If⁡(x)=sin6⁡x+cos6⁡x for x∈ℝ, then f(x) lies in the interval

f(x)=sin6⁡x+cos6⁡x=sin2⁡x+cos2⁡x2−3sin2⁡xcos2⁡xsin2⁡x+cos2⁡x=1−3sin2⁡xcos2⁡x=1−34sin2⁡2x=1−341−cos⁡4x2=58+38cos⁡4x−1≤cos⁡4x≤1⇒58−38≤f(x)≤58+38⇒14≤f(x)≤1

If⁡(x)=sin6⁡x+cos6⁡x for x∈ℝ, then f(x) lies in the interval

f(x)=sin6⁡x+cos6⁡x=sin2⁡x+cos2⁡x2−3sin2⁡xcos2⁡xsin2⁡x+cos2⁡x=1−3sin2⁡xcos2⁡x=1−34sin2⁡2x=1−341−cos⁡4x2=58+38cos⁡4x−1≤cos⁡4x≤1⇒58−38≤f(x)≤58+38⇒14≤f(x)≤1

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$If (x) = \sin^{6} x + \cos^{6} x for x \in ℝ$ , then f(x) lies in the interval

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$[\frac{7}{8}, \frac{5}{4}]$

$[\frac{1}{2}, \frac{5}{8}]$

$[\frac{1}{4}, 1]$

$[\frac{1}{4}, \frac{1}{2}]$

answer is C.

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Detailed Solution

$\begin{array}{l} f (x) = \sin^{6} x + \cos^{6} x = {(\sin^{2} x + \cos^{2} x)}^{2} - 3 \sin^{2} {xcos}^{2} x (\sin^{2} x + \cos^{2} x) \\ = 1 - 3 \sin^{2} {xcos}^{2} x = 1 - \frac{3}{4} \sin^{2} 2 x = 1 - \frac{3}{4} (\frac{1 - \cos 4 x}{2}) = \frac{5}{8} + \frac{3}{8} \cos 4 x \\ - 1 \leq \cos 4 x \leq 1 \Rightarrow \frac{5}{8} - \frac{3}{8} \leq f (x) \leq \frac{5}{8} + \frac{3}{8} \Rightarrow \frac{1}{4} \leq f (x) \leq 1 \end{array}$