RVHS 9758 2023 Prelim P1 Solution

River Valley High School 1 2023 H2MA Prelim Paper 1 1 Solution [6] Inequality P1 Q1 (i) ( ) ( )( ) 2 2 2 4 41 012 21 01 21 xx xx x xx −+ <+− − <−+ Method 1: ( ) ( )( ) ( ) ( )( ) 2 2 21 01 21 Since 2 1 0 for all real values of , 1 210 x xx xx xx − <−+ −≥ − +< 1 or 12xx∴ <− > Method 2: ( ) ( )( ) 2 21 01 21 x xx − <−+ 1 or 12xx∴ <− > (ii) ( ) ( )( ) ( ) ( ) 21 21 2 2 21 012 2 22 1 0 1 2 22 x xx x xx + + − ≤+− − ≤ +− Replace with 2 ,xx 1 2x=− 1x= 1 2x= + + − − o ο ο

River Valley High School 2 112 OR 2 1 OR 2 22 (rejected, since 0 1 2 is positive for all real values of .) x xx x xx x <− > = >= −

River Valley High School 3 2 Solution [6] Complex Numbers P1 Q2 (i) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 22 22 22 22 22 11 *1 i 2 i1 1 i*1 i 2 1i 1 i1i 2 1i 1 2i 1i 11 1 211 1 2 2 11 2 1 4 222 1 2 44 1 z z a a a a a a a a aa a aa a aa a − =+ +− =++ −+ =−+ −+ =−+ −+ = ++ − ++= + −+= + − += + 2 4 30 1 or 3 (rejected since 2) aa aa − += = < (ii) Consider 1arg * n z zw −  ( ) ( ) ( ) 1arg * arg 1 arg * arg arg i arg 1 i 4 2 44 2 zn zw nz z w n n n π π ππ π −=  = −− − = − −−  = −− −  = If 1 * n z zw −  is purely imaginary and negative, then:

River Valley High School 4 1 3 7 11arg , , ,...* 22 2 n z zw ππ π− = or 3 22 kπ π+ , k +∈  Thus, 3 7 11, , , ...2 22 2 3, 7, 11, ... n n π ππ π= = 3 smallest positive values of 3,7,11n=

River Valley High School 5 3 Solution [7] Summation P1 Q3 (i) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) 11 2 22 2 3 2 where 2 121 11 12 1 1 1 11 22 11 2 1 2 n nnu uu n n nn nn n n n n n nn nnn nn nn n nn nn −+ −+ ≥ = −+−+ +− − ++−= −+ +− ++ −= −+ = − = − 2A= (ii) 3 2 1N n nn= −∑ ( ) 3 2 11 2 1 23 12 2 1 22 2 1 2 N n N n nn n nn u uu u uu = −+ = = − = −+ −+ = ∑ ∑ 23 2uu+− 4 3 u u + + 45 3 21 2 ... ... ... 2N NN uu u uu− −− −+ +−+ 21 2NNuu−−+− 1 N N u u − + + ( ) 1 12 1 2 1 2 NN NN uu uuu u + +            −+ = −−+ 1 11 1122 1 11 1 1 22 1 NN NN = −− + + = −+ +

River Valley High School 6 (iii) 3 2 1 11 1 1lim 22 1 11Since 0 as , 0 as 1 It converges. Nn nn NN NNNN ∞ →∞= = −+ −+  → →∞ → →∞ + ∴ ∑ (iv) ( )( )( )3 1 21 N n n nn= −−∑ ( )( )( ) ( ) 1 13 1 Replace with 1.12 11 1 nN n nnn nn += += = ++− +− +∑ ( )( )( ) 1 2 1 3 2 1 11 1 N n N n n nn nn − = − = = −+ = − ∑ ∑ 11 1 1 22 1 NN =−+ − (Alternative, but not recommended) ( )( ) ( )( )( ) ( )( )( ) 3 22 1 12 1 3 11 11 1 ( replaced by 1)12 1 12 NN nn nN n N n n n nn n nnnn n nn n = = −= −= + = =− −+ = −−− = −− ∑∑ ∑ ∑ From (ii), 3 2 1N n nn= −∑ 11 1 1 22 1 NN = −+ + Thus, ( )( )( ) 1 3 1 12 N n nn n + = −−∑ 11 1 1 22 1 NN = −+ + ( )( )( )3 1 12n N nn n= ∴ −−∑ 11 1 1 121 12 NN−− =−+ +

