The value of \(c\) of Lagrange's mean value theorem for \(\mathrm{f}(x)=\sqrt{25-x^2}\) on \([1,5]\) is

\(\begin{aligned}
& \mathrm{f}(x)=\sqrt{25-x^2} \\
& \therefore \quad \mathrm{f}^{\prime}(x=\mathrm{c})=\frac{-2 \mathrm{c}}{2 \sqrt{25-\mathrm{c}^2}} \\
&=\frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}
\end{aligned}\)
Applying Lagrange's mean value theorem, we get
\(\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(1)-\mathrm{f}(5)}{1-5} \\
& \therefore \quad \frac{\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}=\frac{\sqrt{25-1}-\sqrt{25-5^2}}{1-5} \\
& \therefore \quad \frac{-\mathrm{c}}{\sqrt{25-\mathrm{c}^2}}=\frac{-\sqrt{24}}{4}
\end{aligned}\)
\(\begin{array}{ll}
\therefore & 4 \mathrm{c}=\sqrt{24} \cdot \sqrt{25-\mathrm{c}^2} \\
\therefore & 16 \mathrm{c}^2=24\left(25-\mathrm{c}^2\right) \\
\therefore & \mathrm{c}^2=15 \\
\therefore & \mathrm{c}= \pm \sqrt{15}
\end{array}\)
Since \(c=-\sqrt{15}\) does not lie in \([1,5]\) \(c=\sqrt{15}\)