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