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

$f(x) = (1-x) \tan(\cfrac{\pi x}{2}) \quad x=1$

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

$\therefore \underset{x\rightarrow 1}{lt} f(x) = f(1)$

$=\underset{x\rightarrow 1}{lt} (1-x)tan(\cfrac{\pi x}{2})$

$=\underset{x\rightarrow 1}{lt} \sin(\cfrac{\pi x}{2}) \cfrac{1-x}{\cos (\cfrac{\pi x}{2})}$

$=\underset{x\rightarrow 1}{lt} \sin(\cfrac{\pi x}{2})\quad \underset{x\rightarrow 1}{lt}\cfrac{1-x}{\cos (\cfrac{\pi x}{2})}$

$=1. \underset{x\rightarrow 1}{lt}\cfrac{1-x}{\cos (\cfrac{\pi x}{2})}$

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

$= 1. \underset{x\rightarrow 1}{lt} \cfrac{-1}{\cfrac{-\pi}{2} \sin(\cfrac{{\pi}x}{2})} = \cfrac{2}{\pi}$

$\therefore f(1) = \cfrac{2}{\pi}$