Given,

$\alpha \beta <0$

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

Let $\alpha <0$ and $\beta >0$.

It is given that $f\left(x\right)$ has extremum at $x=\alpha , \beta$.

Therefore, $\alpha$ and $\beta$ are two distinct real roots of $f^{\text{'}}\left(x\right)=0$.

But we know that between two distinct real roots of a polynomial, there is at least one real root of its derivative.

Therefore, $f\left(x\right)$ has three distinct real roots $\lambda , \mu$ and $v$ (say) such that $\lambda <\alpha <\mu <\beta <v$.

Thus, first option is correct.

If $f\left(x\right)=0$ has exactly one positive root, then it is evident from the figure that $v>0$ and $\lambda , \mu <0$.

Therefore, $\alpha <\mu <0$

$\Rightarrow f\left(\alpha \right)f\left(0\right)<0$ [$\because \mu$ lies between $\alpha$ and $0$]

$\Rightarrow f\left(\alpha \right)<0$ $\left[\because f\left(0\right)=1>0\right]$

$\Rightarrow f\left(\beta \right)>0$ $\left[\because f\left(\alpha \right)f\left(\beta \right)<0\right]$

Thus, second option is also correct.

If $f\left(x\right)=0$ has exactly one negative real root, then from the figure, we have $\lambda <0$ and $\mu , v>0$.

$\Rightarrow 0<\mu <\beta$

$\Rightarrow f\left(0\right)f\left(\beta \right)<0$ [$\because \mu$ lies between $0$ and $\beta$]

$\Rightarrow f\left(\beta \right)<0$ $\left[\because f\left(0\right)=1>0\right]$

Thus, third option is also correct.