Question

If the Function f\left( x \right) = \left\\{\begin{array}{ll} \- x,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x < 1 \\\ a + {\cos ^{ - 1}}(x + b),\,\,\,\,\,\,\,\,\, 1 \leqslant x \leqslant 2 \\\ \end{array} \right. is differentiable at x = $1$ then $\dfrac{a}{b}$ is equal to :
(A) $\dfrac{{ - \pi - 2}}{2}$
(B) $\dfrac{{\pi + 2}}{2}$
(C) $\dfrac{{\pi - 2}}{2}$
(D) $- 1 - {\cos ^{ - 1}}2$

Answer 1

In this question there are two values that we have to find that are a, b So we have tried to make the two equations for solving this . As we know that if the function is differentiable at any point then it must continue at that point . Hence one equation is made by differentiability and the other one by the use continuity

Complete step-by-step answer:
It is given in the question that f\left( x \right) = \left\\{\begin{array}{ll} \- x,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x < 1 \\\ a + {\cos ^{ - 1}}(x + b),\,\,\,\,\,\,\,\,\, 1 \leqslant x \leqslant 2 \\\ \end{array} \right.
We know that differentiation of ${\cos ^{ - 1}}x = \dfrac{{ - 1}}{{\sqrt {1 - {x^2} }}}$
So, differentiate F(x) with respect to x , we get

f\left( x \right)=\left\\{\begin{array}{ll} \- 1,\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;x < 1 \\\ 0 + \dfrac{{ - 1}}{{\sqrt {1 - {{(x + b)}^2}} }},\,\,\,\,\,\,\,\,\,1 \leqslant x \leqslant 2 \\\ \end{array} \right.

Hence it is given that at x = $1$ the function is differentiable
$\mathop {\lim }\limits_{x \to {1^ + }} F'(x) = \mathop {\lim }\limits_{x \to {1^ - }} F'(x) = F'(1)$
Hence from the above , we get
As we know from above
$\mathop {\lim }\limits_{x \to {1^ - }} F'(x) = - 1,F'(1) = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 + b)}^2}} }}$
Hence ,
$- 1 = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 + b)}^2}} }}$
$\sqrt {1 - {{(1 + b)}^2}} = 1$
On squaring both side
$1 - {(1 + b)^2} = 1$
$(1 + b) = 0$
$b = - 1$
Hence differentiable then it must be continuous at x = $1$
By the continuity
$\mathop {\lim }\limits_{x \to {1^ + }} F(x) = \mathop {\lim }\limits_{x \to {1^ - }} F(x) = F(1)$
$\mathop {\lim }\limits_{x \to {1^ - }} F(x) = - 1$ and $F(1) = a + {\cos ^{ - 1}}0$ ( as we know that $b = - 1$)
Therefore by equating we get
$a + \dfrac{\pi }{2} = - 1$ as we know that ${\cos ^{ - 1}}0 = \dfrac{\pi }{2}$
hence $a = \dfrac{{ - (2 + \pi )}}{2}$ and $b = - 1$
Now
$\dfrac{a}{b} = \dfrac{{(\pi + 2)}}{2}$

So, the correct answer is “Option B”.

Note: Always remember that , if the function is differentiable then it must be continuous as in this question differentiable is given hence from this property we will find out that it is continuous as well from this we will solve the question .In the inverse trigonometric function always remember the range of the function otherwise you did not proceed to the answer As the ${\cos ^{ - 1}}\theta$ have range in $\left[ {0,\pi } \right]$ that is why we use ${\cos ^{ - 1}}0 = \dfrac{\pi }{2}$.