Q54 / 170 · Manipal (MET) 2015 · Mathematics
The value of $c$ prescribed by Lagrange's mean value theorem, when $f(x)=\sqrt{x^2-4}, a=2$ and $b=3$, is
- A$2.5$
- Correct answerB$\sqrt{5}$
- C$\sqrt{3}$
- D$\sqrt{3}+1$
Solution
Clearly, $f(x)=\sqrt{x^2-4}$ is continuous on $[2,3]$ and differentiable on $(2,3)$.
So, by Lagrange's mean value theorem, there exists $c \in(2,3)$ such that
$f^{\prime}(c)=\frac{f(3)-f(2)}{3-2}$
$\begin{aligned} \Rightarrow \quad & \frac{c}{\sqrt{c^2-4}}=\sqrt{5}-0 \\ & {\left[\because f(x)=\sqrt{x^2-4} \Rightarrow f^{\prime}(x)=\frac{x}{\sqrt{x^2-4}}\right] }\end{aligned}$
$\begin{array}{rlrl}\Rightarrow & & c^2 & =5\left(c^2-4\right) \\ \Rightarrow & 4 c^2 & =20 \\ \Rightarrow & & c & =\sqrt{5}\end{array}$
Hence, the correct answer is option: 'B'.