Let f(x) = log(1 + x^2) and A be a constant s | 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 $x 
eq y$,there exists a $c$ between $x$ and $y$ such that $\frac{f(x) - f(y)}{x - y} = f'(c)$.Given $f(x) = \l

Vedclass · Accepted Answer

equal to $1$

Vedclass · Answer

(A) By the Mean Value Theorem,for any $x 
eq y$,there exists a $c$ between $x$ and $y$ such that $\frac{f(x) - f(y)}{x - y} = f'(c)$.Given $f(x) = \log(1 + x^2)$,we find the derivative $f'(x) = \frac{2x}{1 + x^2}$.To find the range of $f'(x)$,we analyze $g(x) = \frac{2x}{1 + x^2}$.Setting $g'(x) = \frac{2(1 + x^2) - 2x(2x)}{(1 + x^2)^2} = \frac{2 - 2x^2}{(1 + x^2)^2} = 0$,we get $x = \pm 1$.The maximum value of $f'(x)$ is $f'(1) = \frac{2(1)}{1 + 1^2} = 1$ and the minimum value is $f'(-1) = \frac{2(-1)}{1 + (-1)^2} = -1$.Thus,$|f'(c)| \leq 1$ for all $c \in \mathbb{R}$.Since $\frac{|f(x) - f(y)|}{|x - y|} = |f'(c)| \leq 1$,the least possible value of $A$ is $1$.

Let $f(x) = \log(1 + x^2)$ and $A$ be a constant such that $\frac{|f(x) - f(y)|}{|x - y|} \leq A$ for all real $x, y$ where $x \neq y$. Then,the least possible value of $A$ is

Similar Questions

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:

For all real values of $a_{0}, a_{1}, a_{2}, a_{3}$ satisfying $a_{0}+\frac{a_{1}}{2}+\frac{a_{2}}{3}+\frac{a_{3}}{4}=0$, the equation $a_{0}+a_{1} x+a_{2} x^{2}+a_{3} x^{3}=0$ has a real root in the interval

The position of a moving car at time $t$ is given by $f(t) = at^{2} + bt + c, t > 0,$ where $a, b,$ and $c$ are real numbers greater than $1.$ Then the average speed of the car over the time interval $[t_{1}, t_{2}]$ is attained at the point

Let $f^{\prime}(0)=-3$ and $f^{\prime}(x) \leq 5$ for all real values of $x$. The $f(2)$ can have possible maximum value as

Let $f(x)=x^3+2x^2-x$ be a real-valued function. Then, the value of Lagrange's constant $C$ in $(-1,2)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module