If f(x) = ₁^x ₂ t t(1+t^2) dt and f( ) + f ( 1 ) = 8 2 for some > 1 , then the value of is

Step-by-step solution

f ( 1 ) = ₁^ 1/ ₂ t t(1+t^2) dt Let t = 1 u dt = - 1 u^2 du f ( 1 ) = ₁^ - ₂ u 1 u (1+ 1 u^2 ) (- 1 u^2 ) du = ₁^ u ₂ u 1+u^2 du Now, f( ) + f ( 1 ) = ₁^ ₂ t t(1+t^2) dt + ₁^ t ₂ t 1+t^2 dt = ₁^ ( 1 t(1+t^2) + t 1+t^2 ) ₂ t dt = ₁^ 1+t^2 t(1+t^2) ₂ t dt = ₁^ ₂ t t dt = 1 2 ₁^ t t dt = 1 2 [ ( t)^2 2 ]₁^ = ( )^2 2 2 Given f( ) + f ( 1 ) = 8 2 ( )^2 2 2 = 8 2 ( )^2 = 16 ( 2)^2 = (4 2)^2 = ( 16)^2 Since > 1 , > 0 = 16 = 16 . Answer: 16