Minimum value of 27cos⁡2x⋅81sin⁡2x is

Let f(x)=27cos⁡2x81sin⁡2x=34sin⁡2x 33cos2x=33cos⁡2x+4sin⁡2xFor minimum value of given function, 3 cos 2 x+4 sin 2 x will be minimum.∴ -32+42≤3cos2x+4sin2x≤32+42⇒ −5≤3cos⁡2x+4sin⁡2x≤5∴ Minimum of 3cos⁡2x+4sin⁡2x=−5so, minf(x)=3−5=1243

Minimum value of 27cos⁡2x⋅81sin⁡2x is

Let f(x)=27cos⁡2x81sin⁡2x=34sin⁡2x 33cos2x=33cos⁡2x+4sin⁡2xFor minimum value of given function, 3 cos 2 x+4 sin 2 x will be minimum.∴ -32+42≤3cos2x+4sin2x≤32+42⇒ −5≤3cos⁡2x+4sin⁡2x≤5∴ Minimum of 3cos⁡2x+4sin⁡2x=−5so, minf(x)=3−5=1243

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Minimum value of $27^{\cos 2 x} \cdot 81^{\sin 2 x}$ is

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$\frac{1}{243}$

-5

1/5

$\frac{1}{27}$

answer is C.

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Detailed Solution

Let $f (x) = 27^{\cos 2 x} 81^{\sin 2 x} = 3^{4 \sin 2 x} 3^{3 \cos 2 x} = 3^{3 \cos 2 x + 4 \sin 2 x}$
For minimum value of given function, 3 cos 2 x+4 sin 2 x will be minimum.
$∴ - \sqrt{3^{2} + 4^{2}} \leq 3 \cos 2 x + 4 \sin 2 x \leq \sqrt{3^{2} + 4^{2}}$
$\Rightarrow - 5 \leq 3 \cos 2 x + 4 \sin 2 x \leq 5$
$∴ Minimum of 3 \cos 2 x + 4 \sin 2 x = - 5$
so, $min f (x) = 3^{- 5} = \frac{1}{243}$