`Iff(x)=int_a^x[f(x)]^(-1)dxa n dint_a^1[f(x)]^(-1)dx=sqrt(2),t h e n` `f(2)=2` (b) `f^(prime)(2)=1/2` `f^(prime)(2)=2` (d) `int_0^1f(x)dx=sqrt(2)`

Options

A. `f(2)=2`
B. `f'(2)=1//2`
C. `f^(-1)(2)=2`
D. `int_(0)^(1)f(x)dx=sqrt(2)`

Answer

A, B, C

Explanation

`-f(x)=int_(a)^(x)1/(f(x))dx`
or `f'(x)=1/(f(x)) . 1-0` or `f(x)f'(x)=1`
or `intf(x)f'(x)dx=int1dx`
or `1/2[f(x)]^(2)=x+c`……………..1
Now given that `int_(a)^(1)[f(x)]^(-1)dx=sqrt(2) `or `f(1)=sqrt(2)`
Thus,from `(1),1/2[f(1)^(2)]=1+c` or `c=0`
or `f(x)=+-sqrt(2x)`
But `f(1)=sqrt(2)` or `f(x)=sqrt(2x)` or `f(2)=2`
Also `f'(x)=1/(sqrt(2x))` or `f'(2)=12 `
`int_(0)^(1)f(x)dx=int_(0)^(1)sqrt(2x)dx=[((2x)^(3//2))/3]_(0)^(1)=((2)^(3//2))/3`
Also `f^(-1)(x)=(x^(2))/2` or `f^(-1)(2)=2`