If $\mathrm{f}(\mathrm{x})$ is a function such that…
If $\mathrm{f}(\mathrm{x})$ is a function such that $\mathrm{f}^{\prime}(\mathrm{x})=\sqrt{\mathrm{f}^2(\mathrm{x})-1}$ and $f(0)=1$, then $f(1)=$
- $\frac{\mathrm{e}^{-2}+1}{2 \mathrm{e}}$
- $\frac{\mathrm{e}^2+1}{2 \mathrm{e}}$
- $\frac{\mathrm{e}^2-1}{2 \mathrm{e}}$
- $\frac{\mathrm{e}^{-2}-1}{2 \mathrm{e}}$
Solution
$
\begin{aligned}
& \text {} \because \mathrm{f}^{\prime}(\mathrm{x})=\sqrt{\mathrm{f}(\mathrm{x}) \quad 1} \\
& \Rightarrow \frac{\mathrm{f}^{\prime}(\mathrm{x})}{\sqrt{\mathrm{f}^2(\mathrm{x})-1}}=1
\end{aligned}
$
Integrating both sides, we get
$
\begin{aligned}
& \Rightarrow \log \left[f(x)+\sqrt{f^2(x)-1}\right]=x+c \\
& \text { At } x=0, \log \left[f(0)+\sqrt{f^2(0)-1}\right]=0+c .....(i)\\
& \Rightarrow \log [1+\sqrt{1-1}]=0+c \Rightarrow 0=c
\end{aligned}
$
Putting the value of $\mathrm{c}$ in $\mathrm{eq}^{\mathrm{n}}$ (i), we get
$
\begin{aligned}
& \Rightarrow \log \left[\mathrm{f}(\mathrm{x})+\sqrt{\mathrm{f}^2(\mathrm{x})-1}=\mathrm{x}+0\right. \\
& \text { At } \mathrm{x}=1: \log \left[\mathrm{f}(1)+\sqrt{\mathrm{f}^2(1)-1}\right]=1 \\
& \Rightarrow \mathrm{f}(1)+\sqrt{\mathrm{f}^2(1)-1}=\mathrm{e} \\
& \Rightarrow \sqrt{\mathrm{f}^2(1)-1}=\mathrm{e}-\mathrm{f}(1) \\
& \Rightarrow \mathrm{f}^2(1)-1=\mathrm{e}^2+\mathrm{f}^2(1)-2 \mathrm{ef}(1) \Rightarrow \mathrm{f}(1)=\frac{1+\mathrm{e}^2}{2 \mathrm{e}}
\end{aligned}
$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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