Given that f(x) is continuously differentiabl | vedclass

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

Question

(D) $(i)$ This statement is true,as every continuous function on a closed interval $[a, b]$ is bounded.$(ii)$ This is true by the Intermediate Value T

Vedclass · Accepted Answer

only $(i), (ii)$ and $(iv)$ are true

Vedclass · Answer

(D) $(i)$ This statement is true,as every continuous function on a closed interval $[a, b]$ is bounded.$(ii)$ This is true by the Intermediate Value Theorem,as $f(a)  0$.$(iii)$ This is not necessarily true,as the maximum or minimum values can occur at the endpoints $a$ or $b$ where the derivative may not be zero.$(iv)$ This is true. By the Mean Value Theorem,there exists a point $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$. Since $f(b) > 0$ and $f(a)  0$,so $f'(c) > 0$.$(v)$ This is not always true. For example,if $f(x) = x$,then $f'(x) = 1 > 0$ for all $x$,so there is no point where $f'(d) Thus,the true statements are $(i), (ii),$ and $(iv)$. The correct option is $(D)$.

Similar Questions

Let $f:(a, b) \rightarrow R$ be a twice differentiable function such that $f(x) = \int_{a}^{x} g(t) \, dt$ for a differentiable function $g(x)$. If $f(x) = 0$ has exactly five distinct roots in $(a, b)$,then $g(x) g'(x) = 0$ has at least:

Given $f(x) = 4 - (\frac{1}{2} - x)^{2/3}$,$g(x) = \begin{cases} \frac{\tan([x])}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}$,$h(x) = \{x\}$,and $k(x) = 5^{\log_2(x + 3)}$. Then,in the interval $[0, 1]$,Lagrange's Mean Value Theorem is $NOT$ applicable to:

Let $f(x) = e^x \cos x + 1$. Which of the following statements is always true?

In the Mean Value Theorem,$f(b) - f(a) = (b - a)f'(c)$. If $a = 4$,$b = 9$ and $f(x) = \sqrt{x}$,then the value of $c$ is:

If the function $f(x) = ax^2 + bx + \sin x$ satisfies all the conditions of Rolle's theorem on $[0, \pi]$ and the slope of the tangent to the curve $y = f(x)$ at $x = \frac{\pi}{4}$ is zero, then $a - b = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module