Discussion

Explanation:

Let y = $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ 

⇒ y = $\sqrt{2+\mathrm{y}}$

⇒ y² = 2 + y

⇒ y² – y – 2 = 0

⇒ y² – 2y + y – 2 = 0

⇒ (y - 2)(y + 1) = 0

Hence, y = 2 or -1

Since $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ cannot be negative.

Hence, $\sqrt{2+\sqrt{2+\sqrt{2+\mathrm{ }\dots \dots \mathrm{\infty }}}}$ = 2

Hence, option (a).

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