Question

The minimum value of $ 2\cos \theta +\frac{1}{\sin \theta } $ $ +\sqrt{2}\tan \theta $ the interval $ \left( 0,\frac{\pi }{2} \right) $ is:

• A. $ 2+\sqrt{2} $
• B. $ 3\sqrt{2} $
• C. $ 2\sqrt{3} $
• D. $ 3+\sqrt{2} $
• E. 7

Answer 1

Answer: (B)

Let $ y=2\cos \theta +\frac{1}{\sin \theta }+\sqrt{2}\tan \theta $ $ \frac{dy}{d\theta }=-2\sin \theta -\cos ec\theta \cot \theta +\sqrt{2}{{\sec }^{2}}\theta $ $ =-2\sin \theta -\frac{\cos \theta }{\sin \theta }.\frac{1}{\sin \theta }+\sqrt{2}\frac{1}{{{\cos }^{2}}\theta } $ For extremum, put, $ \frac{dy}{dx}=0 $ $ \Rightarrow $ $ \theta =\frac{\pi }{4} $ Now, $ \frac{{{d}^{2}}y}{d{{x}^{2}}}=-2\cos \theta -[-\cos e{{c}^{3}}\theta +{{\cot }^{3}}\cos ec\theta ] $ $ +\sqrt{2}.2\sec \theta \tan \theta \sec \theta $ $ =-2\cos \theta +\cos e{{c}^{3}}\theta -{{\cot }^{2}}\theta \cos ec\theta $ $ +2\sqrt{2}{{\sec }^{2}}\theta \tan \theta $ $ >0 $ for $ \theta =\frac{\pi }{4} $ $ \therefore $ y is minimum for $ \theta =\frac{\pi }{4}. $ $ \Rightarrow $ $ \min (y)=2.\frac{1}{\sqrt{2}}+\sqrt{2}+\sqrt{2}(1) $ $ =2\sqrt{2}+\sqrt{2}=3\sqrt{2} $