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 . , ,    = + − = rp r qp q The set of possible positions of R will be the line that is parallel to the vector q (or that it is parallel to OQ) that passes through point P. 5(bii) ( ) 0 0 3 1 3 5 2 5 3 10 8 5 2 4 2 3 5 2 5 x y z x y z −  =  −  =  =  −                − =  − = − − + = −                      − + = − r p q r q p q r q p q The set of possible positions of R will be the plane that has a normal vector q (or that the normal vector is parallel to OQ) and it contains the point P ( )1, 2, 4− .

Suggested Solution A lvl H2 Math P1 6 By factor theorem, if iz k k=+ , ( ) ( ) ( ) ( )( ) ( )( ) ( ) ( ) 2 2 2 2 2 2 2 2 22 i 2 i 8i i 0 1 2i i 2 i 8 i 8 i 0 2i 2 i 8 i 8 0 4 i 2 i 8 i 8 0 2 8 i 4 8 0 k k k k t k k k t k k k t k k k k t k k t k k + + − + + = + + + − − + = + − + + = + − + + = − + + + − = We compare real and imaginary parts; Im: 24 8 0 2k k k− == . Re: ( ) ( ) 2 82 2 8 2 0 t t+ + =  = −− Method 1: ( ) ( ) ( )( )( ) ( ) ( ) ( ) 2 2 2 i 8i 8 0 8i 8i 4 2 i 8 2 2 i 8i 4 4i 2 2 i 12i 4 4i 4 or 4 2i 4 2i 262 2i or i 55 zz z z z z + − − = − + −= + += + +−= + −  + ++  = Hence the other root is 26 i55z = − + . Method 2: Let the other root be 0z . Sum of roots 8i 8 16 i2 i 5 5 −= − = ++ . 0 0 8 162 2i i 55 26 i55 z z + + = + = − + (Note: One may also use product of roots.) Hence the other root is 26 i55z = − + .

Suggested Solution A lvl H2 Math P1 7(i) 12 sin 4 d 2 cos 4 4x x x x c− = + + . 7(ii) ( ) 111 1 222 2 2 2 00 2 ππ π 00 1 0 ππ 2 2 112 sin 4 d cos 4 cos 4 d 44 1 sin 4104 4 2 16 00 . π π π 8 π π 4 π 8 4 x x x x x x x x x x − = − − − −      = − − − +       = − − + − =+  7(iii) ( )   2 π11 222 00 11 22 0 0 1 0 π π π2 π 2 sin 4 d 4 4sin 4 sin 4 d 1 cos84 cos 4 d 2 1 sin 8cos 8 4π π 0c . os 0 π 222 s 4 1 π πin 4π 11 2 π 2 0 0 28 2 9 4 π x x x x x xx x x xx  −− − = − + −= + +    + − +  −   = = + + −  = + − = +  

Suggested Solution A lvl H2 Math P1 8(ai) Let na be the A.P. with first term 4 and common difference d . We have, 15 4, 10aa == . 4 4 10 3 2 d d += = 30 3294 29 4 47.5 2ad = + = + = . 8(aii) Method 1: Number of terms from 21 to 50 50 21 1 30= − + = . And, 21 50 334 20 34, 4 49 77.522aa    = + = = + =       . Thus, ( )21 22 50 30 34 77.5 1672.52aa a++ + = + = . Method 2: ( ) ( ) ( ) ( ) 21 22 50 1 50 1 20 50 3 20 32 4 49 2 4 192 2 2 2 1672.5 aa aa a aa ++ = + − +        = + − +           =   +  + +  8(bi) Let nb be the G.P. with first term 4 and common ratio r. ( ) 4 5 41.6384 4 1.6384 as 0 5b r r r=  =  =  4 20.41 5 S == − 8(bii) ( ) ( ) ( ) 12 19.6 441 5 19.641 5 20 20 0.8 19.6 20 0.8 0.4 0.8 0.02 (Shown) n n n n nb bb   −    − − − + ++  −  Hence,

Suggested Solution A lvl H2 Math P1 ln 0.8 ln 0.02 ln 0.02 17.531 (5 s.f.)ln 0.8 n n  = Least 18n = .

Suggested Solution A lvl H2 Math P1 9(i) Denote , be the angles inclined by the lines 12, 2y x y x= = − with respect to the positive x-axis. As, ( ) 121 2 − = − , the lines are perpendicular and hence, π .2+= Moreover, by definition, ( ) 11t 1, taan 2 n 2 −− − = − = . Therefore, 11 2t 1 πtan2 2an− − − − = . (Shown) 9(ii) Note 0k  . For there to be intersection, the equation 2 1 1 3 4 k xx =++ has real solutions. 2 2 2 1 1 3 4 3 4 0 34 k xx kx x kx kkx += =+ − + − + + = As 0k  and that the equation has real solutions, it’s discriminant 0 . ( ) ( )( ) 2 2 2 3 4 4 0 9 4 16 0 4 16 9 0 91 022 19 22 kk kk kk kk k    − − − −+ −−   −+   −    Together with 0k  , we have 90 2k  . Hence, the required set is 9:0 2kk    . x y O