NJC_8865_2022_Prelim_Solutions_vetted

* © NJC 2022 [Turn over NATIONAL JUNIOR COLLEGE SENIOR HIGH 2 Preliminary Examination H1 MATHEMATICS 8865/01 1 ( ) 2 2 2 2 2 3 1 2 2 1 0 ( 2 1) 4 0 4 4 1 4 0 4 8 1 0 mx mx m x mx mx x mm m m m mm − = − − − − + + = − + −  − + −  − +  31 2 m− or 31 2m+ 2 23e e 70xx+ 23e e 70 0xx− −  Let exu= , then 23 70 0uu− −  Consider 23 70uu−− = 0, then 1 1 4(3)( 70) 2(3)u  − −= u = 5 or u = 28 14 63− =− For 23 70 0uu− −  , 285 or 6uu − 28e 5 or e (reject as e 0)6 x x x −  x > ln5 3(i) Let $x, $y, $z be the amount invested in plan I, II and III respectively. x + y + z = 10000 … (1) y = 2x  2x − y = 0 … (2) 0.014(2) 0.035 8(50)( ) 356010000 zxy+ + = 0.028x + 0.035y + 0.04z = 3560 … (3) From GC, x = 20000, y = 40000, z = 40000 2( 8) 8 4(4)(1) 2(4) 8 4 3 3 182 m m − −  −= = = 

2 © NJC 2022 She deposits $20000 in plan I, $40000 in plan II and $40000 in plan III 3(ii) If she deposits in savings plans II and III, the total interest is 400000.035(40000) 8(50)( ) 300010000+= If she deposits $80000 in savings plan IV, the interest ranges from Min 0.037(80000) = $2960 to Max 0.039(80000) = $3120. The average interest is 3.8%, 0.038(80 000) = $3040 Tammy is advised to deposit in savings plan IV as she can earned $120 more against loss $40 over savings plans II and III. Further the average interest is $40 more than the plan II and III. 4 4 2 1 4 2 1 4 1 13d 619 d 19 6ln 11= 36 6ln 4 9 6ln141 111= 6ln 44 xx xxx xx x  − = − + = − −    − − − − −       −    Or 27.75 − 6ln4 24 31 1 13 1e d 3 d p x xx x + − =−   31 1 3 1 111e 6ln 434 p x+ −  = − ( ) 311 111e 1 6ln 434 p+ − = − By GC, p = 1.0275179 = 1. 0275 (4dp) Method 2 31 333e 1 18ln 4 4 p+ − = − e3p + 1 = 59.2967015 3p + 1 = ln(59.2967015) 3p = 4.082553681 − 1 p = 1. 027517894 = 1.0275 (4dp)

3 © NJC 2022 [Turn over 5(i) 2 5 2( 2) 1 22 12 2 xxy xx x − − −== −− =− − 5(ii) 5(iii) The exact area of the region bounded by the curves C1, C2, and the y-axis 1 1 0 1 1 0 1 1 0 25e 2 d 2 1e 2 2 d 2 1e d 2 x x x x xx xx xx − − − −= + − − = + − + − =+ −    11 0 e ln( 2 ) 1 e ln 2 x x−= − + − = − + − 5(iv) 2 5 1 222 xy xx −= = −−− ( ) 2 d1 d 2 y x x = − When x = 4, y = 1.5 and d1 d4 y x = . ( )11.5 4 4yx− = − 2 4 xy += Equation of tangent is 2 4 xy += or 0.25 0.5yx=+ O x y

4 © NJC 2022 5(v) Enter equation of tangent, 0.25 0.5yx=+ . Find intersection with C2 giving x = 6.0262534 From GC, x < 6.0262534 = 6.03 (3sf) 6(i) 32 2 7 8 20 d 3 14 8d C t t t C ttt = − + + = − + For stationary points on C, d 0d C t = 23 14 8 0tt − + = ( )( )3 2 4 0tt− − = 2 43t or t = = t 0.6 2 3 0.7 d d C t 0.68 0 −0.33 slope t 39. 4 41. d d C t -0.97 0 1.03 slope Or 2 2 d 6 14d C tt =− When 2 3t = , 2 2 d2 6( ) 14 10