YIJC_8865_2022_Prelim_Solutions_vetted

Qn Solutions 1 For ( )2 10 22 knx k x − + + + −   , Discriminant < 0 and 10a=−  2 40b ac− and 10a=−  ( ) ( ) 2 2 2 1 4 1 0 22 2 1 2 2 0 4 1 2 0 knk k k k n k k n + − − −   + + + −  + + −  Let 2 4 1 2 0knk + + − = , then ( )24 4 4 1 2 2 2 4 1 2 2 32 nk n n − − −= =− − + =−  +   Therefore 2 3 2 2 3 2n k n− − +  − + + . Qn Solutions 2 Let $x, $y and $z be the cost of a jigsaw puzzle, a spinning top and a Rubik’s cube respectively. 4 2 5 264.5 34 4 2 49 4 2 4 40 9 3xz x y z xz x y z x y z + + = = + = + + − −  = = Using GC, 30, 16, 22.5x y z= = = The cost of a spinning top is $16.00 −2 − √3 + 2𝑛 −2 + √3 + 2𝑛

Qn Solutions 3(a) ( ) ( ) ( ) ( ) 1 3 2 3 3 32 2 3 23 2 d 2 d 21dd 1 12 1 32 31 xxx x xx xx − − −  =− − = − − − =− (b) ( ) ( )( ) 2 2 2 2 2 4ln ln 4 ln 1 e1e d ln 4 ln 1 ed 4 2e 4 1 e 1 2e 1e x xxx xxx x xx x xx  = − − − −− =+ − =+ − (c) 3 62 2 1 3 2 6 2 1 3 62 1 04 4 11 e d 1 e d 11ln e 2 1 1 1ln 3 e ln1 1 e3 2 2 71 ln 3 e62 x x x xxx xxx x x − −− − +− = + − = − +    = − + − − +       = + −  

Qn Solutions 4(a) (b) ( )ln 3 7y x= + d3 d 3 7 y xx= + At 7 2x= , d 3 6 7d 35 37 2 y x ==  + , 7 35ln 3 7 ln22y     = + =        Equation of tangent to the curve: 35 6 7ln 2 35 2yx    − = −       3535 35ln 6 21 2 3535 6 35ln 21 2 yx yx − = − − = −  (c) Numerical value of area under curve ( ) 1 1 ln 3 7 d 3.8269 (4 .d.p) xx − =+ =  𝑦 𝑥 𝑥 = − 7 3 𝑦 = ln⁡(3𝑥 + 7) (0, ln 7) (−2,0) 𝑂

Qn Solutions 5(a) (i) Let P m be the perimeter of the amusement park and B m be the width of the square. 2224 2 Bx Bx = = 4 2 4 2 10 5 2 2 2 P x x y y x x = + + = = − − ( ) ( ) ( ) 2 2 22 2 22 2 5 2 2 2 2 10 4 2 6 10 4 2 6 A xy x x x x x x x x xx =− = − − − = − − = − + 5(a)ii d 10 8 2 12 0d 10 5 8 2 12 4 2 6 A xxx x = − − = == ++ x 5 4 2 6 −   + = 0.4288 5 4 2 6+ 5 4 2 6 +   + = 0.4290 Sign of d d A x 0.0030818 0 − 0.001581 slope Therefore 5 4 2 6 x= + gives maximum area. 5b(i) When 1, 1.5tC==

1.5 3 3e 3e 1.5 1e 2 1ln 2 1ln ln 22 k k k k k − − − =− = = −= =− = (ii) ln 23 3e tC −=− d 0.25993 0.260 (3 s.f)d C t == (iii) ln 23 3e tC −=− ,3tC→ → The maintenance cost approaches $3000 in a long run. (iv) 5 ln 2 0 3 3e d 10.8t t−−= The total maintenance cost over the 5 months is $10800 Qn Solutions 6(a) 8 12 42 8 12 51 8 6 Case 1 : 4 girls , 2 boys No. of ways C C 4620 Case 2 : 5 girls , 1 boy No. of ways C C 672 Case 3 : 6 girls No. of ways C 28 Total no. of ways 4620 672 28 5320 =  = =  = == = + + = (b) Total no. of ways 3! 3! 2 72=   = Qn Solutions 7(a) P( ) P( ) P( ) 0.65 0.4 0.25 A B A A B  = −  = − =

(b) P( ) 8P( | ) P( ) 9 0.4 8 P( ) 0.45P( ) 9 P( ) 0.55 ABAB B BB B  ==  =  = = (c) P( ) 1 [P( ) P( )] 1 0.95 0.0