Question

$f\left(x\right)=\{\begin{array}{c}a\tan {}^{-1}\left(\frac{1}{x-4}\right)if0\le x<4\\ b\tan {}^{-1}\left(\frac{2}{x-4}\right)if4<x<6\\ \sin {}^{-1}(7-x)+a\frac{\pi }{4}if\sin {}^{-1}(7-x)+a\frac{\pi }{4}\end{array}$ and $f\left(4\right)=\pi /2$ is continuous on $(0,8)$ then

• A. $(1,1)$
• B. $(1,-1)$
• C. $(-1,1)$
• D. $(-1,-1)$

Answer 1

Answer: (C)

$(-1,1)$

According to question,

$\begin{array}{c}continuityatx=4\\ f\left(4^{-}\right)={lim}_{x\to 4^{-}}f\left(x\right)={lim}_{x\to 4^{-}}a{\tan }^{-1}\left(\frac{1}{x-4}\right)\\ =a{\tan }^{-1}(-\alpha )\\ =-\frac{a\pi }{2}\\ f\left(4^{+}\right)={lim}_{x\to 4^{+}}f\left(x\right)={lim}_{x\to 4^{+}}b{\tan }^{-1}\left(\frac{2}{x-4}\right)\\ =b{\tan }^{-1}\left(\alpha \right)=b\frac{\pi }{2}\\ fiscontinuousatx=4\\ \therefore f\left(4^{-}\right)=f\left(4^{+}\right)=f\left(4\right)\\ -\frac{a\pi }{2}=b\frac{\pi }{2}=\frac{\pi }{2}\\ -a=b=1\\ a=-1,b=1\\ \therefore (a,b)=(-1,1)\end{array}$

So, the correct option is C.