Question

$f\left(x\right)=\{\begin{array}{cc}a{\tan }^{-1}\frac{1}{x-4},& 0\le x<4\\ \pi /2& x=4\\ b{\tan }^{-1}\frac{2}{x-4}& 4<x<6\\ {\sin }^{-1}(7-x)+a\pi /4,& 6\le x\le 8\end{array}$

Determine the value of $a$ and $b$ if f(x)is continuous in the interval$[0,8].$

• A. $a=1$
• B. $b=1$
• C. $a=2$
• D. $b=2$

Answer 1

Answer: (A), (B)

$a=1$
$b=1$

At $x=4$

${lim}_{x\to 4^{-}}f\left(x\right)={lim}_{x\to 4^{-}}a{\tan }^{-1}\frac{1}{x-4}=a\frac{\pi }{2}$

${lim}_{x\to 4}f\left(x\right)=f\left(4\right)=\frac{\pi }{2}$

LHL $=$RHL $=f\left(4\right)$ (given it is continuous in $[0,8]$)

$a\frac{\pi }{2}=\frac{\pi }{2}\Rightarrow a=1$

${lim}_{x\to 4^{-}}f\left(x\right)={lim}_{x\to 4^{+}}b{\tan }^{-1}\left(\frac{2}{x-4}\right)=b\frac{\pi }{2}$

$b\frac{\pi }{2}=\frac{\pi }{2}\Rightarrow b=1$