If the function $f(x)\, = \left\{ {\begin{array}{*{20}{c}}{ - x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 1\,\,\,\,}\\{a + {{\cos }^{ - 1}}(x + b),\,\,\,\,\,\,\,\,\,1 \le x \le 2} \end{array}} \right.$ 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
$f\left( x \right) = \left\{ \begin{array}{l} - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x < 1\\ a + {\cos ^{ - 1}}\left( {x + b} \right)\,\,\,1 \le x \le 2 \end{array} \right.$
$f(x)$ is continuous
$ \Rightarrow \mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} a + {\cos ^{ - 1}}\left( {x + b} \right) = f\left( x \right)$
$ \Rightarrow - 1 = a + {\cos ^{ - 1}}\left( {1 + b} \right)$
${\cos ^{ - 1}}\left( {1 + b} \right) = - 1 - a...\left( 1 \right)$
$f(x)$ is differentiate
$ \Rightarrow \text{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) $
$\Rightarrow {\cos ^{ - 1}}\left( 0 \right) = - 1 - a$
$\therefore - 1 - a = \frac{\pi }{2}$
$a = - 1 - \frac{\pi }{2}$
$a = \frac{{ - \pi - 2}}{2}\,\,\,\,\,\,...\left( 3 \right)$
$\therefore \frac{a}{b} = \frac{{\pi + 2}}{2}$
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