Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f\left(x\right)= \left\{\begin{array}{cc}-x& x<1\\ a+{\cos }^{-1}\left(x+b\right)& 1 \le x \le 2\end{array}\right.$

$f\left(x\right)$is continuous at$x=1$as$f\left(x\right)$is differentiable at$x=1$.

$\Rightarrow \underset{x\to 1}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }\left(a+{\cos }^{-1}\left(x+b\right)\right)=f\left(1\right)$

$\Rightarrow$$-1=a+{\cos }^{-1}\left(1+b\right)$

${\cos }^{-1}\left(1+b\right)=-1-a$........$\left(\mathrm{i}\right)$

$f\left(x\right)$is differentiable at$x=1$.

$\Rightarrow$$\mathrm{LHD}=\mathrm{RHD}$

$\Rightarrow$$-1=\frac{-1}{\sqrt{1-{\left(1+b\right)}^2}}$

$\Rightarrow$$1-{\left(1+b\right)}^2=1$

$\Rightarrow$$b= -1$......$\left(\mathrm{ii}\right)$

From$\left(\mathrm{i}\right)$$\Rightarrow$${\cos }^{-1}\left(0\right)= -1-a$

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

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

$a=\frac{-\pi -2}{2}$......$\left(\mathrm{iii}\right)$

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