2023 TJC VJC NYJC H3 Math Prelim Solution

2023 NYJC/TJC/VJC Prelim H3 Math Paper 1 1(i) For all t , 2 1 ( ) 0 n ii i a t b = − 2 2 2 1 1 1 20 n n n i i i i i i i a t a b t b = = =     − +          Note that coefficient of 2t , 2 1 0 n i i a =  since ia is non-zero. For the quadratic function to be non- negative for all values of t, we have discriminant 0 . 2 22 1 1 1 2 22 1 1 1 4 4 0 n n n i i i i i i i n n n i i i i i i i a b a b a b a b = = = = = =     −                            1(ii) Given 0 and 0i i i i i i b b bp q p q a a a    −  −            . Thus 2 2 2 2 0 1 ( )( ) 0 ( ) 0 since 0 ii ii i i i i i i i i i i bbpq aa pa b qa ba pqa p q a b b a   − −       − −  − + +   Taking summation 22 1 22 1 1 1 ( ) 0 ( ) ( ) ( ) ( ) 0 k i i i i i k k k i i i i i i i pqa p q a b b pq a p q a b b = = = =  − + +   − + +      22 1 1 1 () n n n i i i i i i i p q a b b pq a = = =  +  +   1(iii) Since im a M and im b M , thus i i b a mM Mn  From (ii), let mp M= and Mq m= so that 22 1 1 1 1 n n n i i i i i i i m M m M a b b a M m M m = = = = +  +               . By AM-GM inequality, 22 11 22 112 nn ii ii nn ii ii ab ab== == +    Thus 2 2 2 2 1 1 1 1 1 2 n n n n n i i i i i i i i i i i mM a b b a a b Mm = = = = = +  +       . Squaring both sides: 22 22 1 1 1 1 4 n n n i i i i i i i mM a b a b Mm = = = +      (shown)

2(i) d f ( , ) f ( , ) d y x y tx ty x == By letting 1t y = , we have d f ,1d yx xy =  g x y =  (shown). 2(ii) 22 2 2 2d2 e ( 2 )e d xx yy xxy y y x y            = + +  2 2 2 2 2 d 2 e d ( 2 )e x y x y y xy x y y x     = ++ Let 2 2 2 2 2 2ef ( , ) ( 2 )e x y x y xyxy y y x     = ++ .   22 22 2 2 2 2 2 2 2( )( )e 2 ef ( , ) f ( , ) ( ) ( ) 2( ) e ( 2 )e tx x ty y tx x ty y tx ty xytx ty x y ty ty tx y y x                       = = = + + + + Hence, 22 2 2 2d2 e ( 2 )e d xx yy xxy y y x y            = + + is a homogeneous differential equation. 2(iii) Let xu x yu y =  = , dd dd xu uy yy =+ . 222 2 2 2 2 d2 e ( 2 )e d uu uuy u y y y u y y + = + +  223 2 2 d2 e e d uu uuy y y y =+ 2 2 2 e 1 d d 1e u u u uy y = + 2 ln(1 e ) ln ln lnu y c cy+ = + = 2 (1 e )uyA=+ where 1A c= General solution: 2 1e x yyA  =+    . Tangent at (4, 2)− is perpendicular to the line y mx= . When 4x= , 2y=− , d 1 d dd yx m x m y =−  =− Substitute into DE:  442( 8)e ( ) 4 4 2(16) em− − = + + 4 4 1 (1 9e ) 4e m = + .

3(i) ( ) ( )  ( ) ( )f d f f f x x aa t t t x a = = − which on rearranging gives ( ) ( ) ( ) ( ) ( )( ) 0 f f f d 1 f f d 0! x a x a x a t t a t x t t =+ = + −   which implies Taylor’s theorem holds for the case 0n= . 3(ii) Assume Taylor’s theorem ho