Question

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

Answer 1

Answer: (A)

$f(x)=\left\{\begin{array}{cc}-x & x<1 \\ a+\cos ^{-1}(x+b) & 1 \leq x \leq 2\end{array}\right.$
$f\left(x\right)$ is continuous at $x=1$
$\Rightarrow \underset{x \rightarrow 1^{-}}{lim}f\left(\right. x \left.\right)=\underset{x \rightarrow 1^{+}}{lim}\left(\right. a + \left(cos\right)^{- 1} \left(\right. x + b \left.\right) \left.\right)=f\left(\right. 1 \left.\right)$
$\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)
$\left(cos\right)^{- 1} \left(\right. 0 \left.\right)= \, -1-a$
$\therefore $ $-1-a=\frac{\pi }{2}$
$a= \, -1-\frac{\pi }{2}$
$a=\frac{- \pi - 2}{2}$ ......(iii)
$\therefore $ $\frac{a}{b}=\frac{\pi + 2}{2}$