Let f(x)=a x^5+b x^4+c x^3+d x^2+e x, where a, b, c, d, e ∈ R and f(x…

It is given that $\alpha$ is a positive root of $f(x)$ and by inspection, we have $f(0)=0$. Therefore, $x=0$ and $x=\alpha$ are the roots of $f(x)=0$. By Rolle's theorem, $f^{\prime}(x)=0$ has a root $\alpha_1$ between 0 and $\alpha$ i.e. $0<\alpha_1<\alpha$.

So option (a) is correct.

Clearly, $f^{\prime}(x)=0$ is a fourth degree equation in $x$ and imaginary roots always occur in pairs. Since $x=\alpha_1$ is a root of $f^{\prime}(x)=0$.

Therefore, $f^{\prime}(x)=0$ will have another real root, $\alpha_2$ (say).

Now, $\alpha_1$ and $\alpha_2$ are real roots of $f^{\prime}(x)=0$. Therefore, by Rolle's theorem $f^{\prime \prime}(x)=0$ will have a real root between $\alpha_1$ and $\alpha_2$.

Thus, option (b) is correct.

We have seen that $x=0, x=\alpha$ are two real roots of $f(x)=0$. As $f(x)=0$ is a fifth degree equation, it will have at least three real roots. Consequently, by Rolle's theorem $f^{\prime}(x)=0$ will have at least two real roots.

Thus, option (c) is also correct.