Correct option is (B) \(35x^2-12x+1=0\)

The equation which is satisfied by \(x=\frac15\) has a root \(x=\frac15.\)

(A) \(2x^2-7x+6=0\)

\(\Rightarrow\) \(2(\frac15)^2-\frac{7}5+6=0\)

\(\Rightarrow\frac{2-35+150}{25}=0\)

\(\Rightarrow\frac{117}{25}=0\)          (Not satisfy)

Hence, \(x=\frac15\) is not a root of equation \(2x^2-7x+6=0.\)

(B) \(35x^2-12x+1=0\)

\(\Rightarrow\) \(35(\frac15)^2-\frac{12}5+1=0\)

\(\Rightarrow\frac{7}{5}-\frac{12}{5}+1=0\)

\(\Rightarrow\frac{-5}{5}+1=0\)

\(\Rightarrow\) -1+1 = 0

\(\Rightarrow\) 0 = 0

Hence, \(x=\frac15\) is a root of \(35x^2-12x+1=0.\)

(C) \(35x^2+12x-1=0\)

\(\Rightarrow\) \(35(\frac15)^2+\frac{12}5-1=0\)

\(\Rightarrow\frac{7}{5}+\frac{12}{5}-1=0\)

\(\Rightarrow\frac{19-5}{5}=0\)

\(\Rightarrow\frac{14}{5}=0\)         (Not satisfies)

Hence, \(x=\frac15\) is not a root of \(35x^2+12x-1=0.\)

(D) \(10x^2-3x-1=0\)

\(\Rightarrow\) \(10(\frac15)^2-\frac{3}5-1=0\)

\(\Rightarrow\frac{2}{5}-\frac{3}{5}-1=0\)

\(\Rightarrow-\frac{1}{5}-1=0\)

\(\Rightarrow\frac{-6}{5}=0\)         (Not satisfies)

Hence, \(x=\frac15\) is not a root of equation \(10x^2-3x-1=0.\) 

Correct option is B) 35x² – 12x + 1 = 0