Question

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

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

Answer 1

The correct option is A $\frac{2}{\pi }$

$f\left(x\right)=(1-x)\tan \left(\frac{\pi x}{2}\right)x=1$

To be continuous, we can find $f\left(1\right)=k$ to make $f$ continuous at $x=1$

$\therefore \underset{x\to 1}{lt}f\left(x\right)=f\left(1\right)$

$=\underset{x\to 1}{lt}(1-x)tan\left(\frac{\pi x}{2}\right)$

$=\underset{x\to 1}{lt}\sin \left(\frac{\pi x}{2}\right)\frac{1-x}{\cos \left(\frac{\pi x}{2}\right)}$

$=\underset{x\to 1}{lt}\sin \left(\frac{\pi x}{2}\right)\underset{x\to 1}{lt}\frac{1-x}{\cos \left(\frac{\pi x}{2}\right)}$

$=1.\underset{x\to 1}{lt}\frac{1-x}{\cos \left(\frac{\pi x}{2}\right)}$

Now, $\frac{0}{0}$ form,

$=1.\underset{x\to 1}{lt}\frac{-1}{\frac{-\pi }{2}\sin \left(\frac{\pi x}{2}\right)}=\frac{2}{\pi }$

$\therefore f\left(1\right)=\frac{2}{\pi }$