Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

$f\left(x\right)$ is a polynomial in $x$ such that $f\left(x\right)=\left\{{\left(x^2-1\right)}^2 \left(a_0+a_{1}x+a_2{x}^2+a_3{x}^3\right) : a_0, a_1, a_2, a_3\in R\right\}$

So, $f^{\text{'}}\left(x\right)=\left(x^2-1\right) \left[q\left(x\right)\right]$ where $q\left(x\right)$ is polynomial roots of $f^{\prime }\left(x\right)=-1, 1$

So, by Rolle’s theorem, $f\left(1\right)=0=f\left(-1\right)\Rightarrow f\text{'}\left(\alpha \right)=0$, where $\alpha \in \left(-1, 1\right)$

$\Rightarrow$ minimum $m_{f^{\prime }}=3$

Again by Rolle’s theorem, at least one real root lies in $\left(-1, \alpha \right)$ and at least one real root lies in $\left(\alpha , 1\right)$ of $f^{″}\left(x\right)$

So, minimum $m_{f^{″}}=2$
$\Rightarrow$ Minimum possible value of $\left(m_{f^{\prime }}+m_{f^{″}}\right)=5$