Question

The value of $c$ from the Lagrange's mean value theorem for which $f\left(x\right)=\sqrt{25-x^2}$ is $[1,5]$ is

• A. $5$
• B. $1$
• C. $\sqrt{15}$
• D. None of these

Answer 1

The correct option is C

$\sqrt{15}$

Explanation for the correct option:

Find the value of $c$

A function $f\left(x\right)=\sqrt{25-x^2}$ is given with the interval $[1,5]$.

Lagrange's mean value theorem states that, if a function $g$ is continuous on $[a,b]$ then there exists a $c$ value between $a$ and $b$ for which $f\text{'}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$.

Find the derivative of the given at $x=c$.

$$\begin{gathered} f\text{'}(x=c)=\frac{-2c}{2\sqrt{25-c^2}} \\ \Rightarrow f\text{'}(x=c)=\frac{-c}{\sqrt{25-c^2}} \end{gathered}$$

Apply Lagrange's mean value theorem as follows,

$$\begin{gathered} \frac{-c}{\sqrt{25-c^2}}=\frac{\sqrt{25-5^2}-\sqrt{25-1^2}}{5-1} \\ \Rightarrow \frac{-c}{\sqrt{25-c^2}}=\frac{\sqrt{25-25}-\sqrt{25-1}}{4} \\ \Rightarrow \frac{-c}{\sqrt{25-c^2}}=\frac{-\sqrt{24}}{4} \\ \Rightarrow 4c=\sqrt{24}\sqrt{25-c^2} \\ \Rightarrow 16c^2=24\left(25-c^2\right) \\ \Rightarrow 16c^2+24c^2-600=0 \\ \Rightarrow 40c^2-600=0 \\ \Rightarrow 40c^2=600 \\ \Rightarrow c^2=15 \\ \Rightarrow c=\pm \sqrt{15} \end{gathered}$$

Since, $c=-\sqrt{15}$ does not lie between $1$ and $5$.

Therefore, the value of $c$ is $\sqrt{15}$.

Hence, option C is correct .