JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Quadratic Equations and Inequalities

Let $f(x)=x^4-4x^3+12x^2+x-1$.




The discriminant of $x^2-2x+2$ is $4-8=-4<0$, so $f^{\prime \prime }(x)>0$ for all $x$, meaning $f^{\prime }(x)$ is strictly increasing.

Since $f^{\prime }(x)$ is strictly increasing and continuous, it can cross zero at most once. Thus $f(x)$ has at most one local extremum.

Checking values: $f(0)=-1<0$ and $f(1)=1-4+12+1-1=9>0$, so there is a root in $(0,1)$.

For large $x$, $f(x)\to +\infty$, and for large negative $x$, $f(x)\to +\infty$.

Since $f$ has at most one local minimum and $f(0)<0$, the function crosses the $x$-axis exactly twice.

Therefore, there are 2 distinct real roots.