Question

Let $f\left(x\right)=a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x,$ where $a_i^{{}^{\prime }}s$ are real and $f\left(x\right)=0$ has a positive root ${\alpha }_0.$ Then

• A. $f^{\prime }\left(x\right)=0$ has a root such that ${\alpha }_1$ such that $0<{\alpha }_1<{\alpha }_0$
• B. $f^{\prime }\left(x\right)=0$ has at least one real root
• C. $f^{\prime }\left(x\right)=0$ has at two real toots
• D. none of these

Answer 1

Answer: (A), (B)

$f^{\prime }\left(x\right)=0$ has a root such that ${\alpha }_1$ such that $0<{\alpha }_1<{\alpha }_0$
$f^{\prime }\left(x\right)=0$ has at least one real root

$f\left(x\right)=a_5{x}^5+a_4{x}^4+a_3{x}^3+a_2{x}^2+a_{1}x$

Now $f\left(x\right)$ is a polynomial function. Hence f(x) is continuous for all $xϵR$.

Consider Rolle's theorem

$f\left(x\right)$ is continuous on $[0,{\alpha }_0]$ and differentiable on $(0,{\alpha }_0)$

Also

$f\left(0\right)=f\left({\alpha }_0\right)=0$

Therefore, there exists $0<{\alpha }_1<{\alpha }_0$ such that

$f^{\prime }\left({\alpha }_1\right)=0$.

Hence

$f^{\prime }\left(x\right)$ has atleast one real root in the interval of $[0,{\alpha }_0]$