The function f(x)=frac{tan {pi[x-frac{pi}{2}] | vedclass

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

Question

(A) Given, $f(x)=\frac{	an \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.Since $[x-\frac{\pi}{2}]$ is an integer for all $x$, let $[x-\frac{\pi}{2}] = k$, wh

Vedclass · Accepted Answer

continuous for all values of $x$

Vedclass · Answer

(A) Given, $f(x)=\frac{	an \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$.Since $[x-\frac{\pi}{2}]$ is an integer for all $x$, let $[x-\frac{\pi}{2}] = k$, where $k \in \mathbb{Z}$.Then the numerator becomes $	an(\pi k)$, which is equal to $0$ for all integers $k$.Since the denominator $2+[x]^{2}$ is always $\geq 2$ and never zero, the function simplifies to $f(x) = \frac{0}{2+[x]^{2}} = 0$ for all $x \in \mathbb{R}$.$A$ constant function $f(x) = 0$ is continuous and differentiable for all values of $x$.Therefore, the function is continuous for all values of $x$.

The function $f(x)=\frac{\tan \{\pi[x-\frac{\pi}{2}]\}}{2+[x]^{2}}$, where $[x]$ denotes the greatest integer $\leq x$, is

Similar Questions

If $f: [0, 2) \to R$ is defined by $f(x) = \begin{cases} 1 + 2x^k, & 0 \le x < 1 \\ kx, & 1 \le x < 2 \end{cases}$ where $k > 0$ and $f$ is such that $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x)$,then the value of $k^2$ is:

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

Let $f(x) = \begin{cases} x^p \sin \left( \frac{1}{x} \right) + x|x^3|, & x \neq 0 \\ 0, & x = 0 \end{cases}$. Then the complete set of values of $p$ for which $f''(x)$ is continuous at $x = 0$ is:

If $f(x) = \begin{cases} x^2, & \text{when } x \le 1 \\ x + 5, & \text{when } x > 1 \end{cases}$,then

Consider the function $f(x) = \begin{cases} \frac{x+5}{x-2}, & \text{if } x \neq 2 \\ 1, & \text{if } x=2 \end{cases}$. Then,$f(f(x))$ is discontinuous

Vedclass Test Series

Exam Paper Generator

Online Exam Module