Q67 / 160 · AP EAMCET 2014 · Mathematics

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

  • A
    $\frac{9}{2}$
  • B
    $\frac{5}{2}$
  • Correct answer
    C
    $3$
  • D
    $4$

Solution

Given, $f(x)=\sqrt{x-2}, x \in[2,6]$ We know that, by Lagrange's mean value theorem, their exist $c \in(2,6)$ such that $$ \begin{aligned} & f(c)=\frac{f(b)-f(a)}{b-a} \\ & =\frac{1}{2 \sqrt{c-2}}=\frac{f(6)-f(2)}{6-2} \\ \Rightarrow \quad & \frac{1}{2 \sqrt{c-2}}=\frac{\sqrt{6-2}-\sqrt{2-2}}{4} \end{aligned} $$ $\begin{array}{ll}\Rightarrow & \frac{1}{\sqrt{c-2}}=\frac{\sqrt{4}-0}{2} \\ \Rightarrow & \frac{1}{\sqrt{c-2}}=1 \Rightarrow \sqrt{c-2}=1 \\ \Rightarrow & c-2=1 \Rightarrow c=3\end{array}$

Hence, the correct answer is option: 'C'.

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