Question 86

What is the value of $$\frac{1+x}{1-x^4}\div\frac{x^2}{1+x^2}\times x(1-x)$$ ?

Solution

Expression : $$\frac{1+x}{1-x^4}\div\frac{x^2}{1+x^2}\times x(1-x)$$

= $$\frac{1+x}{(1-x^2)(1+x^2)}\times\frac{(1+x^2)}{x^2}\times x(1-x)$$

= $$\frac{1+x}{(1-x)(1+x)}\times\frac{1}{x^2}\times x(1-x)$$

= $$\frac{x}{x^2}=\frac{1}{x}$$

=> Ans - (A)

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