The number of distinct real roots of \(x^4-4 x^3+12 x^2+x-1=0\) is

\(f(x)=x^4-4 x^3+12 x^2+x-1\)
\[
\begin{aligned}
f^{\prime}(x) & =4 x^3-12 x^2+24 x+1 \\
f^{\prime \prime}(x) & =12 x^2-24 x+24 \\
& =12\left(x^2-2 x+2\right) \\
& =12\left\{(x-1)^2+1\right\}>0, \text { for all } x
\end{aligned}
\]
\(\Rightarrow f^{\prime}(x)\) is increasing.
Since, \(f^{\prime}(x)\) is cubic and increasing. \(\Rightarrow f^{\prime}(x)\) has only one real root and two imaginary roots.
\(\therefore f(x)\) cannot have all distinct root.
\(\Rightarrow\) Atmost 2 real roots.
Now, \(f(-1)=15\),
\(f(0)=-1\) and \(f(1)=9\)
\(\therefore f(x)\) must have one root in \((-1,0)\) and other in \((0,1)\).
\(\Rightarrow 2\) real roots.