Question

Find the value of $a$ and $b$ so that the function $f\left(x\right)=a{\tan }^{-1}\left(\frac{1}{x-4}\right)$;$0\le x<4;$ $f\left(x\right)=b{\tan }^{-1}\left(\frac{2}{x-4}\right)$ for $4<x<6$ $f\left(4\right)=\frac{\pi }{2}$ $f\left(x\right)={\sin }^{-1}(7-x)+\frac{cx}{4}$ for $6\le x\le 8$ is continuous in $[0,8]$

Answer 1

Function continuous at $x=4$

$\begin{array}{c}L.H.L|x=4=R.H.L.|x=4=f\left(4\right)\\ \Rightarrow \underset{x\to 4}{lim}a{\tan }^{-1}\left(\frac{1}{x-4}\right)=\underset{x\to 4}{lim}b{\tan }^{-1}\left(\frac{2}{x-4}\right)=\frac{\pi }{2}\\ \Rightarrow a\frac{\pi }{2}=b\frac{\pi }{2}=\frac{\pi }{2}\\ a=b=1\end{array}$