Let f(x) satisfy all the conditions of the Me | vedclass

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

Question

(B) By the Mean Value Theorem,there exists some $c \in (0, 2)$ such that $\frac{f(2) - f(0)}{2 - 0} = f'(c)$.Given $f(0) = 0$,we have $\frac{f(2)}{2}

Vedclass · Accepted Answer

$|f(x)| \le 1$

Vedclass · Answer

(B) By the Mean Value Theorem,there exists some $c \in (0, 2)$ such that $\frac{f(2) - f(0)}{2 - 0} = f'(c)$.Given $f(0) = 0$,we have $\frac{f(2)}{2} = f'(c)$.For any $x \in [0, 2]$,applying the Mean Value Theorem on $[0, x]$,there exists $c_x \in (0, x)$ such that $\frac{f(x) - f(0)}{x - 0} = f'(c_x)$.Since $f(0) = 0$,this simplifies to $\frac{f(x)}{x} = f'(c_x)$,which implies $f(x) = x \cdot f'(c_x)$.Taking the absolute value,$|f(x)| = |x| \cdot |f'(c_x)|$.Given $|f'(x)| \le \frac{1}{2}$ for all $x \in [0, 2]$,we have $|f(x)| \le |x| \cdot \frac{1}{2}$.Since the maximum value of $x$ in $[0, 2]$ is $2$,we get $|f(x)| \le 2 \cdot \frac{1}{2} = 1$.Thus,$|f(x)| \le 1$.

Let $f(x)$ satisfy all the conditions of the Mean Value Theorem in $[0, 2]$. If $f(0) = 0$ and $|f'(x)| \le \frac{1}{2}$ for all $x$ in $[0, 2]$,then:

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 Rolle's theorem is applicable for the function $f(x) = \log \left( \frac{x^2 + a}{x} \right)$ on $[3, 4]$ with $c \in (3, 4)$ such that $f'(c) = 0$, then the value of $f''(c)$ is

The value of $c$ in the Lagrange's mean value theorem for $f(x)=\sqrt{x-2}$ in the interval $[2,6]$ is

If $2a + 3b + 6c = 0$ and $a, b, c \in \mathbb{R}$,then the equation $ax^2 + bx + c = 0$ has at least one root between $0$ and $1$.

If $27a + 9b + 3c + d = 0$,then the equation $4ax^3 + 3bx^2 + 2cx + d = 0$ has at least one root between which of the following?

Vedclass Test Series

Exam Paper Generator

Online Exam Module