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 $\theta =\frac{\pi }{4}.$ $\Rightarrow$ $\min (y)=2.\frac{1}{\sqrt{2}}+\sqrt{2}+\sqrt{2}(1)$ $=2\sqrt{2}+\sqrt{2}=3\sqrt{2}$