CJC_8865_2022_Prelim_Solutions_vetted

1 CATHOLIC JUNIOR COLLEGE H1 MATHEMATICS 2022 JC2 PRELIM EXAMINATION SOLUTION Q1 Solution ( ) 2 10kx k x k+ − +  for all real values of x So, graph has no real roots and is U-shaped Discriminant < 0 2 40b ac− and 0k  ( ) ( )( ) 2 1 4 0k k k− −  23 2 1 0kk− − +  23 2 1 0kk+ −  ( )( )3 1 1 0kk− +  Alternatively, If 23 2 1 0kk+ − = , 22 2 4(3)( 1) 2(3) 11 or 3 k −  − −= =− 1 3k  1 (rej 0)kk−  or Set of values of k is 1: 3kk   Q2 Solution Let the polynomial be 32y ax bx cx d= + + + 2d 32d y ax bx cx = + + At ( )1, 13− , 13 --- (1)a b c d+ + + =− 3 2 0 -------- (2)a b c+ + = At ( )1,7− , 7 --- (3)a b c d− + − + = 3 2 12 --- (4)a b c− + =− Solving (1), (2), (3) and (4), 2a= , 3b= , 12c=− , 6d =− 1− 1 3 k

2 Hence the polynomial is 322 3 12 6x x x+ − − Q3 Solution (a) (i) 2d lnd2 xe xx   + ( )( ) 2d ln ln( 2)d xexx= − + ( )d 2 ln( 2)d xxx= − + 12 2x=− + Alternative Method (Quotient rule) 2d lnd2 xe xx   + ( ) ( ) 22 22 222 2 xx x x e ex e x  +−+=  + 12 2x=− + (a) (ii) d2 1d n xx + 1 2 221 n n xx −    = + −       1 2 22 1 n n xx − =− +  (b) 1 1 2 2 2 2 xx x x x x x x +−= − − − − + − ( ) ( ) 22 2 2 xx xx +−= −− 2 2 xx+−= 11 222 dx x x dx xx = + − −− ( ) 3 3 2 221 332 22 xx C   −= + +           , where C is an arbitrary constant ( ) 3 3 2 2 1 23 x x C= + − + 

3 Q4 Solution (i) ( )ln 2 4 2 6y x x= − − + 1 2d 2 1 2d 2 4 2 y xxx −=− −  11 2x x =− − For stationary points, set d 0d y x = So, 11 02x x −=− 11 2x x =− 2xx=− Method 1 Squaring both sides, ( ) ( ) 2 2 2xx =− 2 44x x x= − + 2 5 4 0xx− + = ( )( )1 4 0xx− − = 1x= or 4x= Since 2x , so 4x= Method 2 20xx− + = Let yx= , then 2 20yy− + = ( )( ) 2 20 1 2 0 yy yy − − = + − = So, 2yx== or 1yx= =− (no solution) Hence 4x= 11 02x x −=− 11 2x x =− When 4x= , ( )ln 2 4 4 2 6 ln 4 4 6 2 ln 4yx=  − − + = − + = + So turning point is ( )4, 2 ln 4+

4 (ii) (iii) At 9x= , d 1 1 1 1 1 1 3 7 4 d 2 9 2 7 3 21 21 9 y xx x −= − = − = − = =−−− and ( )ln 2 9 4 2 9 6 ln14 6 6 ln14y=  − − + = − + = Method 1 Substituting point and gradient into y mx c=+ ( )4ln14 9 21 c= − + ( )4 12ln14 9 ln1421 7c = + = + Hence equation of tangent at 9x= is 4 12 ln1421 7yx=− + + Method 2 Hence equation of tangent at 9x= is ( )4ln14 9 21yx− =− − 44 9 ln1421 21 4 12 ln1421 7 yx yx =− +  + =− + + x y O V.H. ( )ln 2 4 2 6y x x= − − +

5 (iv) Area = area under tangent – area und