R- 0 R is a differentiable function such that: ₁^ x y f(t) d t=y ₁^x f(t) d t+x ₁^y f(t) d t x, y, (R)- 0 and f(1)=1 . Function g is defined as: g(x)=- (e^ f (x^2 )-1 + (e^ f (1 / x^2 )-1 ) ) If I(P)= _P^ 1 / P e^ g(x) d x then, the value of _ P 0⁺ I(P)= a b (a and b R) . Find the value of [a+b] . (You may use _ - ^ e^ -t^2 d t= ) [Note: [k] denotes greatest integer function less than or equal to k .]

Step-by-step solution

₁^ x y f(t) d t=y ₁^ f(t) d t+x ₁^y f(t) d t x, y, (R)- 0 and f(1)=1 Differentiate both sides w.r.t. x y f(x y)=y f(x)+ ₁^y f(t) d t Put x=1 , we get y f(y)=y+ ₁^y f(t) d t ( f(1)=1) Again differentiate w.r.t. y aligned & & f(y)+y f^ (y) & =1+f(y) & & y f^ (y) & =1 & & f^ (y) & = 1 y aligned Hence, f(y)=1+ y g(x)=- (x^2+ 1 x^2 ) Now, we have to find I= ₀^ e^ - (x^2+ 1 x^2 ) d x ...(1) Replace x 1 x aligned I & = ₀^ e^ - (x^2+ 1 x^2 ) 1 x^2 d x 2 I & = ₀^ e^ - (x^2+ 1 x^2 ) (1+ 1 x^2 ) d x 2 I & = ₀^ e^ - (x- 1 x )-