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If the function is differentiable at x = 1, then
$\frac{a}{b}$ is equal to :

Option 1)
$\frac{\pi - 2}{2}$

Option 2)
$\frac{-\pi - 2}{2}$

Option 3)
$\frac{\pi + 2}{2}$

Option 4)
$-1-\cos ^{-1}(2)$

Answers (1)

best_answer

As we learnt in

Rule for continuous -

A function is continuous at x = a if and only if

$L=R=V$

L.H.L R.H.L value at x = a.

- wherein

Where

$L=\lim_{x\rightarrow a^{-}}\:f(x)$

$R=\lim_{x\rightarrow a^{+}}\:f(x)$

$V_{I}=\lim_{x\rightarrow a}\:f(x)$

$f\left ( x \right )=\left\{\begin{matrix} -x &\:\:x< 1 \\ a+cos^{-1}\left ( x+b \right ) &\:\:1\leq x\leq 2 \end{matrix}\right.$

$f1\left ( x \right )=\left\{\begin{matrix} -1\\ 0-\frac{1}{\sqrt{1-\left ( x+b \right )^{2}}} \end{matrix}\right.$

$\therefore -1=-\frac{1}{\sqrt{1-\left ( x+b \right )^{2}}}$

$-1=-\frac{1}{\sqrt{1-\left ( 1+b \right )^{2}}}$

$\therefore$ 1+b=0 b=-1

Now f(x) will be continuous at x=1

$\therefore -1=a+cos^{-1}(1-1)$ $=a+cos^{-1}\0$ $=a+\frac{\pi }{2}$

$\therefore \:a=-1-\frac{\pi }{2}$

$\frac{a}{b}=\frac{-1-\frac{\pi }{2}}{-1}=1+\frac{\pi }{2}$

Option 1)

$\frac{\pi - 2}{2}$

This option is incorrect.

Option 2)

$\frac{-\pi - 2}{2}$

This option is incorrect.

Option 3)

$\frac{\pi + 2}{2}$

This option is correct.

Option 4)

$-1-\cos ^{-1}(2)$

This option is incorrect.

Posted by

divya.saini

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