RVHS 2020 H2 P1 Math Soln V1

Suggested Solution A lvl H2 Math P1 1i 21 21 0 11 5 2 1 2 6 3 −            − = = − −          −−          1ii Let  be the required acute angle. ( ) ( ) 1 2222 22 1 14 15 36 45 4 5 18cos 19cos 49.2 (1 d.p.) cos 1 1 3 6 11 77 711    − −        −      −   = +  − − =  +  =  − + + − = 2 2 22 35 2 3 2 2 5 11 1111 11 xy xy y xy xxy +=++ − + − =++ Differentiate implicitly w.r.t. x. ( ) ( ) 2 5 34 22 2 2 2 2 d d 35dd11 x y y y x x yxxxy y+ = + ++ . When 1, 1xy== and, ( ) ( ) 22 2 2 d d 35dd22 d5 d9 yy xx y x + = + =− Hence the equation of the tangent is; ( )511 9 9 9 5 5 5 9 14. yx yx xy − = − − − = − + +=

Suggested Solution A lvl H2 Math P1 3(i) Method 1 ( ) ( ) ( )( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 22 2 2 2 3cos3f 1 sin 3 1 sin 3 9sin 3 3cos3 3cos3f 1 sin 3 9sin 3 9sin 3 9cos 3f 1 sin 3 9sin 3 9f 1 sin 3 9 1 sin 3f 1 sin 3 9f (Shown)1 sin 3 xx x x x x xx x x x xx x xx x xx x x x  = + + − − = + − − − = + −− = + −+ = + − = + Where 9k =− . 3(ii) ( ) ( ) ( ) ( ) ( ) 22 9f 1 sin 3 9 3cos3 27 cos3f 1 sin 3 1 sin 3 x x x xx xx − = +  == ++ When 0x = , ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 f 0 ln 1 0 0, 3f 0 3, 10 9f 0 9, 10 27f 0 27. 10 = + =  == + − = = − +  == + ( ) 23 23 9 27f 0 3 ... 2! 3! 993 ... 22 x x x x x x x −= + + + + = − + +

Suggested Solution A lvl H2 Math P1 4(i) ( ) ( ) 1 1 3 πi tan i12 3 1 πi tan i2 1 4 2 πi 6 2 1 2 3 3i 1 3 e 2e 1 i 1 1 e 2e 112 cos π isin π 2e66 1z z z − −   − − == = − = − = = + =  = + + +  ( ) ( ) 11 2 3 2 3 2 21 22 zz z z z z= = = 1 1 2 3 23 arg arg arg arg 3 4 6 12 π π π 5π z z z zzz  = − −  = + − = 1 23 51 cos isin .12 5π 2 12 πz zz =+  4(ii) Method 1: As 14 23 1zz zz = and 14 23 zz zz is purely imaginary 14 23 i or izz zz =− . 14 23 2a π π org r 2 zz zz  =  −  . 4 1 4 23 a π π or 22 π 5π π 5π π 11πarg or 2 .12 2 1 rg arg or 12 12 2 zz z z z −+ −=−  −= = −  As 2314 4 2 3 1 21 zzzz zz z z =  == . Hence, 4 π π 11π 11π2 cos isin or 2 cos isin12 12 12 12z      + − + −      =        . Method 2: 2314 4 2 3 1 21 zzzz zz z z =  ==

Suggested Solution A lvl H2 Math P1 As 14 23 zz zz is purely imaginary, ( )14 23 2a 1 πr 2g zz z k z  =  +  , for integer k. ( ) ( ) 1 2 4 4 3 21 π 2 21 π 5π πarg π2 12 12 arg arg zz z k kzk z + = +=  =   + −+  As k is an integer, we either have 4 4 arg , is even12 arg , is odd1 1π 2 π 1 zk zk  =  =− Hence, 4 π π 11π 11π2 cos isin or 2 cos isin12 12 12 12z      + − + −      =        .

Suggested Solution A lvl H2 Math P1 5(a) ( ) ( )2 0  =   = −  = = a b b a b a b ab ab a 0 As both vectors are non-zero, this implies that a is parallel to b. 5(bi) ( )− =rp q0 implies −rp is parallel to q or that − == pr rp0 . , ,    =