Question : If $2\sin(\frac{\pi x}{2})=x^2+\frac{1}{x^2}$, then the value of $(x-\frac{1}{x})$ is:
Option 1: $–1$
Option 2: $2$
Option 3: $1$
Option 4: $0$
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Correct Answer: $0$
Solution : Given: $2\sin(\frac{\pi x}{2})=x^2+\frac{1}{x^2}$ ⇒ $2\sin(\frac{\pi x}{2})=(x - \frac{1}{x})^2 + 2$ Since $\sin\theta$ has a maximum value equal to 1, ⇒ $(x - \frac{1}{x})$ = 0 Hence, the correct answer is 0.
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