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# Engineering

If the function f defined on \left [ \frac{\pi }{6} ,\frac{\pi }{3}\right ] by

f\left ( x \right )=\left\{\begin{matrix} \frac{\sqrt{2}cosx-1}{cotx-1},& x\neq \frac{\pi }{4}\\ k,& x=\frac{\pi }{4} \end{matrix}\right.

is continuous , then k is equal to :

  • Option 1)

    2

  • Option 2)

    \frac{1}{2}

  • Option 3)

     1

  • Option 4)

     \frac{1}{\sqrt{2}}

Answers (1)
S solutionqc

Given that f is continuous at  x=\frac{\pi }{4}

\therefore f\left ( \frac{\pi }{4} \right )k=\lim_{x\rightarrow \frac{\pi }{4}}\frac{\sqrt{2}cos\left ( x \right )-1}{cot\left ( x \right )-1}

   \left ( \frac{0}{0} \right ) form

=\lim_{x\rightarrow \frac{\pi }{4}}\frac{\sqrt{2}\left ( -sin\left ( x \right ) \right )}{-cosec^{2}\left ( x \right )}=\frac{\sqrt{2}\left ( -\frac{1}{\sqrt{2}} \right )}{-\left ( \sqrt{2} \right )^{2}}

=\frac{1}{2}


Option 1)

2

Option 2)

\frac{1}{2}

Option 3)

 1

Option 4)

 \frac{1}{\sqrt{2}}

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