JEE Advanced Mathematics Question (2024) — Solution

Step-by-step solution

f(x+y)=f(x)+f(y) ...(1) f ( nx )= nf ( x ) n N ...(2) Now put y=-x in eq.(1) aligned & f ( x )+ f (- x )= f (0) \ f (0)=0\ \\ & f (- x )=- f ( x ) aligned f is odd function from eq. (2) from eq. (2) aligned & f(-n x)= nf (-x) \\ & f (- nx )=-n f( x ) \\ & aligned f ( mx )= mf ( x ) m Z ...(3) from eq. (2) and eq. (3) f ( nx )= nf ( x ) n Z ...(4) Now put x= p q where p , q Z , q 0 f ( np q )= nf ( p q ) n Z put n = q f(p)=q f ( p q ) pf (1)= qf ( p q ) from eq.(4) Let f (1)= a then pa = qf ( p q ) f ( p q )= ap q f ( x )= ax x Q Now, f ( -3 5 )= a ( -3 5 )=12 a =-20 f ( x )=-20 x x Q ...(5) From the given functional equation it is not possible to find a unique function for irrational values of ' x ', there are infinitely many such functions satisfying given functional equation for irrational values of x, but in this problem we finally need the function at rational values of ' x ' only. So, for rational values of x we are getting a unique function mentioned in (5). Now, g(x+y)=g(x) g(y) n ( g ( x + y )= n ( g ( x ))+ ( g ( y )) Let ( g ( x ))= h ( x ) array ll & h ( x + y )= h ( x )+ h ( y ) \\ & h ( x )= kx x Q array g ( x )= e ^ kx x Q ...(6) and g ( -1 3 )= e ^ - K 3 =2 K =-3 2 K = ( 1 8 ) g ( x )= e ^ ( 1 8 ) x = ( 1 8 )^ x =2^ -3 x x Q aligned & Now, f ( 1 4 )=-5, g(-2)=2^6=64 \\ & g(0)=1 aligned aligned & So ( f ( 1 4 )+ g (-2)-(8) g(0) ) \\ & =(-5+64-8)(1)=51 \\ & aligned