RVHS 9758 2025 Prelim P1 Solutions

River Valley High School 1 2025 H2MA Prelim Paper 1 1 Solution [4] Complex Number (a) Method 1 ( ) ( ) ( ) ( ) 32 f 1 0 1 1 1 0 0 a b c d a b c d −= − + − + − + = − + − + = 0b c d a− + − = ------------------- (1) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 32 f 4 0 4 4 4 0 52 47 15 8 4 0 i a i b i c i d a i b i c i d += + + + + + + = + + + + + + = Comparing real parts, 15 4 52 0b c d a+ + + = ----------------- (2) Comparing imaginary parts, 8 47 0b c a+ + = -------------------(3) Solving (1), (2) and (3) using the GC, 7ba=− , 9ca= and 17da= Method 2 Since all the coefficients of ( )f x are real, 4i+ and 4i− are conjugate roots of ( )f0 x = . ( ) ( ) ( )     32 2 32 32 f 4 i 4 i 1 8 17 1 7 9 17 7 9 17 x ax bx cx d a x x x a x x x a x x x ax ax ax a = + + + = − + − − +     = − + + = − + + = − + + 7ba=− , 9ca= and 17da=

River Valley High School 2 2 Solution [6] Inequality ( )( ) ( )( ) ( ) ( )( ) ( )( ) ( ) ( )( ) 2 2 2 2 95 11 9 5 1 011 9 4 5 011 44 011 2 011 x xx xx xx xx xx xx xx x xx +−+ − + − +− − − − + +− ++ +− + +− 1x− or 1x  , 2x− (Alternatively, 2x− or 21 x−  − or 1x  ) 2 9 e 5 .1 e e 1 x xx +−+ Replace x with ex in the previous inequalitry, e1x − or e1x  (no solution since e0x  for x ) 0x (Note e2x − is always true) Thus 0x . Alternative presentation: e2x − or 2 e 1x−  − or e1x  (no solution since e0x  for x ) 0x Thus 0x .

River Valley High School 3 3 Solution [6] Complex Numbers ( ) 2 1 2 1 7 0z i z i− + + + = In the above equation, 1 , 1 2i, 1 7ia b c= =− − = + ( ) ( )( ) ( ) 2 2 4 2 1 2i 1 2i 4 1 1 7i 21 1 2i 4i 3 4 28i 2 1 2i 7 24i= 2 b b acz a −  −= +  − − − += +  − − −= +  − − We want to find 7 24i−− . Let 7 24i i , ,x y x y− − = +  ( ) ( ) ( ) 2 22 7 24i i ---(*) 7 24i 2 i xy x y xy − − = + − − = − + Comparing real and imaginary parts, 22 7 ---(1) 2 24 12 ---(2) xy xy x y − =− =− −= Substitute (2) into (1): ( )( ) 2 2 2 2 42 42 22 12 7 144 7 144 7 7 144 0 16 9 0 yy yy yy yy yy − − =− − =− − =− − − = − + = 22 22 16 0 or 9 0 16 9 4 (reject as ) yy yy yy − = + = = =− =  When 4, 3yx= =− . When 4, 3yx=− =

River Valley High School 4 The square roots of 7 24i−− are 3 4i− and 3 4i−+ . Hence, ( )1 2i 3 4i 2z +  −= 1 2i 3 4i 1 2i 3 4i or 22 2 i or 1 3i zz zz + + − + − +== = − =− +

River Valley High School 5 4 Solution [6] Summation (a) ( )( ) ( )( ) ( )( ) ( ) 1 1 11 1 2 1 12 1 1 1 1 2 1 1 111 ( 1) 1 (1)(1 1) 111 22 11 22 N r kN k N k rr kk kk N N N = −= −= + = ++ = − + − + = + = − − + + + = − − + =− +    (b) 11 1 1 1 11 118 ( 1) ( 1) 1 1 18 ( 1) ( 1) ( 1) 1189 ( 1) ( 1) k r k r kk r r r k rr r r r r r r r r r r r r r r  = + =  = = =  == =++  −= + + + =++      As n→ , 1 01n →+ . Hence 1 1 1( 1)r rr  = =+ . ( ) 1 1 18 1 9 ( 1) 18 ( 1) 9 k r k r rr rr = = = + =+   181 19 8 19 8 k k k k −= + =+ =

