Question

Let $f\left(x\right)={(2-\frac{x}{a})}^{\tan \left(\frac{\pi x}{2a}\right)},x\ne a$. The value which should be assigned to $f$ at $x=a$ so that it is continuous everywhere is

• A. $\frac{2}{\pi }$
• B. $e^{-2/\pi }$
• C. $2$
• D. $e^{2/\pi }$

Answer 1

The correct option is D $e^{2/\pi }$

For $f$ to be continuous, we must have
$f\left(a\right)=\underset{x\to a}{lim}{(2-\frac{x}{a})}^{\tan \left(\frac{\pi x}{2a}\right)}$
$\Rightarrow logf\left(x\right)=\tan \left(\frac{\pi x}{2a}\right)log(2-\frac{x}{a})$
$=\frac{log(2-\frac{x}{a})}{\cot \left(\frac{\pi }{2}\frac{x}{a}\right)}\left(\frac{0}{0}form\right)$
$\underset{x\to a}{lim}logf\left(x\right)=\underset{x\to a}{lim}\frac{\frac{-1/a}{2-\frac{x}{a}}}{-\frac{\pi }{2a}{\csc }^2\left(\frac{\pi }{2}\frac{x}{a}\right)}$
(L'Hospital's Rule)
$=\frac{2}{\pi }\underset{x\to a}{lim}\frac{1}{2-\frac{x}{a}}\times \underset{x\to a}{lim}{\sin }^2\left(\frac{\pi }{2}\frac{x}{a}\right)$
$=\frac{2}{\pi }$
Therefore, $\underset{x\to a}{lim}f\left(x\right)=e^{2/\pi }$
Hence, option 'D' is correct.