Search any question & find its solution
Question:
Answered & Verified by Expert
The value of $x$ in the interval $[4,9]$ at which the function $\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$ satisfies the mean value theorem is
Options:
Solution:
2725 Upvotes
Verified Answer
The correct answer is:
$\frac{25}{4}$
(i) $f(x)=\sqrt{x}$ is continuous in $[4,9]$
(ii) $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}$
Thus $\mathrm{f}(\mathrm{x})$ is differentiable in $(4,9)$
(iii) $\mathrm{f}(4) \neq \mathrm{f}(9)$. All the three conditions of $\mathrm{LMV}$ theorem satisfied then there exist at least one $\mathrm{c} \in(4,9)$ such that.
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}
$$
(ii) $\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{2 \sqrt{\mathrm{x}}}$
Thus $\mathrm{f}(\mathrm{x})$ is differentiable in $(4,9)$
(iii) $\mathrm{f}(4) \neq \mathrm{f}(9)$. All the three conditions of $\mathrm{LMV}$ theorem satisfied then there exist at least one $\mathrm{c} \in(4,9)$ such that.
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.