TMJC 9758 2023 Prelim P1 Solution

Page 1 of 21 2023 H2 MATH (9758/01) JC 2 PRELIMINARY EXAMINATION – SUGGESTED SOLUTIONS Qn Solution 1 Graphing Techniques (a) Finding asymptotes for 1C : ( ) ( ) 2 2 22 2 032 2 23 x y yx − −= = ±− O

Page 2 of 21 (b) 2C is a circle centred at (2,0) with radius k. For 1C and 2C to intersect exactly twice, 3k = . O k

Page 3 of 21 Qn Solution 2 Techniques of Integration (a) ( ) ( ) 22 22d e 2 2ed xx xxxx ++ = + ( ) ( ) ( ) 22 2 11 22 00 12 0 30 3 11e d 2 1e d 2 1 e2 1 ee2 1 e12 xx xx xx xx xx ++ + += + =  = − = − ∫∫ (b) dsin cos d xxt t t= ⇒= ( ) ( ) ( ) ( ) ( ) 3 2 2 3 2 2 3 2 2 3 2 2 2 1 d 1 1 cos d 1 sin 1 cos d cos 1 cos d cos 1 d cos sec d tan , 1 x x tt t tt t tt t t t tt tC C x C x − = − = = = = = +∈ = + − ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ ∫  x 1 t

Page 4 of 21 Qn Solution 3 Arithmetic and Geometric Series (a) ( ) ( )( ) ( ) ( )( ) 2 5 1 3 1002 2 5 1 3 100 02 n n n n +− ≤ +− − ≤ Using GC, ( ) ( )( ) ( ) ( )( ) When 7, 2 5 1 3 100 2 02 When 8, 2 5 1 3 100 24 02 nnn nnn = + − − = −≤ = +− − => Maximum number of squares Student A can form using the 100 cm wire is 7. (b) The circumference of the circles follow a geometric progression with common ratio 2 .3 ( ) ( ) ( ) 2 11 12 100 Total circumference of 12 circles 22 2100 2 2 2 ... 233 3 221 3 100 21 3 xx x x x ππ π π π =  = + + ++     −  = − Using GC, ( ) 5.3464 5.35 3 s.f. x= =

Page 5 of 21 Qn Solution 4 Graphing and Transformation (a) 1y x= Re place byx xa − 1y xa= − Replace by 3 yy a 3ay xa= − Replace by 2yy − 32 ay xa= + − Note a > 0. 1. Translation of a units in the positive x-direction. 2. Stretch by factor 3a parallel to the y-axis. 3. Translation of 2 units in the positive y-direction. OR 1. Stretch by factor 3a parallel to the x-axis. 2. Translation of a units in the positive x-direction. 3. Translation of 2 units in the positive y-direction. (b) y x y = 2 x = a

Page 6 of 21 (c) 323 1 323 1 x x += − += ±− 323 1x+= − or 323 1x+= −− 3 11x =− 3 51x =−− 13x−= ( )5 13x− −= 4x= 2 5x= For 323 1x+< − , from the graph in part (b), 2 5x< or 4x> y x y = 2 x = 1 y = 3

Page 7 of 21 Qn Solution 5 Complex Numbers (a) Since the coefficients of the polynomial are real, 3i+ is a root implies that 3i− is also a root. A quadratic factor is: ( ) ( )3i 3izz  −+ −−   ( ) ( )3i 3izz  =−− −+   ( ) ( ) 2 2 3iz= −− 2 23 31zz= − ++ 2 23 4zz= −+ Let zk= be the third root. ( )( )32 8 23 4z za z z zk− += − + − Comparing coefficient of 2z : 0 23 23 k k = −− =− Therefore 3 i or 3 i or 2 3z= + −− . Alternative method (sub in 3i+ to find a first) (Not recommended in this question) ( ) 3P8z z za=−+ Since 3iz= + is a root, ( )P 3i 0+= . ( ) ( ) 3 3i 8 3i 0 a+ − ++= ( ) ( ) ( ) ( )( ) ( ) 32 23 3 33i33i i 8 38 i 0 a+++ − − + = 83a= Since the coefficients of the polynomial are real, 3i+ is a root implies that 3i− is also a root. A quadratic factor is: ( ) ( )3i 3izz  −+ −−   ( ) ( )3i 3izz  =−− −+   ( ) ( ) 2 2 3iz= −− 2 23 31zz= − ++ 2 23 4zz= −+ By comparing constant term, ( )( ) 32 8 83 23 4 23zz z z z−+ = − + + 3 i or 3 i or 2 3z= + −−

Page 8 of 21 (b) 5arg 6 5arg 6 n w wn π π = = For nw to be purely imaginary, arg ,2 5 62 63 5 nw kk n k kn π π ππ π = +∈ = + +=  Using GC, the smallest three positive integers of n, 3, 9,15n= .

Page 9 of 21 Qn Solution 6 Functions and Equations and Inequality (a) ( ) 2 2 2 17a bc− − += ⇒ 4 2 17a bc− += ---(1) 2 11 3 22 4a bc + += ⇒ 11 3 42 4a bc+ += ---(2) ( ) 2 55 3a bc + += ⇒ 25a + 5b + c = 3 ---(3) Using GC, a = 1, 5b=− , c = 3. 2 53yx x=−+ (b) Let 2f ( ) 5 3, 0y xx x x= =−+ ≤ ( ) 2 22 2 53 55 322 13 5 42 5 13 24 5 13 24 5 13 since 024 yx x yx yx xy xy xy x =−+   = −− +     +=−  −= ± + = ±+ = −+ ≤ 1 5 13f () 24xx− = −+

Page 10 of 21 (c) (d) 0 π 33 3 cos22 2 330 cos 322 x x x ≤≤ −≤ ≤ ≤+ ≤ Since [ ] [ )hgR 0,3 0, 6 =D= ⊆ , therefore the function gh exists. [ ] [ ] [ ] hg h h ghD RR 0, 0,3 0, 4π  →  → Restricted domain of g [ ]ghR 0, 4= Qn Solution 7 Differentiation (a) 22 22 1xy ab+= Differentiate w.r.t x: ( )( ) 22 22 2 2 2 2d 0d 2d 2 d d since 0 shownd x yy ab x yy x bx a y bx yx ay +=   =− =−≠ (b) At ( )cos , sin ,Pa b θθ ( ) ( ) 2 2 cosd cos d sin sin bayb x ab a θ θ θθ= −= − Equation of tangent at ,P y x ( )2,5− ( )4,3− ( )6,3 ( )8, 3 ( )9, 4