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