Question

If $f\left(x\right)=\left[x\right]\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{\left[x+1\right]}}+\sin \left(\mathrm{\pi }\left[\mathrm{x}+1\right]\right)}{1+\left[x\right]}$, then

• A. $f\left(x\right)$ is continuous on $\mathrm{ℝ}$.
• B. $f\left(x\right)$ is continuous on $\mathrm{ℝ}$, but not differentiable on $\mathrm{ℝ}$
• C. $f\left(x\right)$ is discontinuous at every integer.
• D. None of these.

Answer 1

The correct option is C

$f\left(x\right)$ is discontinuous at every integer.

Explanation for the correct option:

Step 1: Simplify the given expression

The given function is $f\left(x\right)=\left[x\right]\frac{\sin {\displaystyle \frac{\mathrm{\pi }}{\left[x+1\right]}}+\sin \left(\mathrm{\pi }\left[\mathrm{x}+1\right]\right)}{1+\left[x\right]}$.

Let $\left[x+1\right]=n$, where $n\in \mathrm{\mathbb{N}}$.

Thus, $\sin \left(\mathrm{\pi }\left[\mathrm{x}+1\right]\right)=\sin \left(\mathrm{\pi n}\right)$

$\Rightarrow \sin \left(\mathrm{\pi }\left[\mathrm{x}+1\right]\right)=0$.

So, $f\left(x\right)=\frac{\left[x\right]\sin {\displaystyle \frac{\mathrm{\pi }}{\left[x+1\right]}}}{1+\left[x\right]}$.

Step 2: Check for continuity

Let us assume that $n$ is an integer.

Thus, $\underset{x\to n^{-}}{\lim }f\left(x\right)=\underset{x\to n^{-}}{\lim }\frac{\left[x\right]\sin {\displaystyle \frac{\mathrm{\pi }}{\left[x+1\right]}}}{1+\left[x\right]}$

$$\begin{gathered} \Rightarrow \underset{x\to n^{-}}{\lim }f\left(x\right)=\frac{\left(n-1\right)\sin {\displaystyle \frac{\mathrm{\pi }}{n}}}{n} \\ \Rightarrow \underset{x\to n^{-}}{\lim }f\left(x\right)=\frac{n-1}{n}\sin \frac{\mathrm{\pi }}{n} \end{gathered}$$

Thus, $\underset{x\to n^{+}}{\lim }f\left(x\right)=\underset{x\to n^{+}}{\lim }\frac{\left[x\right]\sin {\displaystyle \frac{\mathrm{\pi }}{\left[x+1\right]}}}{1+\left[x\right]}$

$$\begin{gathered} \Rightarrow \underset{x\to n^{+}}{\lim }f\left(x\right)=\frac{n\sin {\displaystyle \frac{\mathrm{\pi }}{n+1}}}{1+n} \\ \Rightarrow \underset{x\to n^{+}}{\lim }f\left(x\right)=\frac{n}{n+1}\sin \frac{\mathrm{\pi }}{n+1} \end{gathered}$$

Thus, $\underset{x\to n^{-}}{\lim }f\left(x\right)\ne \underset{x\to n^{+}}{\lim }f\left(x\right)$.

Therefore, $f\left(x\right)$ is discontinuous at every integer.

Hence, option$\left(C\right)$ i.e. $f\left(x\right)$ is discontinuous at every integer is the correct option.