Question

# If $$y = \dfrac{2}{\sin \theta + \sqrt{3}\cos \theta}$$, then the minimum value of $$y$$ that is greater than zero is

A
1
B
2
C
13+1
D
12

Solution

## The correct option is B $$1$$Y will be minimum if $$\sin\theta +\sqrt{3}\cos\theta$$ is maximum$$\dfrac{d}{d\theta}\sin\theta +\sqrt{3}\cos\theta =0$$$$\dfrac{d}{d\theta}\sin\theta +\sqrt{3}\dfrac{d}{d\theta}\cos\theta =0$$$$\cos\theta -\sqrt{3}\sin\theta =0$$dividing $$\cos\theta$$ to the whole term we get$$\dfrac{\cos\theta}{\cos\theta}-\sqrt{3}\dfrac{\sin\theta}{\cos\theta}=0$$$$1-\sqrt{3}\tan\theta =0$$$$+\sqrt{3}\tan\theta =+1$$$$\tan\theta =\dfrac{1}{\sqrt{3}}$$$$\theta =30^o$$$$\dfrac {d(1-\sqrt{3}\tan\theta )}{dx}=-\sqrt3\sec^2 \theta<0$$$$\Rightarrow$$ at $$\theta =30^o$$ we will get the maximun value of $$\sin\theta +\sqrt{3}\cos\theta$$Now, $$y=\dfrac{2}{\sin\theta +\sqrt{3}\cos\theta}$$$$=\dfrac{2}{\sin 30^o+\sqrt{3}\cos 30^o}$$$$=\dfrac{2}{\dfrac{1}{2}+\sqrt{3}\dfrac{\sqrt{3}}{2}}=\dfrac{2}{\dfrac{1}{2}+\dfrac{3}{2}}$$$$=\dfrac{2}{\dfrac{4}{2}}=\dfrac{4}{4}=1$$so, $$y=1$$.Mathematics

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