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BITSAT 2011
Exam
BITSAT
Year
2011
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2011 · Mathematics · Application of Derivatives

The minimum value of (f(x)= ^4 x+ ^4 x) in the interval ( (0,...

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

Options

  1. A(12\frac{1}{2})
  2. B2
  3. C\$\sqrt{2}
  4. D1

Correct answer

A. (12\frac{1}{2})

Step-by-step solution

Let (y=\sin ^4 x+\cos ^4 x) dydx=4sin⁡3xcos⁡x+4cos⁡3x(−sin⁡x)=4sin⁡xcox⁡(sin⁡2x−cos⁡2x) and  =(2sin⁡2x)(−cos⁡2x)=−sin⁡4x and ∴dydx=0⇒sin⁡4x=0⇒4x=0,π,2π,3π\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) \\ \text{\text{ and } } =(2 \sin 2 x)(-\cos 2 x)=-\sin 4 x \\ \text{ and } \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}$
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