Question

Let $f(x) = \begin{cases} a (x) \sin \frac{\pi \ x }{2} & \text{for } x \neq 0 \\ -(n+1)/2 & \text{for} x = 0 \end{cases} $ where $\alpha (x) $ is such that $\displaystyle\lim_{x \to 0} |\alpha (x) | = \infty $ Then the function $f(x)$ is continuous at $x = 0$ if $\alpha (x) $ is chosen as

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

Answer 1

The Correct Option is B

Given,
$f(x) =
\begin{cases}
\alpha(x) \sin\,\frac{\pi\,x}{2} & \text{for $X \neq 0$ } \\[2ex]
1 & \text{for $x=0$}
\end{cases}\,...(i)$
For $f(x)$ to be continuous at $x=0$
$\displaystyle\lim _{x \rightarrow 0} f(x)=f(0)$
From E (i), $f(0)=1$
$\therefore$ For $f(x)$ to be continuous at $x=0$
$\displaystyle\lim _{x \rightarrow 0} \alpha(x) \sin \frac{\pi x}{2}=1$
The above limit is equal to 1, when
$\alpha(x)=\frac{2}{\pi x} $
i.e. $\displaystyle\lim _{x \rightarrow 0} \frac{\sin \frac{\pi X}{2}}{\frac{\pi x}{2}}=1$
$\left[\because \displaystyle\lim _{x \rightarrow 0} \frac{\sin \theta}{\theta}=1\right]$