Question

If the function $f\left(x\right)=\{\begin{array}{cc}-x,& x<1\\ a+{\cos }^{-1}(x+b),& 1\le x\le 2\end{array}$ is differentiable at $x=1$, then $\frac{a}{b}$ is equal to:

• A. $\frac{-\pi -2}{2}$
• B. $\frac{\pi +2}{2}$
• C. $\frac{\pi -2}{2}$
• D. $-1-{\cos }^{-1}\left(2\right)$

Answer 1

The correct option is B $\frac{\pi +2}{2}$

$f^{\prime }\left(x\right)=$$\{\begin{array}{c}-1,x<1\\ \frac{-1}{\sqrt{1-{(x+b)}^2}}:1\le x\le 2\end{array}$

$LHD=RHD$

$\therefore -1=\frac{1}{\sqrt{(}1-(1+b{)}^2)}$

$\therefore \sqrt{1-(1+b{)}^2}=1$

$\therefore b=-1$

$LHL=RHL$

$\therefore -1=a+co{s}^{-1}(x+b)$

$\therefore -1-a=co{s}^{-1}(1-1)$

$\therefore a=-1-\frac{\pi }{2}$

$\therefore \frac{a}{b}=\frac{\pi +2}{2}$