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$$\begin{aligned} f^{\prime}(x) & =4 x^3-12 x^2+24 x+1 \\ f^{\prime \prime}(x) & =12 x^2-24 x+24 \\ & =12\left(x^2-2 x+2\right) \\ & =12\left\{(x-1)^2+1\right\}>0, \text{\text{ for } all } x \end{aligned}$$$\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.$