If f (x) = x – x2 + x3 – x4 + .... ∞ for | x | < 1, then f –1(

1.	If f (x) = x – x2 + x3 – x4 + .... ∞ for \| x \| < 1, then f –1(x) is equal to(a) \(\frac{1+x}{x}\)(b) \(\frac{x}{1+x}\)(c) \(\frac{1-x}{x}\)(d) \(\frac{x}{1-x}\)
Answer» Answer: (d) = \(\frac{x}{1-x}\) f (x) = x – x²+ x³– x⁴ + ..... ∞, \| x \| < 1 This is an infinite G.P. with a = x, r = – x. ∴ f (x) = S_∞ = \(\frac{a}{1-r}\) = \(\frac{x}{1+x}\) Let y = f(x) = \(\frac{x}{1+x}\) ⇒ y (1 + x) = x ⇒ y + xy = x ⇒ y = x (1 – y) ⇒ y = \(\frac{y}{1-y}\) ⇒ \(f^{-1}(x) = \frac{x}{1-x}\)

If f (x) = x – x2 + x3 – x4 + .... ∞ for | x | < 1, then f –1(x) is equal to(a) \(\frac{1+x}{x}\)(b) \(\frac{x}{1+x}\)(c) \(\frac{1-x}{x}\)(d) \(\frac{x}{1-x}\)

Answer»

Answer: (d) = \(\frac{x}{1-x}\)

f (x) = x – x²+ x³– x⁴ + ..... ∞, | x | < 1

This is an infinite G.P. with a = x, r = – x.

∴ f (x) = S_∞ = \(\frac{a}{1-r}\) = \(\frac{x}{1+x}\)

Let y = f(x) = \(\frac{x}{1+x}\)

⇒ y (1 + x) = x

⇒ y + xy = x

⇒ y = x (1 – y)

⇒ y = \(\frac{y}{1-y}\)

⇒ \(f^{-1}(x) = \frac{x}{1-x}\)