Question : If $\frac{1}{x^2+a^2}=x^2-a^2$, then the value of $x$ is:
Option 1: $\left(1-a^4\right)^\frac{1}{4}$
Option 2: $a$
Option 3: $\left(a^4-1\right)^\frac{1}{4}$
Option 4: $\left(a^4+1\right)^\frac{1}{4}$
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Correct Answer: $\left(a^4+1\right)^\frac{1}{4}$
Solution :
Given, $\frac{1}{x^2+a^2}=x^2-a^2$
$⇒1=(x^2-a^2)(x^2+a^2)$
$⇒1=x^4-a^4$
$⇒x^4 = 1+a^4$
$\therefore x=(1+a^4)^\frac{1}{4}$
Hence, the correct answer is $(a^4+1)^\frac{1}{4}$.
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