Question

Let $\left[t\right]$ denotes the greatest integer $\le t$ and $\underset{x\to 0}{lim}x\left[\frac{4}{x}\right]=A.$ 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 $\sqrt{A+1}$

$f\left(x\right)=\left[x^2\right]\sin \pi x$
It is continuous $\forall x\in Z.$
$f\left(x\right)$ is discontinuous at points where $\left[x^2\right]$ is discontinuous i.e. $x^2\in Z$ with an exception that $f\left(x\right)$ is continuous as $x$ is an integer.
$\therefore$ Points of discontinuity for $f\left(x\right)$ would be at $x=\pm \sqrt{2},\pm \sqrt{3},\pm \sqrt{5},.......$
Also, it is given that $\underset{x\to 0}{lim}x\left[\frac{4}{x}\right]=A$ (indeterminate form $(0\times \infty )$)
$\Rightarrow \underset{x\to 0}{lim}x(\frac{4}{x}-\left\{\frac{4}{x}\right\})$
$\Rightarrow 4-\underset{x\to 0}{lim}x\left\{\frac{4}{x}\right\}=A$
$\Rightarrow A=4$
$\sqrt{A+5}=3$
$\sqrt{A+1}=\sqrt{5}$
$\sqrt{A+21}=5$
$\sqrt{A}=2$
$\therefore$ Point of discontinuity for $f\left(x\right)$ is $x=\sqrt{5}$