Question

If the function
$f(x) = \begin{cases} -x & \quad x < 1 \\ a + \cos^{-1} (x + b), & \quad \leq x \leq 2 \end{cases}$
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. $-1 - \cos^{-1}(2)$

Answer 1

Answer: (C)

$f(x) = \begin{cases} -x & \text{x < 1} \\[2ex] a+cos^{-1}(x+b) & \text{1$\le x\le2$} \end{cases}$
f (x) is continuous
$\Rightarrow $ $\displaystyle \lim_{x \to 1^-}\,f (x)=$$\displaystyle \lim_{x \to 1^-}$$a+cos^{-1}\left(x+b\right)=f\left(x\right)$
$\Rightarrow -1=a+cos^{-1}\left(1+3\right)$
$cos^{-1}\left(1+b\right)=-1-a ....\left(1\right)$
f (x) is differentiate
$\Rightarrow $ LHD = RHD
$\Rightarrow $$-1=\frac{-1}{\sqrt{1-\left(1+b\right)^{2}}}$
$\Rightarrow 1-\left(1+b\right)^{2}=1$
$\Rightarrow b=-1 ...\left(2\right)$
From $\left(1\right) \Rightarrow cos^{-1}\left(0\right)=-1-a$
$\therefore -1-a=\frac{\pi}{2}$
$a=-1-\frac{\pi}{2}$
$a=\frac{-\pi-2}{2} ...\left(3\right)$
$\therefore \frac{a}{b}=\frac{\pi+2}{2}$