Let [t] denote the greatest integer leq t. Th | vedclass

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

Question

(A) The function is defined as $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in (-2, 2)$.
We analyze the points of discontinu

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The function is defined as $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in (-2, 2)$.
We analyze the points of discontinuity by checking the behavior of $[x]$ and $[x+1]$ at integer values $x \in \{-1, 0, 1\}$.
For $x \in (-2, -1)$,$[x] = -2$ and $[x+1] = -1$. Thus,$f(x) = -2|x^2-1| + \sin(\pi/1) - (-1) = -2|x^2-1| + 1$.
For $x \in [-1, 0)$,$[x] = -1$ and $[x+1] = 0$. Thus,$f(x) = -1|x^2-1| + \sin(\pi/2) - 0 = -|x^2-1| + 1$.
For $x \in [0, 1)$,$[x] = 0$ and $[x+1] = 1$. Thus,$f(x) = 0|x^2-1| + \sin(\pi/3) - 1 = \frac{\sqrt{3}}{2} - 1$.
For $x \in [1, 2)$,$[x] = 1$ and $[x+1] = 2$. Thus,$f(x) = 1|x^2-1| + \sin(\pi/4) - 2 = |x^2-1| + \frac{1}{\sqrt{2}} - 2$.
The function is discontinuous at points where the floor functions change values,which are $x = -1, 0, 1$. Checking the limits at these points confirms discontinuity. Therefore,there are $3$ points of discontinuity.

Let $[t]$ denote the greatest integer $\leq t$. The number of points where the function $f(x)=[x]|x^{2}-1|+\sin \left(\frac{\pi}{[x]+3}\right)-[x+1]$ for $x \in(-2,2)$ is not continuous is:

Similar Questions

If $f: R \rightarrow R$ is defined by $f(x) = \begin{cases} \frac{x + 2}{x^2 + 3 x + 2}, & x \in R - \{-1, -2\} \\ -1, & x = -2 \\ 0, & x = -1 \end{cases}$ then $f$ is continuous on the set

Find the values of $k$ so that the function $f$ is continuous at the indicated point.
$f(x) = \begin{cases} \frac{k \cos x}{\pi - 2x}, & \text{if } x \neq \frac{\pi}{2} \\ 3, & \text{if } x = \frac{\pi}{2} \end{cases}$ at $x = \frac{\pi}{2}$

If $f(x) = \begin{cases} \frac{\cos(ax) - \cos(bx)}{x^2}, & x \neq 0 \\ \frac{1}{2}(b^2 - a^2), & x = 0 \end{cases}$ where $a$ and $b$ are real and distinct constants,then:

The function $f(x) = p[x + 1] + q[x - 1],$ where $[x]$ is the greatest integer function,is continuous at $x = 1$ if:

If $f(x) = \lim_{n \rightarrow \infty} \left( \frac{\log(2+x) - x^{2n} \sin x}{1+x^{2n}} \right)$ for $0 \leq x \leq \frac{\pi}{2}$, then at $x=1$, $f(x)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module