Question

In the mean value theorem $f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(c\right)$, if $a=4,b=9$and $f\left(x\right)=\sqrt{x}$, then the value of $c$is

• A. $8.00$
• B.
• C.
• D.

Answer 1

The correct option is D

Determine the value of $c$

We have, $f\left(x\right)=\sqrt{x}$

Therefore, differentiating it with respect to $x$, we get

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{1}{2\sqrt{x}} \\ \therefore f\text{'}\left(c\right)=\frac{1}{2\sqrt{c}};[\text{Substitue}x\text{with}c] \end{gathered}$$

$$\begin{gathered} \because f\left(b\right)-f\left(a\right)=(b-a)f\text{'}\left(c\right)\text{[Given]} \\ \Rightarrow \sqrt{b}-\sqrt{a}=(b-a)f\text{'}\left(c\right)[\because f(x)=\sqrt{x}] \\ \Rightarrow f\text{'}\left(c\right)=\frac{\sqrt{b}-\sqrt{a}}{(b-a)};[\text{Substitue}a=4,b=9,\text{given]} \\ \Rightarrow f\text{'}(c)=\frac{\sqrt{9}-\sqrt{4}}{(9-4)} \\ \Rightarrow \frac{1}{2\sqrt{c}}=\frac{3-2}{5};\left[\because f\text{'}\left(c\right)=\frac{1}{2\sqrt{c}}\right] \\ \Rightarrow \sqrt{c}=\frac{5}{2} \\ \Rightarrow c={\left(\frac{5}{2}\right)}^2 \\ \Rightarrow c=6.25 \end{gathered}$$
Hence, option D is the correct answer.