Question

Let $f\left(x\right)=a{x}^3+b{x}^2+cx+1$ have extrema at $x=\alpha ,\beta$ such that $\alpha \beta <0$ and $f\left(\alpha \right)f\left(\beta \right),0$. Then the equation $f\left(x\right)=0$ has.

• A. Three equal real roots
• B. One positive root if $f\left(\alpha \right)<0$ and $f\left(\beta \right)>0$
• C. One negative root if $f\left(\alpha \right)>0$ and $f\left(\beta \right)>0$
• D. Three distinct real roots

Answer 1

Answer: (B), (C), (D)

Three distinct real roots

As $f\left(x\right)$ has extrema at $x=\alpha ,\beta$ hence
$\therefore f\left(\alpha \right)=0,f^{\prime }\left(\beta \right)=0,\alpha \beta =0$
Applying Rolle's theorem, between any two real root of $f\left(x\right)=0$, at lease one real root of $f^{\prime }\left(x\right)=0$ must lie. Hence $f\left(x\right)=0$ has atleast two real roots, say $x_1$ & $x_2$. So, the thired root $x_3$ is also real because imaginary roots always occur in conjugate pairs.