Q77 / 120 · VITEEE 2019 · Mathematics
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
- A$\frac{13}{4}$
- B$\frac{17}{4}$
- C$\frac{21}{4}$
- Correct answerD$\frac{25}{4}$
Solution
(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.
$$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{1}{5} \Rightarrow c=\frac{25}{4}
$$
Hence, the correct answer is option: 'D'.