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if \(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } }\), then y(2) =1. 4 or 12. 4 only3. 1 only4. undefined |
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Answer» Correct Answer - Option 2 : 4 only \(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } } \) \(\Rightarrow y - x = \sqrt {x + \sqrt {x + \ldots } }\) \(\Rightarrow y - x = \sqrt y\) ⇒ y = (y –x)2 ⇒ y2 + x2 – 2xy – y = 0 at x = 2, we get, y2 – 5y + 4 = 0 ⇒ (y - 4) (y - 1) = 0 ⇒ y = 4, y = 1 But At x=2, from the given equation \(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } }\), the value of y will be greater than 2. So at x=2, y=1 is not possible. |
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