RVHS 23 Prelim H2MA P1 Std Soln

River Valley High School 1 2023 H2MA Prelim Paper 1 Markers Comments 1 Solution [6] Inequality P1 Q1 (i) ( ) ( )( ) 2 2 2 4 4 1 012 21 01 2 1 xx xx x xx −+ +− − −+ Many students did not factorize the denominator correctly, missing out the negative sign to ( )( )1 2 1xx−+ , which resulted in errors to solutions Method 1: ( ) ( )( ) ( ) ( )( ) 2 2 21 01 2 1 Since 2 1 0 for all real values of , 1 2 1 0 x xx xx xx − −+ − − +  1 or 12xx −  Some students still incorrectly wrote ( ) 2 2 1 0 x− or stated always positive. Method 2: ( ) ( )( ) 2 21 01 2 1 x xx − −+ 1 or 12xx −  (ii) ( ) ( )( ) ( ) ( ) 21 21 2 2 21 01 2 2 2 2 1 0 1 2 2 2 x xx x xx + + − +− −  +− Replace with 2 ,xx Most students could identify the correct replacement to use. However, most students missed out  in the question and thus missed out the solution 1x=− . Many students also did not give the reason for rejection correctly i.e. 2x > 0 for all real x 1 2x=− 1x= 1 2x= + + − − o  

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 x x x xx x −  =  =−

River Valley High School 3 2 Solution [6] Complex Numbers P1 Q2 (i) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 22 22 22 11 * 1 i 2 i1 1 i * 1 i 2 1i 1 i 1 i 2 1i 1 2i 1 i 11 1 21 1 1 2 2 1 1 2 1 4 2 2 2 1 2 4 4 1 z z a a a a a a a a a a a a a a a a a − =+ +− =++ −+ =−+ −+ =−+ −+ = ++ − + + = + − + = + − + = + 2 4 3 0 1 or 3 (rejected since 2) aa aa − + = = Some students incorrectly applied properties of modulus. Common errors include: • ( ) ( ) 2 i 1 1 1aa+ − = − − • 11zz− = − • ( ) 11 * 1 i 2 z z − =+ ( ) 11 * 1 i 2 z z − = + It is necessary to state both roots for a before rejecting the one which doesn’t satisfy the criteria. (ii) Consider 1arg * n z zw −  ( ) ( ) ( ) 1arg * arg 1 arg * arg arg i arg 1 i 4 2 4 4 2 zn zw n z z w n n n      −=  = − − − = − − −  = − − −  = If 1 * n z zw −  is purely imaginary and negative, then: Most students attempted to find 1arg * n z zw −  using properties of argument, however only some were successful. Note: It is incorr ect to assume w = 1 + i just because arg(w) = 4  Many students were not able to identify correctly the possible values of the arg ument for purely imaginary and negative.

River Valley High School 4 1 3 7 11arg , , ,...* 2 2 2 n z zw   − = or 3 22 k + , k + Thus, 3 7 11, , , ...2 2 2 2 3, 7, 11, ... n n    = = 3 smallest positive values of 3,7,11n=

River Valley High School 5 3 Solution [7] Summation P1 Q3 (i) ( ) ( )( ) ( ) ( ) ( ) ( )( ) ( ) 11 2 2 2 2 3 2 where 2 1 2 1 11 1 2 1 1 1 11 22 11 2 1 2 n n nu u u n n n n n n n n n n n n n n n n n n n n n nn nn −+− +  = − +−+ + − − + + −= −+ + − + + −= −+ = − = − 2A= Most students were able to get the re sult ex cept for those who made careless mistakes in algebraic. There were a few students who adopted an incorrect approach to substitute in n = 1 to deduce A. That is not correct as that only gives a particular case and does not show that the result is applicable for all n at least 2 (ii) 3 2 1N n nn= − ( ) 3 2 11 2 1 2 3 12 2 1 22 2 1 2 N n N n n n n nn u u u u u u = −+ = = − = − + −+ =   23 2uu+− 4 3 u u + + 45 3 2 1 2 ... ... ... 2N N N uu u u u− − − −+ + − + 21 2NNuu−−+− 1 N N u u − + + ( ) 1 1 2 1 2 1 2 NN NN uu u u u u + +            −+ = − − + 1 1 1 112 2 1 1 1 1 1 2 2 1 NN NN = − − + + = − + + Most students were able to carry out method of differences. A few missed out the 1/2 in their working. Some students committed the error of writing the last few rows in terms of n instead of N.

