Question

Let $f\left(x\right)=\{\begin{array}{c}\alpha \left(x\right)\sin \frac{\pi x}{2}forx\ne 0\\ 1forx=0\end{array}$

where $\alpha \left(x\right)$ is such that ${lim}_{x\to 0}\left|\alpha \right(x\left)\right|=\infty$
Then the function $f\left(x\right)$ is continuous at $x=0$ if $\alpha \left(x\right)$ is chosen as

• A. $\frac{2}{\pi x}$
• B. $\frac{1}{x^2}$
• C. $\frac{2}{\pi x^2}$
• D. $\frac{1}{x}$

Answer 1

The correct option is A $\frac{2}{\pi x}$

Given
$f\left(x\right)=\{\begin{array}{c}\alpha \left(x\right)\sin \frac{\pi x}{2}forx\ne 0\\ 1forx=0\end{array}$ ..... $\left(1\right)$
For $f\left(x\right)$ to be continuous at $x=0$
$\underset{x\to 0}{lim}f\left(x\right)=f\left(0\right)$
From Eq (i), $f\left(0\right)=1$
$\therefore$ For $f\left(x\right)$ to be continuous at $x=0$
$\underset{x\to 0}{lim}\alpha \left(x\right)\sin \frac{\pi x}{2}=1$
The above limit is equal to $1$, when
$\alpha \left(x\right)=\frac{2}{\pi x}$
i.e., $\underset{x\to 0}{lim}\frac{\sin \frac{\pi x}{2}}{\frac{\pi x}{2}}=1[\because \underset{x\to 0}{lim}\frac{\sin \theta }{\theta }=1]$