) 2 2 2 2 2 2 A 1 rea of 1 2 21 2 1 2 1 2 2 2 2 3 21 2 31 2 13 2 a b a b a a b a a b b b 0 a b a b 0 ab ab ABC              + + + += +  +  + +  + +  += −  + + +  + =  = +=  +  13 Area of 3 2 (shown)1Area of 2 ab ab ABC ABO    + + ==  Method 2 2 2 a b c 0 a b c   + + = =− − ( ) ( ) 1 Are 1 2 1 22 2 1 2 1 2 a of triangle ab b c b 0 c b bc bc ABO      =  = − −  = −  = =

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1Area of 2 1 2 1 22 1 32 1 332 1 32 1 32 1 32 a b c b b c b c b b c c b b c c c b b c b b c 0 0 b c bc bc ABC BA BC       = = −  − = − − −  − = − −  − = −  −  −  − −  − = −  − − −  = − −  =  + 1 3Area of 3 2 (shown)1Area of 2 bc bc ABC ABO    ++ ==  (ii) For A, B and C to be collinear, the area of ABC must be 0, so 3=− . (iii) Method 1 Since D is on AB, by Ratio Theorem, ( )1 abOD k k= − + when ( ): : 1AD DB k k=− . Since D lies on the line OC, ( )2abmOD mOC = =− + , in which the coefficient of b is twice the coefficient of a. ( )21 2 3 kk k =− = So 22: : 1 2 :133AD DB = − = Method 2 2  +=− abc As  varies, C moves along the same line passing through the origin with direction vector ( )2ab+ . The line cuts AB at the point D when 3=− from (ii) answer. 22 33 a b a bOD ++=− =− By the Ratio Theorem, D divides AB such that : 2:1AD DB=