Question

if \(y = x + \sqrt {x + \sqrt {x + \sqrt {x + \ldots \infty } } }\), then y(2) =
1. 4 or 1
2. 4 only
3. 1 only
4. undefined

Answer 1

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.