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)

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