Which of the following functions satisfies th | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Rolle's theorem states that for a function $f(x)$ to satisfy the theorem on $[a, b]$,it must satisfy three conditions: $1$. $f(x)$ is continuous o

Vedclass · Accepted Answer

$f(x) = x + \frac{1}{x}$ in $[\frac{1}{3}, 3]$

Vedclass · Answer

(D) Rolle's theorem states that for a function $f(x)$ to satisfy the theorem on $[a, b]$,it must satisfy three conditions: $1$. $f(x)$ is continuous on $[a, b]$. $2$. $f(x)$ is differentiable on $(a, b)$. $3$. $f(a) = f(b)$. Checking option $A$: $f(x) = |\operatorname{sgn}(x)|$ is discontinuous at $x = 0$,so it fails. Checking option $B$: $f(x) = 3x^2 - 2$ on $[2, 3]$. Here $f(2) = 3(4) - 2 = 10$ and $f(3) = 3(9) - 2 = 25$. Since $f(2) 
eq f(3)$,it fails. Checking option $C$: $f(x) = |x - 1|$ on $[0, 2]$. $f(0) = |0 - 1| = 1$ and $f(2) = |2 - 1| = 1$. Thus $f(0) = f(2)$. However,$f(x)$ is not differentiable at $x = 1$,which is inside $(0, 2)$,so it fails. Checking option $D$: $f(x) = x + \frac{1}{x}$ on $[\frac{1}{3}, 3]$. $f(\frac{1}{3}) = \frac{1}{3} + 3 = \frac{10}{3}$ and $f(3) = 3 + \frac{1}{3} = \frac{10}{3}$. Thus $f(\frac{1}{3}) = f(3)$. Since $f(x)$ is a rational function with a non-zero denominator in the interval,it is continuous and differentiable on the interval. Therefore,it satisfies Rolle's theorem.

Which of the following functions satisfies the conditions of Rolle's theorem on the given interval?

Similar Questions

Let $f$ be continuous on $[1, 5]$ and differentiable in $(1, 5).$ If $f(1)=-3$ and $f'(x) \ge 9$ for all $x \in (1, 5)$,then which of the following is true?

The value of $c$ for which Rolle's theorem holds for the function $f(x)=x^3-3x^2+2x$ in the interval $[0,2]$ is:

If the equation $a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x = 0$,where $a_1 \neq 0$ and $n \geq 2$,has a positive root $x = \alpha$,then the equation $n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \dots + a_1 = 0$ has a positive root which is:

For which real value of $K$ does the equation $2x^3 + 3x + K = 0$ have two real roots in the interval $[0, 1]$?

Let $f(x) = \begin{cases} x^\alpha \ln x, & x > 0 \\ 0, & x = 0 \end{cases}$. Rolle's theorem is applicable to $f$ for $x \in [0, 1]$ if $\alpha = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module