Let f(x) satisfy the requirements of Lagrange | vedclass

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

Question

(B) According to Lagrange's Mean Value Theorem,for any $x \in (0, 2]$,there exists a $c \in (0, x)$ such that $f'(c) = \frac{f(x) - f(0)}{x - 0}$.Sinc

Vedclass · Accepted Answer

$|f(x)| \leqslant 1$

Vedclass · Answer

(B) According to Lagrange's Mean Value Theorem,for any $x \in (0, 2]$,there exists a $c \in (0, x)$ such that $f'(c) = \frac{f(x) - f(0)}{x - 0}$.Since $f(0) = 0$,we have $f'(c) = \frac{f(x)}{x}$.Given that $|f'(x)| \leqslant \frac{1}{2}$ for all $x \in [0, 2]$,it follows that $|f'(c)| \leqslant \frac{1}{2}$.Substituting this into our equation,we get $\left| \frac{f(x)}{x} \right| = |f'(c)| \leqslant \frac{1}{2}$.This implies $|f(x)| \leqslant \frac{|x|}{2}$.Since $x \in [0, 2]$,the maximum value of $|x|$ is $2$,so $|f(x)| \leqslant \frac{2}{2} = 1$.Thus,$|f(x)| \leqslant 1$.

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

Similar Questions

Let $f(x)$ and $g(x)$ be two differentiable functions in $R$ such that $f(2) = 8, g(2) = 0, f(4) = 10$,and $g(4) = 8$. Then which of the following is true?

Let $f: R \rightarrow R$ be a thrice differentiable function such that $f(0)=0, f(1)=1, f(2)=-1, f(3)=2$ and $f(4)=-2$. Then,the minimum number of zeros of $(3 f^{\prime} f^{\prime \prime} + f f^{\prime \prime \prime})(x)$ is....................

If $f(x) = \sin^2 x + x \sin 2x \log x$,then $f(x) = 0$ has

Suppose $f$ is differentiable for all $x$. If $f(1) = -2$ and $f'(x) \geq 2$ for all $x \in [1, 6]$,then:

If from the Mean Value Theorem,$f'({x_1}) = \frac{f(b) - f(a)}{b - a}$,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module