ASRJC H3 Math 2024 Prelim Solutions

1 2024 ASRJC H3 Math Prelim Solutions 1 (a) 2d sin sin 0d xxx for (0, ).x π 2 2 d1ln 0d xxx for (0, ).x (i) Given that A , B and C are angles of a triangle, then π0 2A , π0 2B and π0 2C . Applying Jensen inequality, 3 11 11 sin( ) sin33 n kk kk xx ==     ( )11sin sin sin sin33 A B C A B C + +  + +    ( )11sin sin sin sin π33 A B C + +    1 π3sin sin sin sin3 3 2 A B C + +  =  33sin sin sin 2A B C+ +  (Shown) (ii) Let f ( ) lnxx . Then 1 2 1 2 11ln ln ln ln nna a a a a ann 1 12 12ln( ) ln nn n a a aa a a n 1 12 12()n n n a a a a a an (b) (i) 1 1 2 3 5 9 2 1( ) ...... .1 2 4 8 2 n nPn − − +=      Using AM-GM inequality, ( ) ( ) ( ) 1 1 1 1 1 11 11 2 3 5 9 2 1 .....2 3 5 9 2 1 1 2 4 8 2.....1 2 4 8 2 1 1 1 1 11 1 1 1 ..... 11 2 4 8 2() 1 1 1 11 ..... 2 4 8 2() n n n n n n n n n n Pn n n Pn n − − − − − − ++ + + + + +               + + + + + + + + + +                          + + + + + +              

2 ( ) ( ) ( ) 1 1 1 1 111 2 11 2() 12 2() 2() 2( ) 1 n n n n n n n Pn n n Pn n nPn n Pn n −  −  + −  +−  + + (ii) ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 23 23 2 2( ) 1 ( 1)( 2)... 2 12 ( 1) 2 ( 1)( 2) 2 21 ..... 2! 3! ! 1 1 11 2 2 2 ..... 22! 3! ! 1 1 11 2 2 2 ..... 2 ......2! 3! ! e n n n n Pn n n n nn n n n nn n n n n n n n + −−− − −       = + + + + +                + + + + +  + + + + + +  2 ( )( ) ( ) ( )( ) 1 2 2 2 2 1 1 1 2 1 2 22 2 1 1 2 2 2 1 2 2 2 1 44 44 4 4 nn n n n n n n n n n n n n n n n n n n n n n n n n n n FF w w w w w w w w w w w w w w w w w w w w w w w w w w − − − − − − − − − − − − − − − − − − − = + − − − + = − − + = − + − − = − + − ( )( )1 2 2 1 4n n n n n n nF F w w w w w− − − −− = − + − ----------------------(1) (a) Let nw be nu . Then 21 40n n nu u u −−+ − = . So we have 1 0nnFF −−= for 2n by result above. ie 1nnFF −= for 2n ( )( ) 2 2 2 2 1 1 0 1 0 4 2 1 4 2 1 3F u u u u= + − = + − =− Therefore 3nF =− for all 1.n So we have 22 11 4 3 for 1.n n n nu u u u n−−+ = − 

3 (b)(i) Let nw be nv . ( ) 22 11 2 11 2 1 1 1 4( )(1) 3 4 4 0 20 2 vv vv v v + = − − + = −= = 22 11 4 3 for 1.n n n n nF v v v v n −−= + − =−  From (1), we have 2 0nnvv −−= or 21 40n n nv v v −−+ − = for 2.n . (ii) Since 1,2,1,2,…. satisfies 2 0nnvv −−= for 2.n , then from (1), is constantnF . Also since 0 1v = and 1 2v = , then 3nF =− . So the sequence satisfies (*). (iii) Take the sequence 1,2,7,2,1,2,7,2,….. that satisfies 2 0nnvv −−= for odd 2n and 21 40n n nv v v −−+ − = for even 2n with period 4. Then by (1), is constantnF and 3nF =− . So the sequence satisfies (*). 3 For all real values of t, ( ) 2 f ( ) g( ) 0t x x + Thus we have ( ) 2 f ( ) g( ) d 0 b a t x x x+ ( ) ( ) 222 f ( ) 2 f ( )g( ) g( ) d 0 b a t x t x x x x + +  ( ) ( ) 222 f ( ) d 2 f ( )g( ) d g( ) d 0 b b b a a a t x x t x x x x x + +    From above, we have ( ) ( ) ( )( ) ( )( ) ( ) ( )( ) ( )( ) 2 2 22 2 22 f ( ) g( ) d 0 2 f ( )g( ) d 4 f ( ) d g( ) d f ( )g( ) d f ( ) d g( ) d b a b b b a a a b b b a a a t x x x x x x x x x x x x x x x x x +          Equality holds when ( ) 2 f ( ) g( ) d 0 b a t x x x+= . Since f and g are continuous this means that we must have f ( ) g( ) 0t x x += for all real x, i.e. f is a scalar multiple of g.

