⇒Application Of Derivative

Correct Option is (A)

$\begin{array}{rl}\mathrm{f}(x)& =x\sqrt{x+6}\\ \therefore {\mathrm{f}}^{\prime }(x)& =x\left(\frac{1}{2\sqrt{x+6}}\right)+\sqrt{x+6}(1)\\ & =\frac{x}{2\sqrt{x+6}}+\sqrt{x+6}\end{array}$

Since $\mathrm{f}(x)$ satisfies all the conditions of Rolle's Theorem,

There exists $c\in (-6,0)$ such that

$\begin{array}{rl}& f^{\prime }(c)=0\\ & \Rightarrow \frac{c}{2\sqrt{c+6}}+\sqrt{c+6}=0\\ & \Rightarrow c+2c+12=0\\ & \Rightarrow c=-4\end{array}$

Correct Option is (C)

$\begin{array}{rl}& \mathrm{f}(x)=\log x\\ & \Rightarrow {\mathrm{f}}^{\prime }(x)=\frac{1}{x}\end{array}$

By Lagrange's Mean value theorem,

$\begin{array}{rl}& {\mathrm{f}}^{\prime }(\mathrm{c})=\frac{\mathrm{f}(\mathrm{e})-\mathrm{f}(1)}{\mathrm{e}-1}\\ & \Rightarrow \frac{1}{\mathrm{c}}=\frac{\log \mathrm{e}-\log 1}{\mathrm{e}-1}\\ & \Rightarrow \frac{1}{\mathrm{c}}=\frac{1}{\mathrm{e}-1}\\ & \Rightarrow \mathrm{c}=\mathrm{e}-1\end{array}$

Correct Option is (A)

$\begin{array}{rl}& \mathrm{f}(x)=\sqrt{25-x^2}\\ & \therefore {\mathrm{f}}^{\prime }(x=\mathrm{c})=\frac{-2\mathrm{c}}{2\sqrt{25-{\mathrm{c}}^2}}\\ & =\frac{-\mathrm{c}}{\sqrt{25-{\mathrm{c}}^2}}\end{array}$

Applying Lagrange's mean value theorem, we get

$\begin{array}{ll}& {\mathrm{f}}^{\prime }(\mathrm{c})=\frac{\mathrm{f}(1)-\mathrm{f}(5)}{1-5}\\ \therefore & \frac{\mathrm{c}}{\sqrt{25-c^2}}=\frac{\sqrt{25-1}-\sqrt{25-5^2}}{1-5}\\ \therefore & \frac{-c}{\sqrt{25-c^2}}=\frac{-\sqrt{24}}{4}\end{array}$

$\begin{array}{ll}\therefore & 4c=\sqrt{24}\cdot \sqrt{25-c^2}\\ \therefore & 16c^2=24(25-c^2)\\ \therefore & c^2=15\\ \therefore & c=\pm \sqrt{15}\\ & \text{ Since }c=-\sqrt{15}\text{ does not lie in }[1,5]\\ & c=\sqrt{15}\end{array}$