MHT CET · Maths · Application of Derivatives
The function \(\mathrm{f}(x)=\sin ^4 x+\cos ^4 x\) is increasing in
- A \(0 < x < \frac{\pi}{8}\)
- B \(\frac{\pi}{4} < x < \frac{\pi}{2}\)
- C \(\frac{3 \pi}{8} < x < \frac{5 \pi}{8}\)
- D \(\frac{5 \pi}{8} < x < \frac{3 \pi}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{\pi}{4} < x < \frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(\therefore \mathrm{f}(x) =\sin ^4 x+\cos ^4 x \)
\( \therefore \mathrm{f}^{\prime}(x) =4 \sin ^3 x \cos x-4 \cos ^3 x \sin x \)
\( =4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \)
\( =+2 \sin 2 x \cos 2 x \)
\( =-\sin 4 x\)
\(\therefore \) If \(\mathrm{f}(x)\) is increasing, then \(\mathrm{f}^{\prime}(x)>0\)
\( \text {i.e.,} -\sin 4 x>0 \Rightarrow \pi < 4 x < 2 \pi\)
\(\Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2}\)
\( \therefore \mathrm{f}^{\prime}(x) =4 \sin ^3 x \cos x-4 \cos ^3 x \sin x \)
\( =4 \sin x \cos x\left(\sin ^2 x-\cos ^2 x\right) \)
\( =+2 \sin 2 x \cos 2 x \)
\( =-\sin 4 x\)
\(\therefore \) If \(\mathrm{f}(x)\) is increasing, then \(\mathrm{f}^{\prime}(x)>0\)
\( \text {i.e.,} -\sin 4 x>0 \Rightarrow \pi < 4 x < 2 \pi\)
\(\Rightarrow \frac{\pi}{4} < x < \frac{\pi}{2}\)
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