Question : If $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$, then the value of $(a^{2}-ax)$ is:
Option 1: 1
Option 2: 2
Option 3: –1
Option 4: 0
Correct Answer: –1
Solution : To solve $(a^{2}-ax)$, we first need to calculate the value of $a$: $a= \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}} \times \frac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}+\sqrt{x-2}} = \frac{2x+2\sqrt{x^2-4}}{4}= \frac{x+\sqrt{x^2-4}}{2}$ ⇒ $2a=x+\sqrt{x^2-4}$ ⇒ $2a-x=\sqrt{x^2-4}$ Squaring both sides, we have, ⇒ $4a^2+x^2-4ax=x^2-4$ ⇒ $a^2-ax=-1$ Hence, the correct answer is –1.
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