River Valley High School 6 (iii) 3 2 1 1 1 1 1lim 2 2 1 11Since 0 as , 0 as 1 It converges. Nn n n N N NNNN  →= = − + −+  → → → → +   Students should note that to explain convergence, there is a need to explicitly explain 11 0 , 0 as 1 NNN→ → →+ It is not sufficient to just state the limit of the sum is ¼ . Also some students wrote as n → but we are supposed to consider N instead. Lastly, a number of students incorrectly stated 3 1 nn− 0→ as n → , this does not explain that the sum converges. (iv) ( )( )( )3 1 21 N n n n n= −− ( )( )( ) ( ) 1 13 1 Replace with 1.1 2 1 1 1 nN n nnn n n += += =+ + − + − + ( )( )( ) 1 2 1 3 2 1 11 1 N n N n n n n nn − = − = = −+ = −   1 1 1 1 2 2 1 NN = − + − Students who succeeded in identifying the correct replacement tended to make mistake in not updating the upper limit, resulting in error in solutions. There were some students who replaced n by n – 1 by going from (ii) to (iv) but lacked the understanding in knowing what that implied and how to get the correct solution. Have included that as an alternative method for students to see the difference. But this method is more cumbersome and not recommended! (Alternative, but not recommended)

River Valley High School 7 ( )( ) ( )( )( ) ( )( )( ) 3 22 1 12 1 3 11 11 1 ( replaced by 1)12 1 12 NN nn nN n N n n n n n n nnn n n n n n == −= −= + = =− − + =− −− = −−    From (ii), 3 2 1N n nn= − 1 1 1 1 2 2 1 NN = − + + Thus, ( )( )( ) 1 3 1 12 N n n n n + = −− 1 1 1 1 2 2 1 NN = − + + ( )( )( )3 1 12n N n n n=  −− 1 1 1 1 121 12 NN−− = − + + 1 1 1 1 2 2 1 NN = − + −

River Valley High School 8 4 Solution [13] Complex Numbers P1 Q4 (i) From: 432 4 6 0x x x ax b− + − + = Given that 0xx= is a root, then: 432 0 0 0 04 6 0x x x ax b− + − + = ---eqn (1) Consider applying conjugate on both sides: ( ) ( ) 432 0 0 0 04 6 * 0 *x x x ax b− + − + = ( ) ( ) ( ) ( ) ( ) 432 0 0 0 0* 4 * 6 * * * 0x x x ax b− + − + = Since coefficients are all real, then ( ) ( )* and *a a b b== ( ) ( ) ( ) ( ) 432 0 0 0 0* 4 * 6 * * 0x x x a x b− + − + = Therefore, 0 *x is a root as well. Alternatively, Substitute 0xx = into LHS of eqn (1): ( ) ( ) ( ) ( ) 4 3 2 0 0 0 0* 4 * 6 * *x x x a x b− + − + ( ) ( ) ( ) ( ) 432 0 0 0 0* 4 * 6 * *x x x a x b= − + − + ( ) 4 3 2 0 0 0 04 6 *x x x ax b= − + − + since a, b are real ( )0 * 0== Thus 0 *x is also a root. This question was not well attempted and there is one misconception that 00*xx =− Some of the students verify that 0 *x is the root without much explanation. There is a difference between “Show” and “Verify” (ii) Using Remainder Theorem: Since 2ix=− is a root of the equation, ( ) ( ) ( ) ( ) 432 2 i 4 2 i 6 2 i 2 i 0 ab− − − + − − − + = ( ) ( )7 24i 4 2 11i 6 3 4i 2 i 0 a a b− − − − + − − + + = 3 4i 2 i 0a a b− − + + = Comparing the real and imaginary parts, 3 2 0 and 4 0 4 and 5 a b a ab − + = − + = == 432 4 6 4 5 0x x x x− + − + = The responses from t he students were mixed. The students could have used their GC to evaluate some of the terms (or to check their answers) to prevent the careless mistakes in evaluating the terms. Some students used part (i) answer to obtain 2ix=+ to obt ain the other roots but there are slips in their calculation which led to incorrect answers. Students shou ld learn to present their workings clearly and

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