Correct option is (D) x = 1

\(\sqrt{5x-1}+\sqrt {x-1} = 2\)       __________(1)

\(\Rightarrow\) \(\sqrt{5x-1}=2-\sqrt {x-1}\)

\(\Rightarrow\) \((\sqrt{5x-1})^2=(2-\sqrt {x-1})^2\)     (By squaring both sides)

\(\Rightarrow\) 5x - 1 = 4 + (x - 1) \(-4\sqrt{x-1}\)    \((\because(a-b)^2=a^2+b^2-2ab)\)

\(\Rightarrow\) 5x - 1 - 4 - x + 1 \(=-4\sqrt{x-1}\)

\(\Rightarrow\) 4x - 4 \(=-4\sqrt{x-1}\)

\(\Rightarrow\) \((4x-4)^2=16(x-1)\)

\(\Rightarrow\) \(16(x-1)^2=16(x-1)\)

\(\Rightarrow\) \((x-1)^2=x-1\)

\(\Rightarrow\) \((x-1)^2-(x-1)=0\)

\(\Rightarrow\) (x - 1) (x - 1 - 1) = 0

\(\Rightarrow\) (x - 1) (x - 2) = 0

\(\Rightarrow\) x - 1 = 0 or x - 2 = 0

\(\Rightarrow\) x = 1 or x = 2

Put x = 2 in equation (1), we obtain

\(\sqrt{5x-1}+\sqrt {x-1} = 2\)

\(\Rightarrow\) \(\sqrt{10-1}+\sqrt{2-1} = 2\)

\(\Rightarrow\) 3+1 = 2

\(\Rightarrow\) 4 = 2            (Not satisfied)

Hence, x = 2 is not a solution of given equation.

Put x = 1 in equation (1), we obtain

\(\sqrt{5x-1}+\sqrt {x-1}\) \(=\sqrt{5-1}+\sqrt{1-1}\)

= 2+0 = 2    (Satisfied)

Hence, x = 1 is a solution of given equation.

Correct option is D) x = 1

Solution :-