Question

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

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

Answer 1

Answer: (A)

$f(x)=\{\begin{array}{cc}-x& x<1\\ a+{\cos }^{-1}(x+b)& 1\le x\le 2\end{array}$
$f\left(x\right)$ is continuous at $x=1$
$\Rightarrow \underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }(a+{\left(cos\right)}^{-1}(x+b))=f(1)$
$\Rightarrow$ $-1=a+{\left(cos\right)}^{-1}\left(1+b\right)$
${\left(cos\right)}^{-1}\left(1+b\right)=-1-a$ ........(i)
$f\left(x\right)$ is differentiable at $x=1$
$\Rightarrow$ LHD = RHD
$\Rightarrow$ $-1=\frac{-1}{\sqrt{1-{\left(1+b\right)}^2}}$
$\Rightarrow$ $1-{\left(1+b\right)}^2=1$
$\Rightarrow$ $b=-1$ ......(ii)
From (i)
${\left(cos\right)}^{-1}(0)=-1-a$
$\therefore$ $-1-a=\frac{\pi }{2}$
$a=-1-\frac{\pi }{2}$
$a=\frac{-\pi -2}{2}$ ......(iii)
$\therefore$ $\frac{a}{b}=\frac{\pi +2}{2}$