The rational expression A = \((\frac{x+1}{x-1}\) - \(\frac{x-1}{x+1

1.	The rational expression A = \((\frac{x+1}{x-1}\) - \(\frac{x-1}{x+1}\) - \(\frac{4x}{x^2+1})\) is multiplied with the additive inverse of B = \(\frac{1-x^2}{4x}\) to get C, then C = ……………A) 2 B) \(\frac{2x}{x^4-1}\)C) \(\frac{32x^2}{x^4-1}\)D) 1
Answer» Correct option is (A) 2 \(C=A\times-B\) \(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right)(\frac{x^4-1}{4x})\) \(=\left(\frac{(x+1)^2(x^2+1)-(x-1)^2(x^2+1)-4x(x^2-1)}{(x-1)(x+1)(x^2+1)}\right)(\frac{x^4-1}{4x})\) \(=\left(\frac{(x^2+1)(x^2+2x+1-(x^2-2x+1))-4x(x^2-1)}{(x^2-1)(x^2+1)}\right)(\frac{x^4-1}{4x})\) \(=\frac{4x(x^2+1-x^2+1)}{x^4-1}\times\frac{x^4-1}{4x}\) = 2 Correct option is A) 2

The rational expression A = \((\frac{x+1}{x-1}\) - \(\frac{x-1}{x+1}\) - \(\frac{4x}{x^2+1})\) is multiplied with the additive inverse of B = \(\frac{1-x^2}{4x}\) to get C, then C = ……………A) 2 B) \(\frac{2x}{x^4-1}\)C) \(\frac{32x^2}{x^4-1}\)D) 1

Answer»

Correct option is (A) 2

\(C=A\times-B\)

\(=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x}{x^2+1}\right)(\frac{x^4-1}{4x})\)

\(=\left(\frac{(x+1)^2(x^2+1)-(x-1)^2(x^2+1)-4x(x^2-1)}{(x-1)(x+1)(x^2+1)}\right)(\frac{x^4-1}{4x})\)

\(=\left(\frac{(x^2+1)(x^2+2x+1-(x^2-2x+1))-4x(x^2-1)}{(x^2-1)(x^2+1)}\right)(\frac{x^4-1}{4x})\)

\(=\frac{4x(x^2+1-x^2+1)}{x^4-1}\times\frac{x^4-1}{4x}\)

= 2

Correct option is A) 2