River Valley High School 7 11 1 1 22 1 NN =−+ −

River Valley High School 8 4 Solution [13] Complex Numbers P1 Q4 (i) From: 432 46 0x x x ax b− + − += Given that 0xx= is a root, then: 432 0000 46 0x x x ax b− + − += ---eqn (1) Consider applying conjugate on both sides: ( ) ( ) 432 0000 4 6 * 0*x x x ax b− + −+= ( ) ( ) ( ) ( ) ( ) 432 0000*4 *6 * * * 0x x x ax b− + − += Since coefficients are all real, then ( ) ( )* and *aa bb= = ( ) ( ) ( ) ( ) 432 0000* 4* 6* * 0x x x ax b− + − += Therefore, 0 *x is a root as well. Alternatively, Substitute 0xx ∗= into LHS of eqn (1): ( ) ( ) ( ) ( ) 432 0000* 4* 6* *x x x ax b− + −+ ( ) ( ) ( ) ( ) 432 000 0*4 *6 * *x x x ax b=−+ −+ ( ) 432 0000 46 *x x x ax b= − + −+ since a, b are real ( )0* 0= = Thus 0 *x is also a root. (ii) Using Remainder Theorem: Since 2ix= − is a root of the equation, ( ) ( ) ( ) ( ) 432 2i 4 2i 6 2i 2i 0 ab−− −+ −− − + = ( ) ( )7 24i 4 2 11i 6 3 4i 2 i 0 aa b−− − − + − − + += 3 4i 2 i 0aa b−− ++= Comparing the real and imaginary parts, 3 2 0 and 4 0 4 and 5 ab a ab − += −+= = = 432 4 6 4 50xxxx− + − +=

River Valley High School 9 Using Factor Theorem: ( ) ( ) ( ) 432 2 4 6 45 2i 2i xxxx x x x Ax B − + −+ ≡−− −+ + +     Comparing constant term: ( )( )5 2i2i 1BB= + −⇒= Substitute 2x= : ( ) ( ) ( ) 16 32 24 8 5 22 i 22 i4 2 542 1 0 AB AA − + −+ ≡−− −+ + +     =+ +⇒ = For 2 0x Ax B+ +=⇒ 2 10x += i or ix= − The roots are: 2 i, 2 i, i, ix=− +− • (iii) 43 2 234 4 432 6 4 10 14 6 0 Divide throughout by , 1111 46 0 by ay y y y y ay by y abyyyy − + − += −+ − + =    − + − +=       Replace 1 with x y , Then 1 2 i, 2 i, i, i 2 12 1i , i , i , i5 55 5 y y =− +− = +−−

River Valley High School 10 5 Solution [7] Abstract Vectors P1 Q5 (i) By Ratio Theorem, ( ) 3 13 13 44 13 Since 44 1 4 OC OAOS OS OC OA OS OC OA AC AC OB OS += + = + = ++ = + = = = + ab a b ab                            (ii) 11 1 42 4 MS OS OM= − = + −= − a b ba b       : 11 , 24 MSl OM MS λ λλ = + = +− ∈ r r b ab      ( )11 22 BN ON OB= − = + + −=−a ab ba b      : 1 , 2 BNl OB BN µ µµ = + = +− ∈ r rb a b      (iii) At T, 11 1 24 2 λµ +−= +−   b ab b ab