Question

Let $f(x)=\{\begin{array}{ll}a(x)\sin \frac{\pi x}{2}& \text{for }x\ne 0\\ -(n+1)/2& \text{for}x=0\end{array}$
where $\alpha (x)$ is such that $\underset{x\to 0}{\lim }|\alpha (x)|=\infty$
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

Answer: (B)

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