Question

Let $\left[t\right]$ denotes the greatest integer $\le t$ and $\begin{array}{l}\underset{x\to 0}{lim}x\left[\frac{4}{x}\right]=A\end{array}$Then the function, $f\left(x\right)=\left[x^2\right]\sin \pi x$ is discontinuous, when $x$ is equal to:

• A. $\sqrt{(A+1)}$
• B. $\sqrt{A}$
• C. $\sqrt{(A+5)}$
• D. $\sqrt{(A+21)}$

Answer 1

The correct option is A

$\sqrt{(A+1)}$

Explanation for The correct option:

Finding the required value of $x$

The given function $f\left(x\right)=\left[x^2\right]\sin \pi x$

It is continuous $\forall x\in Z$ as $\sin \pi x$ as is continuous at $Z$.

So, the function $f\left(x\right)$ is discontinuous at points where$\left[x^2\right]$ is discontinuous i.e. $x^2\in Z$ is an exception point that $f\left(x\right)$ is continuous as $x$ is an integer.

Therefore, the points of discontinuity for $f\left(x\right)$ be

$x=\pm \surd 2,\pm \surd 3,\pm \surd 5,\dots \dots$

And given that

$\begin{array}{l}\underset{x\to 0}{lim}x\left[\frac{4}{x}\right]=A\end{array}$

We get indeterminate form of $(0\times \infty )$

Therefore,

$\begin{array}{l}4-\underset{x\to 0}{lim}\left[\frac{4}{x}\right]=A\end{array}$ Since $\left[x\right]$ is a greatest integer function

$$\begin{gathered} \Rightarrow A=4 \\ \Rightarrow \sqrt{(A+5})=3 \\ \Rightarrow \sqrt{(A+1)}=\sqrt{5} \end{gathered}$$

thus the points of discontinuity for the function is $\Rightarrow \sqrt{(A+1)}$

Hence, the correct option is (A)