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Q. Assertion–Reason Type Question
Statement-1: \( \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} = 3 \)
Statement-2: \( \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} = x,\ x>0 \)
✏️ Solution
Let \( y = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} \)
Then \( y = \sqrt{6+y} \)
Squaring: \( y^2 = 6 + y \)
\( y^2 – y – 6 = 0 \)
\( (y-3)(y+2) = 0 \Rightarrow y = 3 \) (positive)
So Statement-1 is TRUE
For Statement-2:
Let \( y = \sqrt{x+\sqrt{x+\cdots}} \Rightarrow y = \sqrt{x+y} \)
\( y^2 = x + y \Rightarrow y^2 – y – x = 0 \)
\( y = \frac{1+\sqrt{1+4x}}{2} \neq x \)
So Statement-2 is FALSE
Correct Option: (c)
\( \boxed{\text{(c)}} \)