Q73 / 125 · BITSAT 2011 · Mathematics

The minimum value of \(f(x)=\sin ^4 x+\cos ^4 x\) in the interval \(\left(0, \frac{\pi}{2}\right)\) is

  • Correct answer
    A
    \(\frac{1}{2}\)
  • B
    2
  • C
    \(\sqrt{2}\)
  • D
    1

Solution

Let \(y=\sin ^4 x+\cos ^4 x\) \(\begin{aligned} & \frac{d y}{d x}=4 \sin ^3 x \cos x+4 \cos ^3 x(-\sin x) \\ & =4 \sin x \operatorname{cox}\left(\sin ^2 x-\cos ^2 x\right) \\ & =(2 \sin 2 x)(-\cos 2 x)=-\sin 4 x \\ & \therefore \frac{d y}{d x}=0 \Rightarrow \sin 4 x=0 \Rightarrow 4 x=0, \pi, 2 \pi, 3 \pi \end{aligned}\) or \(\mathrm{x}=0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3 \pi}{4}, \ldots \ldots \ldots, \Rightarrow \mathrm{x}=\frac{\pi}{4}\)

Hence, the correct answer is option: 'A'.

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