Question

For a polynomial $g\left(x\right)$ with real coefficients, let $m_g$ denote the number of distinct real roots of $g\left(x\right)$. Suppose $S$ is the set of polynomials with real coefficients defined by $S=\left\{{(x^2–1)}^2(a_0+a_{1}x+a_2{x}^2+a_3{x}^3):a_0,a_1,a_2,a_3\in R\right\}$

For a polynomial $f$, let $f_1$ and $f_2$ denote its first and second order derivatives, respectively.

Then the minimum possible value of $(m_{f1}+m_{f2})$, where $f\in S$, is _____

Answer 1

Given: $f\left(x\right)={(x^2–1)}^{2}h\left(x\right);h\left(x\right)=a_0+a_{1}x+a_2{x}^2+a_3{x}^3$

Now, $\begin{array}{rcl}f\left(1\right)& =& f(–1)\\ & =& 0\end{array}$

$\Rightarrow$ $f\text{'}\left(\alpha \right)=0,\alpha \in (–1,1)$ [Rolle’s Theorem]

Also,

$\begin{array}{rcl}{f}_1\left(1\right)& =& f\text{'}(–1)\\ & =& 0\end{array}$

$\Rightarrow$ $f_1\left(x\right)=0$ has at least $3$ roots, $–1,\alpha ,1$with $–1<\alpha <1$

$\Rightarrow$ $f_2\left(x\right)=0$ will have at least $2$ roots, say $\beta ,\gamma$such that $–1<\beta <\alpha <\gamma <1$ [Rolle’s Theorem]

So, $min\left(m_{f2}\right)=2$

So minimum of $(m_{f1}+m_{f2})=5$

Thus, the minimum possible value of $\mathbf{(}{\mathit{m}}_{\mathbf{f}\mathbf{1}}\mathbf{}\mathbf{+}\mathbf{}{\mathit{m}}_{\mathbf{f}\mathbf{2}}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{}$