Let $f(x)=a x^3+5 x^2+c x+1$ be a polynomial function. If f(x) has extrema at $x=\alpha, \beta$ such that $\alpha \beta<0$ and $f(\alpha) f(\beta)<0$. Then, the equation f(x) = 0 has

all the above

We have,

$\alpha \beta<0$

$\Rightarrow \alpha$ and $\beta$ are of opposite signs.

Let $\alpha<0$ and $\beta>0$.

It is given that $f(x)$ has extrema at $x=\alpha, \beta$. Therefore, $\alpha$ and $\beta$ are two distinct real roots of $f^{\prime}(x)=0$. But, we know that between two distinct real roots of a polynomial there is at least one real root of its derivative. Therefore, $f(x)$ has three distinct real roots $\lambda, \mu$ and $v$ (say) such that

$\lambda<\alpha<\mu<\beta<v$

Thus, option (a) is correct.

If $f(x)=0$ has exactly one positive root, then it is evident from Figure, that $\nu>0$ and $\lambda, \mu<0$

Therefore,

$\alpha<\mu<0$

$\Rightarrow f(\alpha) f(0)<0$               [∵ $\mu$ lies between $\alpha$ and 0]

$\Rightarrow f(\alpha)<0$               [∵ f(0) = 1 > 0]

$\Rightarrow f(\beta)>0$                 [∵ $f(\alpha) f(\beta)<0$]

Thus, option (b) is correct.

If f(x) = 0 has exactly one negative real root, then from Figure, we have

$\lambda<0$ and $\mu ~~v>0$

$\Rightarrow 0<\mu<\beta$

$\Rightarrow f(0) f(\beta)<0$           [∵ $\mu$ lies between 0 and $\beta$]

$\Rightarrow f(\beta)<0$          [∵ f(0) = 1 > 0]

Thus, option (c) is correct.