4 (i) Setting f ( ) 1x = and g( ) xxe= in(*) Since f is not a scalar multiple of g, we have ( ) ( ) ( ) ( ) ( ) ( )( )( ) ( )( ) 2 2 2 22 2 1 1 2 1 2 1 2 b b bxx a a a b a b a b a b a b a b a b a e dx dx e dx e e b a e e e e b a e e e e e e b a e e  −  − − −  − − + −  − +    Choosing 0a= and bt= gives ( )111 2 11 12 tt t t e t e e te −  + − + (ii) Setting f ( ) 1x = and g( ) sinxx= and 0a= and 2b = , (*) becomes   21 1 1 2 2 2 0 0 0 21 1 2 2 00 1 2 0 sin 1 sin 1sin cos 22 sin 2 xdx dx xdx xdx x xdx                          − =        Setting ( ) cosf x x= and ( ) 1 4( ) sing x x= and 0a= and 2b = , (*) becomes ( ) ( ) 21 1 1 1 22 2 2 4 0 0 0 21 115 2 224 00 0 1 12 2 0 0 1 2 0 cos sin cos sin 41 sin 1 cos 2 sin52 16 1 sin 2 sin25 2 2 16 1 . sin25 2 2 sin x x dx x dx xdx x x dx xdx xx xdx xdx x                            +       −            1 2 0 64 25dx   Combining, 1 2 0 64 sin .25 2 x dx    

5 4 (i) 2 2 d d d dg( ) ( ) g'( )d d d d y u y yu x y g x x yx x x x= +  = + + ( ) ( ) 2 2 2 2 2 2 d f ( ) h( )d dd ( ) g'( ) f ( ) h( )dd d d d ( ) g'( ) f ( ) g( ) h( ) d d d dd g( ) f ( ) g '( ) f ( )g( ) h( ) dd u x u xx yy g x x y x u xxx y y yg x x y x x y xx x x yy x x x x x y xxx +=  + + + =  + + + + =   + + + + = (ii) 22g( ) f ( ) 2 ( ) 2 ( )x x f x g x xx+ = +  = + − 4g '( ) f ( )g( ) 24g '( ) 2 ( ) g( ) x x x x x g x x xx +=  + + − = This is the first order differential equation satisfied by g( ).x If f ( ) , nx kx= then 2g( ) 2 nx kx x= + − and 1 2 2g'( ) nx knx x −=− − Subst into the above first order de for g(x), 1 2 1 2 1 2 2 2 2 2 4 2 2 2 2 4 n n n n n n n knx kx kxx x x knx kx kx k x x − + + + + − − + + − = − − + + − = Case 1: 2 2 2 0n n n+ = +  = . Rejected as there is a constant term -2 that cannot be eliminated Case 2: 2 2 1 1n n n+ = +  =− . Comparing coefficients of terms: 2 4 2 Checking constant terms: 1 2 4 4 0 x k k LHS RHS =  = =− + + − = = So 2f ( ) ,x x= 22g( ) 2 2x xx= + − =

6 ( ) 2 2 dUsing f ( ) h( ), we haved d 2 6 d d2 3d ln 3 2ln 3 3 u x u xx u ux x x u uxx u x C Au x Au x += += =− − − = + −= =− Using d g( )d y x y ux+= and we are given 4y= and 5dy dx =− at 1x= ( )5 2 4 3 uu− + =  = at 1x= . 0 ( ) 3A u x =  = . 2 2 22 dUsing g( ) , we haved d 23d d 32d 1 ln 3 22 ln 3 2 2 32 When 1, 4 5 35 22 x x y x y ux y yx y yx y x D y x E y Be x y B e ye − −+ += += =− − − = + − =− + −= = =  =− =+

7 5 ( )2 1 2 2.3... 1rp− = − Claim: All prime factors of 21r− are greater than p. Suppose there exists a prime factor 11 such that 2p p p  . Then 11| (2 -1) and | 2(2.3..... )p r p p which implies 1 |1p . (contradiction) Now ( )2 1 2 2.3... 1 1( 4) 3( 4)r p mod mod− = − −  ie 21r− is of the form 43n+ for some integer n. Since 21r− is odd, the only possible prime factors are of the form 4 1 or 4 3kk++ for integers k. But ( )( ) ( )1 2 1 2 1 24 1 4 1 4 4 1k k k k k k+ + = + + + ie of the form 41k+ Hence there must exists a prime factor of 21r− of the form 43k+ ie congruent to 3 modulo 4 Therefore there exists a prime factor q of 21r− such that qp and ( )3 mod 4q . Suppose there is a finite number of primes of the form 43n+ and p is the largest prime of this form. From result above, 21r− has a prime factor q such that qp and ( )3 mod 4q ie q is of the form 43k+ for some integer k (contradiction) Hence there is an infinite number of primes of the form 43n+ , where n is an integer. 214 3 3 kn ++= where k is an integer ( ) ( ) 24 3 3 1 3 9 1kkn= − = − ( )3 9 1 4 k n − = Since ( ) ( )( ) 129 1 9 1 9 9 ... 1k k k −−− = − + + + which is divisible by 4, therefore ( )3 9 1 4 k n − = is always an integer for all non-negative integer k. Therefore ( )3 9 1 4 k n − = where k