Let f(x) be continuous on [0, 5] and differen | vedclass

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

Question

(A) By the Mean Value Theorem,for any $x \in (0, 5]$,there exists a $c \in (0, x)$ such that $f^{\prime}(c) = \frac{f(x) - f(0)}{x - 0}$.Given $f(0) =

Vedclass · Accepted Answer

$|f(x)| \leq 1$

Vedclass · Answer

(A) By the Mean Value Theorem,for any $x \in (0, 5]$,there exists a $c \in (0, x)$ such that $f^{\prime}(c) = \frac{f(x) - f(0)}{x - 0}$.Given $f(0) = 0$,we have $f^{\prime}(c) = \frac{f(x)}{x}$.Taking the absolute value,$|f^{\prime}(c)| = \left|\frac{f(x)}{x}\right| = \frac{|f(x)|}{|x|}$.Since $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x \in (0, 5)$,it follows that $|f^{\prime}(c)| \leq \frac{1}{5}$.Therefore,$\frac{|f(x)|}{x} \leq \frac{1}{5}$,which implies $|f(x)| \leq \frac{x}{5}$.Since $x \in [0, 5]$,the maximum value of $\frac{x}{5}$ is $\frac{5}{5} = 1$.Thus,$|f(x)| \leq 1$ for all $x \in [0, 5]$.

Let $f(x)$ be continuous on $[0, 5]$ and differentiable in $(0, 5)$. If $f(0) = 0$ and $|f^{\prime}(x)| \leq \frac{1}{5}$ for all $x$ in $(0, 5)$,then which of the following is true for all $x$ in $[0, 5]$?

Similar Questions

Consider the function $f(x)=2x^3-3x^2-x+1$ and the intervals $I_1=[-1,0]$,$I_2=[0,1]$,$I_3=[1,2]$,$I_4=[-2,-1]$. Then,

If the function $f(x) = 2x^2 + 3x + 5$ satisfies the Lagrange's Mean Value Theorem $(LMVT)$ at $x = 3$ on the closed interval $[1, a]$,then the value of $a$ is equal to:

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0$. Then:

Consider the function $f(x) = 8x^2 - 7x + 5$ on the interval $[-6, 6]$. The value of $c$ that satisfies the conclusion of the Mean Value Theorem is:

The value of $c$ for the function $f(x) = \log x$ on $[1, e]$ if Lagrange's Mean Value Theorem $(LMVT)$ is applied,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module