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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.

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divya.saini

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