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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
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B
2
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C
13+1
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D
12
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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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