If 2/3 = (x - 1/y) + (x^2 - 1/y^2)+..... to ∞ and xy = 2, then the value of x and y under the condition x < 1 are

Question

If \(\frac{2}{3}\) = \(\bigg(x-\frac{1}{y}\bigg)\) + \(\bigg(x^2-\frac{1}{y^2}\bigg)\)+ ..... to ∞ and xy = 2, then the value of x and y under the condition x < 1 are

(a) \(x=\frac{1}{3}, y = 6\)

(b) \(x=\frac{1}{2}, y = 4\)

(c) \(x=\frac{1}{4}, y = 8\)

(d) \(x=\frac{1}{6}, y = 12\)

Answer 1

(b) \(x=\frac{1}{2}, y = 4\)

Given \(\bigg(x-\frac{1}{y}\bigg)\)+ \(\bigg(x^2-\frac{1}{y^2}\bigg)\) + ..... to ∞ = \(\frac{2}{3}\)

⇒ (x + x² + x³ + .... to ∞) – \(\bigg(\frac{1}{y}+\frac{1}{y^2}+....to\,\infty\bigg)\) = \(\frac{2}{3}\)

⇒ \(\frac{x}{1-x}-\frac{\frac{1}{y}}{1-\frac{1}{y}}\) = \(\frac{2}{3}\) ⇒ \(\frac{x}{1-x}-\frac{1}{y-1}\) = \(\frac{2}{3}\)

⇒ \(\frac{\frac{2}{y}}{1-\frac{2}{y}}-\frac{1}{y-1}\) = \(\frac{2}{3}\)             (∵ xy = 2 ⇒ x = \(\frac{2}{y}\))

⇒ \(\frac{2}{y-2}\) - \(\frac{1}{y-1}\) = \(\frac{2}{3}\) ⇒ \(\frac{2y-2-y+2}{y^2-3y+2}\) = \(\frac{2}{3}\)

⇒ 3y = 2(y² – 3y + 2) 

⇒ 2y² – 9y + 4 = 0 ⇒ 2y² – 8y – y + 4 = 0 

⇒ (2y – 1) (y – 4) = 0.

⇒ y = \(\frac{1}{2}\) or y = 4

When y = \(\frac{1}{2}\), \(x\) = 4 and y = 4 , \(x\) = \(\frac{1}{2}\)

∵ \(x\) < 1 ∴ y = 4, x = \(\frac{1}{2}\) is true.