Let f(x) be a real-valued function. If f^{pri | vedclass

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

Question

(D) Given that $f^{\prime}(x)$ is a constant for all $x \in R$. Let $f^{\prime}(x) = a$,where $a$ is a constant. Integrating both sides with respect t

Vedclass · Accepted Answer

$f(x)$ is continuous on $R$

Vedclass · Answer

(D) Given that $f^{\prime}(x)$ is a constant for all $x \in R$. Let $f^{\prime}(x) = a$,where $a$ is a constant. Integrating both sides with respect to $x$,we get $f(x) = ax + b$,where $b$ is an arbitrary constant. Given $f(0) = 2$,substituting $x = 0$ in $f(x) = ax + b$,we get $a(0) + b = 2$,which implies $b = 2$. Given $f^{\prime}(0) = 1$,since $f^{\prime}(x) = a$ for all $x$,we have $a = 1$. Thus,the function is $f(x) = x + 2$. Since $f(x) = x + 2$ is a polynomial function,it is continuous for all real numbers $x \in R$.

Let $f(x)$ be a real-valued function. If $f^{\prime}(x)$ is a constant for all $x \in R$,$f(0)=2$,and $f^{\prime}(0)=1$,then

Similar Questions

The values of $a$ and $b$ such that the function $f(x) = \begin{cases} -2 \sin x, & -\pi \leq x \leq -\frac{\pi}{2} \\ a \sin x + b, & -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \cos x, & \frac{\pi}{2} \leq x \leq \pi \end{cases}$ is continuous in $[-\pi, \pi]$,are

Let $f(x) = \begin{cases} x \sin \left( \frac{1}{x} \right) \sin \left( \frac{1}{x \sin \left( \frac{1}{x} \right)} \right), & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then $f(x)$ is:

Let $f(x) = \begin{cases} \frac{\tan^2 \{x\}}{x^2 - [x]^2} & \text{for } x > 0 \\ 1 & \text{for } x = 0 \\ \sqrt{\{x\} \cot \{x\}} & \text{for } x < 0 \end{cases}$ where $[x]$ is the greatest integer function and $\{x\}$ is the fractional part function of $x$,then:

Discuss the continuity of the $sine$ function.

For $a, b > 0$,let $f(x) = \begin{cases} \frac{\tan((a+1)x) + b \tan x}{x}, & x < 0 \\ \frac{\sqrt{ax + b^2x^2} - \sqrt{ax}}{b \sqrt{a} x \sqrt{x}}, & x > 0 \end{cases}$ be a continuous function at $x = 0$. Then $\frac{b}{a}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module