Question

Let $\alpha$ and $\beta$ are the roots of $(m^2+1)x^2-3x+(m+1{)}^2=0$. If sum of roots is maximum then $|{\theta }^3-{\beta }^3|$

• A. $8\sqrt{5}$
• B. $8\sqrt{3}$
• C. $10\sqrt{5}$
• D. $4\sqrt{3}$

Answer 1

The correct option is A $8\sqrt{5}$

$(m^2+1)-3x+(m+1{)}^2$
$\alpha +\beta =\frac{3}{m^2+1}$
$\alpha \beta =\frac{(m+1{)}^2}{m^2+1}$
$\therefore (\alpha +\beta )=$ is maximum $\therefore m^2+1$ is min
$\therefore m=0$
$\alpha +{\beta }^{=}3,\alpha \beta =1$
$|{\theta }^3-{\beta }^3|$ $=\left|\right(\alpha -\beta \left)\right({\alpha }^2+{\beta }^2+\alpha \beta \left)\right|$
$=\left|\sqrt{(\alpha +\beta {)}^2-4\alpha \beta }\right|$
$\left|\right(\alpha +\beta {)}^2-\alpha \beta |$
$\sqrt{5}.8=8\sqrt{5}$