If $f(x)=\int\limits_0^x\{f(t)\}^{-1} d t$, and $\int\limits_0^1\{f(t)\}^{-1} d t=\sqrt{2}$, then f(x) =
Answer & explanation
Correct answer: option 1
We have,
$f(x)=\int\limits_0^x\{f(t)\}^{-1} d t$
$\Rightarrow f'(x)=\{f(x)\}^{-1}$
$\Rightarrow f'(x) f(x)=1$
$\Rightarrow 2 f(x) f'(x)=2 \Rightarrow\{f(x)\}^2=2 x+C \Rightarrow f(x)=\sqrt{2 x+C}$
Now,
$f(1)=\sqrt{2} \Rightarrow \sqrt{2}=\sqrt{2+C} \Rightarrow C=0$
∴ $f(x)=\sqrt{2 x}$