Let `f` be a real valued functional defined on the interval `(-1,1)` such that `e^(-x)f(x)=2+int_(0)^(x)sqrt(t^(4)+1)dt`, for all `x epsilon(-1,1)` and let `f^(-1)` be the inverse function of `f`. Then `f^(-1))'(2)` is equal to

Options

A. 1
B. `1//3`
C. `1//2`
D. `1//e`

Answer

B

Explanation

We have,
`e^(x)f(x)=2+overset(x)underset(0)int sqrt(t^(4)+1)dt " "`……….(i)
Differentiating both sides w.r. to x, we get
`rArr- e^(-x)f(x)+e^(-x)f'(x)=sqrt(x^(4)+1) " "`.......(ii)
Now,
`fof^(-1)(x)=x`
`rArr (d)/(dx)(fof^(-1)(x))=1`
`rArr (d)/(dx){f(f^(-1)(x))}=1`
`rArr f'(f^(-1)(2))(f^(-1))'(2)=1" "`[Puttingx=2]
`rArr f'(0)(f^(-1))(2)=1 " "[:' f(0)=2 rArrf^(-1)(2)=0]`
`rArr (f^(-^(N)1))'(2)=(1)/(f'(0))["Putting " x=0 in (ii),
-f(0)+f'(0)=1 rArr f'(0)=3]`
`rArr (f^(-1))'(2)=(1)/(3)`