RRB NTPC Mains or Stage 2 Model Paper 3 with solutions

Concept:For an infinite nested square root, set the whole repeating expression equal to a variable and form an equation.Explanation:Let $x = \sqrt{6+\sqrt{6+\sqrt{6+\cdots}}}$.Because the pattern continues infinitely, the inner radical is also equal to $x$.So, $x = \sqrt{6+x}$.Squaring both sides gives $x^2 = 6+x$.Rearrange: $x^2 - x - 6 = 0$.Factorise: $(x-3)(x+2)=0$.Thus, $x=3$ or $x=-2$.A square root value cannot be negative, so $x=3$.Answer:$\sqrt{6+\sqrt{6+\sqrt{6+\cdots}}} = 3$Hence, option A is correct.