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If $\sin \theta=x+\frac{a}{x}$ for all $x \in R-\{0\}$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$a \leq \frac{1}{4}$
Given equation is
$\sin \theta=x+\frac{a}{x}, x \in R-\{0\}$
$\Rightarrow x^{2}+a=x \sin \theta$
$\Rightarrow x^{2}-x \sin \theta+a=0$
Now, discriminant $=\sqrt{\sin ^{2} \theta-4 a}$
For $x$ to be real root, discriminant $\geq 0$
$\Rightarrow \sqrt{\sin ^{2} \theta-4 a} \geq 0$
$\Rightarrow \sin ^{2} \theta-4 a \geq 0 \Rightarrow \sin ^{2} \theta \geq 4 a$
$\Rightarrow \frac{1}{\sin ^{2} \theta} \leq \frac{1}{4 a} \Rightarrow a \leq \frac{\sin ^{2} \theta}{4}$
$\Rightarrow a \leq \frac{1}{4}\left(\because \sin ^{2} \theta\right.$ lies between 0 and 1$)$
$\sin \theta=x+\frac{a}{x}, x \in R-\{0\}$
$\Rightarrow x^{2}+a=x \sin \theta$
$\Rightarrow x^{2}-x \sin \theta+a=0$
Now, discriminant $=\sqrt{\sin ^{2} \theta-4 a}$
For $x$ to be real root, discriminant $\geq 0$
$\Rightarrow \sqrt{\sin ^{2} \theta-4 a} \geq 0$
$\Rightarrow \sin ^{2} \theta-4 a \geq 0 \Rightarrow \sin ^{2} \theta \geq 4 a$
$\Rightarrow \frac{1}{\sin ^{2} \theta} \leq \frac{1}{4 a} \Rightarrow a \leq \frac{\sin ^{2} \theta}{4}$
$\Rightarrow a \leq \frac{1}{4}\left(\because \sin ^{2} \theta\right.$ lies between 0 and 1$)$
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