MI 9758 2023 Prelim P1 Solutions

1 © Millennia Institute 9758/01/PU3/Prelim/23 Solution 2023 PU3 H2 MATHEMATICS PRELIMINARY EXAMINATION Paper 9758/01 Qn Solution 1 [4] ( ) 3f x ax bx c= ++ At ( )1,10 , ( ) ( ) 3 10 1 1 10 a bc abc = ++ ⇒ ++= At ( )2,12− , ( ) ( ) 3 12 2 2 8 2 12 a bc a bc =− +−+ ⇒ − − += Since that ( )f x is divisible by 2x− , ( )f2 0 = . ( ) ( ) 3 0 2 2 82 0 a bc a bc = ++ ⇒ + += Using GC, 7 19, , 633a bc= −= = 2 [4] 2 4 2 2 2 34 23 2 d 3d d 13 d d 11 d2 11 22 yxx x y x xx y Cx xx y Cx Dxx = + = + = − −+ =+ ++ When x = 1, d1 3 0 10 .d2 2 y CCx = ⇒− − + = ⇒ = When x = 1, 23 1 1 3 23 4 .6 222 6 3y DD= ⇒+++= ⇒= Hence, 2 1 134 .2 2 23 xy xx=+ ++ 3(i) [3] Since 2ix= −+ is a root of 32 70x ax x b+ − += , it satisfies the equation.

2 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution ( ) ( ) ( ) ( ) ( ) 23 0 12 4i 3 4i 0 12 3 4 4 i 0 By comparing real and imaginary parts, 12 3 0 12 3 44 0 2i 7 2 4 4 2 1 15 iiab ab ab a ab ab aa a b + += ++ − += + ++ − = + += ⇒ = − − − = ⇒= ⇒= ∴ − + −−+ =− −+ Alternative method (remove from examiner report) Since all the coefficients of the equation are real, 2ix= −+ is a root means that 2ix= −− is also a root. ( ) ( ) [ ] ( ) ( ) [ ] ( ) [ ] ( )( ) 32 2 2 2 7 2i 2i 2i 2i 2i 45 x a x xb x x xC x x xC x xC x x xC + − + = −−+ −−− +     = +− ++ +     =+− + = ++ + By comparing: Constant term: 5bC= Coefficient of x: 7 4 5 3CC−= + ⇒ = − Coefficient of x2: 4aC= + So, 15, 1ba= −= 3(ii) [3] 32 7 15 0xx x+−−= Since all the coefficients of the equation are real, 2ix= −+ is a root means that 2ix= −− is also a root. Quadratic factor: ( ) ( ) ( ) ( ) ( ) 2 2 2 2i 2i 2i 2i 2i 45 xx xx x xx −−+ −−−     = +− ++     = +− =++ ( )( ) ( )( ) 32 2 2 7 15 0 45 0 By comparing coefficients or long-division, 5 15 3 45 30 xx x x x xc cc xx x +−−= ++ += = − ⇒= − ++ −=

3 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution The last root is 3x= . Hence the other roots are 2xi= −− and 3x= 4(i) [4] ( )( ) ( ) ( )( ) 2 2 22 2 14 25 14 2 5 05 14 2 10 5 05 2 74 05 21 4 05 x xx x xx x x x xx x xx x xx x − ≥−− −−− − ≥− −− −−+ ≥− −− ≥− +− ≥− 1 4 or 52 xx−≤≤ > 4(ii) [2] To solve 2 14 25 x xx − ≥−− , replace x with |x| 1 4 or 52 0 4 or 5 4 4 or 5 or 5 xx xx xxx −≤ ≤ > ⇒≤ ≤ > − ≤ ≤ <− > 5(i) [3] 5(ii) [3] From graphing calculator, the two graphs in (i) intersect at (2.52611, ‒ 2.26990) (5 d.p.) and (3.75707, ‒3.24224) (5 d.p.) 3.75707 2.52611 2 2 lnArea of = 3cos ln dcos 0.94708 0.95 units . (2 d.p.) xR xxxx  − ≈ = ∫ − − + + 2 ln( ) cos( ) xy x= y x ,02 π  3 ,02 π  2x π= 3 2x π= 3ln( ) cos( )y xx= O

4 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 5(iii) [2] ( ) 3.75707 2 2 2.52611 3 2 lnVolume of = 3cos ln dcos 16.968 17.0 units . (3 s.f