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 holds for  0nk + =   . That is, ( ) ( ) ( )( ) ( ) ( )( ) 1 1 f 1f + f d !! k r x rk k a r ax x a t x t t rk + = = − −  . Using integration by parts with ( )( ) ( ) 1 df ; d kk vu t x t x + = = − : ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) 111 12 11 12 1f d f 1 1 + f d 1 1 f 1 1 + f d 1 x xkkkk aa x kk a kk x kk a t x t t t x t k t x t t k a x a k t x t t k +++ ++ ++ ++ − =− − + − + =− + − +    ( ) ( )( )( ) ( )( )( ) ( )( )( )  ( )( )( ) ( )( ) ( ) ( ) ( ) ( )( )( ) ( )( )( ) ( ) ( )( )( ) 1 1112 1 1 11 2 1 12 1 ff ! 1 1 1 + f + f d ! 1 1 f ! f 1 + + f d 1 ! 1 ! f 1+ f d ! 1 ! k r r r x kkkk a k r r r k x kk k a k r x rk k ar ax x a r a x a t x t t k k k a xa r a x a t x t t kk a x a t x t t rk = ++++ = + ++ + + ++ = =− −− ++ =− −− ++ = − − +       which establishes the theorem for the case 1nk=+ and hence proves Taylor’s theorem. 3(iii) For x close to a, 0xa− . Then ( ) ( ) ( )( ) ( )( ) 2fff f + + 1! 2! aax a x a x a   −−= approximately. Substitute ( )f sinxx= , 1.6x= , π 2 a= into the above equation: 22 π π π 1 π π 1 πsin1.6 sin cos 1.6 sin 1.6 1 1.6 2 2 2 2 2 2 2 2  + − − − = − −                .

3(iv) Since ˆ0 xx where ˆx is some finite positive integer, for ˆ2nx , ˆˆ2 terms 2 terms ˆ2 ˆ !! ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ1 2 3 2 2 1 2 2 1 2 ˆ ˆ ˆ ˆ where ˆ1 2 3 2 nn x n x nx xx nn x x x x x x x x x x n k x x x xk x − − =     ++    =   The last inequality holds since ˆ ˆ ˆ 1, , ,ˆˆ2 1 2 2 2 0 x x x x x n  ++  . 3(v) For 0x= , lim lim 0 0! n nn x n→ → == . For each fixed x + , k is finite since ˆx is a finite number. As n→ , ˆ2 1 02 nx k −  → and 0 lim 0 lim 0!! nn nn xx nn→ →    = . For each fixed x − , let ,x y y +=−  . ( )1lim lim lim 0! ! ! n nnn n n n yxy n n n→ → → −= = = by the above result. We therefore conclude that , lim 0 ! n n xx n→   = . 3(vi) By (v), For all x , ( ) ( )( ) ( ) ( )( ) ( ) 1 1 11flim R lim f lim 0 1 ! 1 ! n n nn nn n n c xx x c nn + + ++ → → → = = = ++ . By Taylor’s theorem, ( ) ( )e T Rx nn xx=+ . For each x , letting n→ and putting 0a= in particular, gives ( ) ( ) 00 0 e lim T lim R lim !! n rr x nnn n n rr xxxx rr  → → → == = + = =  . Remark: In the above computation, we put in 0a= . For any arbitrary a , Taylor’s theorem gives ( ) ( ) ( ) ( ) 0 0 0 replace with 0 ee lim T lim R lim ! e on dividing by e! e as before! rn a x nnn n n r r x a a r rx a x x r xaxx r xa r x r → → → =  − = − = −= + = −= =    So the choice of a is arbitrary.

4(i) Since p is prime and 11 kp −  , ( )gcd 1, 1kp−= . There exist ,ab  such that ( )11pa k b+ − = . For any m , ( )( ) ( )( )1 1 1p a m k k b mp− − + − + = . Choose m large enough such that :0ky b mp= +  and ( ):1kx a m k= − − . Hence, there exist ,kkxy  with 0ky  such that ( )11kkpx k y+ − = . 4(ii) Since ( ) ( )1 | 1 kk k k y−− , ( )10kku k k y= −  (mod 1k− ). From part (i), ( )11kkpx k y+ − = implies ( )11 kky− (mod p), and so, ( )1kku k k y k= −  (mod p). 4(iii) Suppose 11 ji uu ij −− (mod p) for some 2 i j p   . Hence, in modulo p, we have the following. ( ) ( ) ( ) ( ) ( ) 11 1 1 by (ii), mod ij k j u i u j i i j u k p ij i ij j ij −  − −  −  −  −  Since 2 i j p   , ij= and so for each 2 kp , each 1 ku k− is distinct in modulo p. 4(iv) Let 1 1v = and for each 21 kp  − , let kv be the remainder when 1 ku k− is divided by p. Note that 1 p p u py pp =− (mod p), and so, we define pvp= . Only 1 1v  (mod p). From part (i), for all 2 kp , ( )11 kky− (mod p) and it is clear that 1ky  . Hence, 1kkv ky= (mod p) for all 2 kp . Together with the result in part (ii), we see that all kv ’s, 1 kp , are unique and form the set  1, 2, , p . ( ) ( ) ( ) ( )( ) ( ) 23 12 ...... 1! 2 3 ... mod by part i1! mod k k u u uv v v k k pk kp = −    − = This shows that all 1v , 12vv , 1 2 3v v v , …, and 12 ... pv v v leave different remainders when divided by p. Alternative to working in grey: ( )1 1 mod1 k k k k k k uv ky y px y pk= = = + −  +− Hence ( )1 modkvp since :ky b mp=+ in (i) and b is not a multiple of p Or

From (i), there exist ,kkxy  , :0ky b mp= +  such that ( )11kkpx k y+ − = . Hence ( )11 kky− (mod p). 1kkky y+ (mod p) ( )1 mod p since :ky b mp=+ in (i) and b is not a multiple of p 4(v) A permutation is 1, 2, 7, 5, 4, 10, 3, 9, 8, 6, 11. Check that 1v , 12vv , 1 2 3v v v , …, and 1 2 11...v v v in modulo 11 are unique: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. 1 1v = ( ) ( ) ( ) ( ) 2 12 2 1 ! 2 mod112 1 ! 2 mod11 uvv = −  − = 2 2v= ( )2 7 3 mod11 3 7v= ( ) ( ) 2 7 5 3 5 mod11 4 mod11      4 5v= ( )