Question

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

Answer 1

Answer: (C)

$f(x)=\{\begin{array}{ll}-x& \text{x < 1}\\ a+co{s}^{-1}(x+b)& \text{1≤x≤2}\end{array}$
f (x) is continuous
$\Rightarrow$ $\underset{x\to 1^{-}}{\lim }f(x)=$$\underset{x\to 1^{-}}{\lim }$$a+co{s}^{-1}\left(x+b\right)=f\left(x\right)$
$\Rightarrow -1=a+co{s}^{-1}\left(1+3\right)$
$co{s}^{-1}\left(1+b\right)=-1-a....\left(1\right)$
f (x) is differentiate
$\Rightarrow$ LHD = RHD
$\Rightarrow$$-1=\frac{-1}{\sqrt{1-{\left(1+b\right)}^2}}$
$\Rightarrow 1-{\left(1+b\right)}^2=1$
$\Rightarrow b=-\mathrm{1...}\left(2\right)$
From $\left(1\right)\Rightarrow co{s}^{-1}\left(0\right)=-1-a$
$\therefore -1-a=\frac{\pi }{2}$
$a=-1-\frac{\pi }{2}$
$a=\frac{-\pi -2}{2}...\left(3\right)$
$\therefore \frac{a}{b}=\frac{\pi +2}{2}$