What should be added to \(\cfrac{4x}{x^2-1}\) to get \(\cfrac{x+1}{

1.	What should be added to \(\cfrac{4x}{x^2-1}\) to get \(\cfrac{x+1}{x-1}\) ?A) \(\cfrac{x-1}{x^2-1}\)B) \(\cfrac{x-1}{x+1}\)C) \(\cfrac{x^2-1}{x+1}\)D) \(\cfrac{x^2+1}{x+1}\)
Answer» Correct option is (B) \(\frac{x-1}{x+1}\) Let p(x) should be added to \(\frac{4x}{x^2-1}\) to get \(\frac{x+1}{x-1}.\) i.e., \(\frac{4x}{x^2-1}+p(x)\) \(=\frac{x+1}{x-1}\) \(\Rightarrow\) p(x) \(=\frac{x+1}{x-1}-\frac{4x}{x^2-1}\) \(=\frac{(x+1)^2-4x}{x^2-1}\) \(=\frac{x^2+2x+1-4x}{(x-1)(x+1)}\) \(=\frac{x^2-2x+1}{(x-1)(x+1)}\) \(=\frac{(x-1)^2}{(x-1)(x+1)}\) \(=\frac{x-1}{x+1}\) Correct option is B) \(\cfrac{x-1}{x+1}\)

What should be added to \(\cfrac{4x}{x^2-1}\) to get \(\cfrac{x+1}{x-1}\) ?A) \(\cfrac{x-1}{x^2-1}\)B) \(\cfrac{x-1}{x+1}\)C) \(\cfrac{x^2-1}{x+1}\)D) \(\cfrac{x^2+1}{x+1}\)

Answer»

Correct option is (B) \(\frac{x-1}{x+1}\)

Let p(x) should be added to \(\frac{4x}{x^2-1}\) to get \(\frac{x+1}{x-1}.\)

i.e., \(\frac{4x}{x^2-1}+p(x)\) \(=\frac{x+1}{x-1}\)

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

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

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

\(=\frac{x-1}{x+1}\)

Correct option is B) \(\cfrac{x-1}{x+1}\)