River Valley High School 6 5 Solution [7] Integration by parts (a) ( ) ( ) 42 2 2 2 1 12dd 216 4 1 sin24 xx xx x x x c− = − − =+   (b) 2 1 2 4 1dLet sin , 2 4 d du 1 d2 16 xvux x x vxx x − ==   == − 2 2 2 2 11 4 2 2 3 1 4 2 2 3 1 4 22 14 22 14 sin d sin d2 4 4 4 2 16 1sin d4 4 2 16 1 1 4sin d4 4 2 4 16 11 24sin 16 144 2 1sin 164 4 4 x x x x x x xx x x x x x x x x x x x xx xc xx xc −− − − − −    = −     −    =−  −  −= − −   − − = − − + = + − +    Alternative 1

River Valley High School 7 2 1 2 4 dLet sin , 4 d 2 1 du 2 d4 1 16 x v xu x x xvx x − ==   == − 2 1sin d24 xx x−   = 2 2 2 1 4 1 2sin d4 4 4 1 16 xx x x x x −  − −  = 3 22 1 4 1 1 4sin d4 4 2 1 16 xxx x x − − + −  = 4 22 1 11 16sin 14 4 2 2 x xx C− − ++ = 2 2 4 1sin 14 4 16 x x x C−  + − + Alternative #2 Consider ( ) 1sin d xx−  ( )1 2 dLet sin , 1 d du 1 d 1 vux x vxx x −== == −

River Valley High School 8 ( ) ( ) ( ) ( ) ( ) ( ) 11 2 1 2 1 112 2 12 sin d sin d 1 12sin d 2 1 1 2sin 1 1 2 sin 1 xx x x x x x xx x x x x x x c x x x c −− − −+− − =− − −= − −  − = + − + = + − +   Then ( ) ( ) 1 1 2sin d sin 1f f x f f f c−−  = + − + where f is a function in x . 22 2 2 2 11 22 14 sin d sin 12 4 4 4 4 1sin 164 4 4 x x x x x xc xx xc −− −      = + − +           = + − +  6 Solution [6] Abstract Vectors mn is a vector perpendicular to both m and n . Thus, ( ) 0.  =m m n (a) ( )1 2OM = a + c MR is perpendicular to p, so MR is parallel to .a × b ( ) ( ) ( ) ( ) , for 1 2 MR OR OM OR    =   − =  = +  ab ab a + c a b Thus, R is a set of points on the line given by the equation ( ) ( )1 ,.2 = +  r a + c a b (b) : , , .p    =r a + b At point of intersection,

River Valley High School 9 ( ) ( ) ( ) 22 1 (*)2 Scalar product (*) with , 11 22 11 ( 2)22 1 2         +  = +  +   = +  + − = =− a + c a b a + b a a a c a a b a a b ( ) 2 Scalar product (*) with , 11 22 1 (4)2 2       +  +   =  + = = b b a b c b a b b a b Position vector of the point of intersection is 1 2.2− a + b Alternative Since a, b are perpendicular unit vectors parallel to plane p, ab would be perpendicular to a and b (and thus to p). Thus, OM =    +a + b a b for some , , .   Since X is the foot of perpendicular from M to p, OX = a + b where ( ) ( ) 1 2 11 22 11(1) 222 1 2 OM = =+ =+ = + − =− a a c a a a c a

River Valley High School 10 ( )1 2 11 22 11(0) (4)22 2 OM = =+ =+ =+ = b a c b a b c b Position vector of the point of intersection is 1 2.2− a + b 7 Solution [7] Differentiation and Applications (a) ( ) 22 4 d d d2 2 4 d d d d2 4 2 0 d x y y xy y y yx y y x x x x yy x x y x + − = + − = + − − + − = ( ) d2 4 2 d yy x y x x− − = − (b) Area of triangle OMN, A ( )( )1 3 1.52 yy== (c) 1.5 d d d d d d d1.5 d 21.5 24 Ay A A y x y x y x yx yx = = =  −=  −− (d) At stationary point, d 0d 21.5 024 2 (*) A x yx yx yx =  − = −− = −−− ( ) ( ) ( ) 22 2 2 4 2 2 3 8 0 x x x x x xx + − = −=