Question

Let $f(x)=\sqrt{x}+5$ for $1 \leq x \leq 9$. Then the value of $c$ whose existence is guaranteed by the Mean Value Theorem is
Given, $f(x)=\operatorname{sqrt} x+5$ for 1 leq $x$ leq 9

• A. 2
• B. 3
• C. 4
• D. 5
• E. 6

Answer 1

Answer: (C)

Given, $f(x)=\sqrt{x}+5$ for $1 \leq x \leq 9$
$\Rightarrow f^{\prime}(x)=\frac{1}{2 \sqrt{x}}$
Since, $f(x)$ satisfied Mean Value theorem,
$ \therefore f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}, \text { where } c \in(1,9) $
$ \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{3-1}{9-1}$
$ \Rightarrow \frac{1}{2 \sqrt{c}}=\frac{2}{8}=\frac{1}{4} $
$ \Rightarrow \sqrt{c}=2 \Rightarrow c=4$