Question : What is the value of $\frac{1+x}{1-x^{4}}\div \frac{x^{2}}{1+x^{2}}\times x(1-x)\; ?$
Option 1: $\frac{1}{x}$
Option 2: $x^2-1$
Option 3: $x+1$
Option 4: $x$
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Correct Answer: $\frac{1}{x}$
Solution :
$\frac{1+x}{1-x^{4}}\div \frac{x^{2}}{1+x^{2}}\times x(1-x)\;$
= $\frac{1+x}{(1+x)(1-x)(1+x^{2})}\div \frac{x^{2}}{1+x^{2}}\times x(1-x)\;$
= $\frac{1}{(1-x)(1+x^{2})}×\frac{1+x^{2}}{x^2}\times x(1-x)\;$
= $\frac{1}{(1-x)x^{2}}\times x(1-x)\;$
= $\frac{1}{x}$
Hence, the correct answer is $\frac{1}{x}$.
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