Question

If the function $f\left(x\right)=\frac{\left(1 - x\right)}{2}tan \frac{\pi x}{2}$ is continuous at $x=1,$ then $f\left(1\right)$ is equal to

• A. $\frac{1}{\pi }$
• B. $\frac{\pi }{2}$
• C. $0$
• D. $\pi $

Answer 1

Answer: (A)

$f(1)=\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}\left(\frac{1-x}{2}\right) \tan \left(\frac{\pi x}{2}\right)$
Put $x=1+h$
$=\lim _{h \rightarrow 0} \frac{(1-(1+h))}{2} \tan \left(\frac{\pi}{2}(1+h)\right)$
$=\lim _{h \rightarrow 0} \frac{(-h)}{2} \tan \left(\frac{\pi}{2}+\frac{\pi h}{2}\right)$
$=\lim _{h \rightarrow 0} \frac{(-h)}{2}\left(-\cot \frac{\pi h}{2}\right)$
$=\lim _{h \rightarrow 0} \frac{\frac{\pi}{2} \cdot h}{\tan \left(\frac{\pi h}{2}\right)} \cdot \frac{1}{\pi}=\frac{1}{\pi}$