Practicing Success
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) = |
$\sqrt{2 x}$ $\sqrt{2 \log _e x}$ $\sqrt{3 x-1}$ none of these |
$\sqrt{2 x}$ |
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, $f(1)=\sqrt{2} \Rightarrow \sqrt{2}=\sqrt{2+C} \Rightarrow C=0$ ∴ $f(x)=\sqrt{2 x}$ |