Question

Let $f(x)=\sqrt{x}+5$ for $1\le x\le 9$. Then the value of $c$ whose existence is guaranteed by the Mean Value Theorem is
Given, $f(x)=\mathrm{sqrt}x+5$ 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 $1\le x\le 9$
$\Rightarrow f^{\prime }(x)=\frac{1}{2\sqrt{x}}$
Since, $f(x)$ satisfied Mean Value theorem,
$\therefore f^{\prime }(c)=\frac{f(b)-f(a)}{b-a},\text{ where }c\in (1,9)$
$\Rightarrow \frac{1}{2\sqrt{c}}=\frac{3-1}{9-1}$
$\Rightarrow \frac{1}{2\sqrt{c}}=\frac{2}{8}=\frac{1}{4}$
$\Rightarrow \sqrt{c}=2\Rightarrow c=4$