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.) xS xx xxπ  −  ≈ = ∫ 6(i) [3] ( ) ( ) ( ) ( ) ( ) f 1 1 11f where D , ,21 2 2 Let f 21 2 21 21 f 21 f 21 xx x yx xy x xy y x xy y yx y yy y xx x − −   =− = −∞ − ∪ − ∞  +    = =− + += − += − =− + =− + ⇒= − + Domain of f–1 = Range of f = 11,, 22   −∞ − ∪ − ∞     Or Domain of f–1 = 1\ 2 −  6(ii) [1] Method 1 - Hence Since ( ) ( ) 1ff 21 xxx x − = −= + and the domains 1f fDD −= are the same, ( ) ( )( ) 21f ff x xx −= = Method 2 - Otherwise ( ) ( ) 2ff 21 21 21 21 21 212 21 xx x x x x x xx x xx x = − +  − +=− − + + += × +−+ + = x y O

5 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 6(iii) [2] ( ) ( )( ) ( ) ( ) ( ) 2023 2022 2023 f ff f 5f 5 f5 11 xx x = = = =− 6(iv) [2] ( ) ( )g 1g where D ,xa ax= = ∞ Range of g = 2 10, a   Since gfRD⊆ , composite function fg exists. 7(i) [4] 2 18iz = Let izxy= + ( ) 2 22 22 2 2 4 1 2 i 18i 2 i 18i By comparison of real and imaginary parts, 0 92 18 9 0 81 0 3 when 3, 3 when 3, 3 3 3i, 3 3i xy x xy y xy xy y x x x x x xy xy z z += + −= −= = ⇒= −= −= =± = = = −= − = + = −− x y O 2 1,a a   1 2− 0 2 1 a

6 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 7(ii) [4] 2 18iz = ( ) 22 18, arg = 2zz π⇒= ( ) 1i3 22, arg = 3 w ww π = −− ⇒= − * 2 2 * 21 18 9 ww z z  = = = ( ) ( ) * 2 2arg arg arg 2 32 7 6 5 6 w wzz ππ π π  = −−  = −− − = =− * 5i 6 2 1 e9 w z π− = 8a(i) [3] ( ) 22(3 2 ) (3 2 ) 9 | | 12 4 | |−− = −+ba ba b a b a ( ) 2 223 2 9(2) 12 4(1)⇒− = − +b a ab ( ) ( ) 2 34 40 12⇒= − ab 40 34 1 12 2 −⇒= =ab 8a(ii) [3] Now, cosθ=ab a b where θ is the angle between a and b Hence, 0.5 1cos (1)(2) 4θ = = =ab ab  . 224 1 15sin .44θ −⇒= = Hence, 15 15sin (1)(2) . 42θ×= = =a b ab Geometrical Interpretation of ×ab : (1) the perpendicular (or shortest) distance from B to the line OA OR 1 4 2241− θ

7 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution (2) area of parallelogram with adjacent sides OA and OB / twice the area of triangle OAB OR (3) height of triangle OAB with OA as the base. 8(b) [3] By Ratio Theorem, 2 3OP += ba … (1) 2OM += bc … (2) OBMN forms a rhombus 2 (3)2 ON BM OM = = − += − −= −−− b bc b cb   P is a mid-point of MN ( ) 1 2OP OM ON⇒= +  --- (4) Method 1: Equating two different expressions for OP  Put (1), (2), (3) into (4) 21 3 22 2 1 2 OP OP + +−   = = +      = ba bc cb c   423 2 4 3 . (shown) ⇒+= ⇒+−= bac a b c0 Method 2: Equating two different expressions for ON  From (4), 2ON OP OM= −  22 [from (1), (2) and (3)]2 32 25 2 3 62 − ++    ⇒= −         −⇒ =+− cb ba bc cb a b c

8 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 25 3 6 222 24 33 2 4 3 . (shown) ⇒ + +−−= ⇒ + −= ⇒+−= a bbcc 0 ab c0 a b c0 9(a) i-iii [5] (i) Horizontal asymptote: 2yb= x-intercept: ( )1, 0a− (ii) Horizontal asymptote: yb= x-intercept: ( ),0a− (iii) Horizontal asymptote: 1y b= x-intercept: inconclusive (it would result in a vertical asymptote at xa= ) 9(b) (i) [3] 9(b) (ii) [2] ( ) 5 3 f dxx −∫ refers to the area under the curve from 3x=− to 5x= . Method 1 ( ) 31 1 5 3 10 2132 d 22f d 4 13 d2 . 5 3 7 d 2xxxxx xx x − −= ++ = + +−∫∫ ∫ ∫ Method 2 ( ) 0 5 1 3 12 (2f d )(1 4.7 ) 5 3 d 111+ (1) 2 22 x xxx −   = ++ =  ∫∫ x y O ( )3, 0 1 2− 4

9 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 10 (a)(i) [2] ( ) ( ) ( ) 2 12 2 242 d 1 2esin e e (2) .d 1e1e x xx xxx − = = −− 10 (a)(ii) [3] ( ) ( ) ( ) ( ) ( ) ( ) 24 2 4 4 44 1 12 4 4 2 1 4 2 12 12 4 12 4 2e e 2e e d + d 1e 1e 1e 1sin e 4e 1 e d [From ]4 1e1= sin e 14 2 1sin e 2 1 e 4 1sin e 1 e 2 xx x x x xx x xx x x xx xx xx x c c c −− − − − + = − −− = −− −  −−+    = − −+  = − −+ ∫∫ ∫ (i) 10 (b)(i) [1] ( ) 22 2 2 2 sin cos sin 1 sin sin sin . (shown) nn nn θθ θ θ θθ −− − = − = − 10 (b)(ii) [4] ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 0 2 1 0 212 2 0 0 2 22 0 2 2 0 22 2 00 sin d sin sin d sin cos cos 1 sin cos d [0 0] 1 sin cos d 1 sin sin d [from ] 1 sin d 1 sin d n n nn n nn nn n n n nn π π ππ π π ππ θθ θ θθ θ θ θ θ θθ θ θθ θ θθ θθ θθ − −− − − − =  = − −− −  =−+− = −− =− −− ∫ ∫ ∫ ∫ ∫ ∫∫ (i) ( ) 22 2 00 22 2 00 sin d 1 sin d 1sin d sin d . (shown) nn nn nn n n ππ ππ θθ θθ θθ θθ − − = − −= ∫∫ ∫∫

10 © Millennia Institute 9758/01/PU3/Prelim/23 Solution Qn Solution 10 (b) (iii) [2] [ ] 22 3 00 2 0 31sin d sin d3 2 cos3 2[0 ( 1)]3 2 3 ππ π θθ θθ θ −= = − = −− = ∫∫