If a > 0, then a + a + a + .... is - Vidyadip

MCQ

If $a > 0,$ then $\sqrt {a + \sqrt {a + \sqrt {a + ....\infty } } } $ is

A
$\frac{1}{2}\sqrt {(4a - 1)} $
✓
$\frac{1}{2}[1 + \sqrt {(4a + 1)} ]$
C
$\frac{1}{2}[1 - \sqrt {(4a - 1)} ]$
D
$\frac{1}{2}[1 \pm \sqrt {(4a + 1)} ]$

✓

Answer

Correct option: B.

$\frac{1}{2}[1 + \sqrt {(4a + 1)} ]$

b
(b) $x = \sqrt {a + \sqrt {a + \sqrt {a + .....\infty } } } $

$ \Rightarrow $$x = \sqrt {a + x} $$ \Rightarrow $${x^2} - x - a = 0$$ \Rightarrow $$x = \frac{{1 \pm \sqrt {1 + 4a} }}{2}$

As $a > 0$, $x > 0$; $\therefore $ $+ve$ sign should be considered.

$2B + C = - 1$ $x = \frac{1}{2}(1 + \sqrt {4a + 1} )$.

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