If f(x) is continuous at x=π4 where f(x)=cos⁡x−sin⁡xcos⁡2x for x≠π4 then fπ4=

Given function f(x)=cosx−sinxcos2x Given that the above function is continuous at x=π4Hence, fπ4=limx→π4cosx-sinxcos2xUse the LH rule to get the valuefπ4=limx↔π4-sinx-cosx-2sin2x =222 =12

If f(x) is continuous at x=π4 where f(x)=cos⁡x−sin⁡xcos⁡2x for x≠π4 then fπ4=

Given function f(x)=cosx−sinxcos2x Given that the above function is continuous at x=π4Hence, fπ4=limx→π4cosx-sinxcos2xUse the LH rule to get the valuefπ4=limx↔π4-sinx-cosx-2sin2x =222 =12

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$If f (x) is continuous at x = \frac{π}{4} where f (x) = \frac{\cos x - \sin x}{\cos 2 x} for$ $x \neq \frac{π}{4} then f (\frac{π}{4}) =$

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$\frac{1}{\sqrt{2}}$

$- \frac{1}{\sqrt{2}}$

$\sqrt{2}$

- $\sqrt{2}$

answer is A.

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Detailed Solution

Given function $f (x) = \frac{cosx - sinx}{\cos 2 x}$

Given that the above function is continuous at $x = \frac{π}{4}$

Hence, $f (\frac{π}{4}) = \lim_{x \to \frac{π}{4}} \frac{\cos x - \sin x}{\cos 2 x}$

Use the LH rule to get the value

$\begin{array}{rcl} f (\frac{π}{4}) & = & \lim_{x \leftrightarrow \frac{π}{4}} \frac{- \sin x - \cos x}{- 2 \sin 2 x} \\ = & \frac{\frac{2}{\sqrt{2}}}{2} \\ = & \frac{1}{\sqrt{2}} \end{array}$