Question

If $\alpha ,\beta$ are roots of the equation $x^2+(3-\lambda )x-2=0,$ then the value of $\lambda$ for which the value of ${\alpha }^2+{\beta }^2$ is minimum is-

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is D $3$

Let $\alpha$ and $\beta$ be the root of the equation

$x^2+(3-\lambda )x-2=0$

$\Rightarrow \alpha +\beta =-(3-\lambda )=\lambda -3$

$\alpha \beta =-2$

${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$

$=(\lambda -3{)}^2-2\times -2$

$=(\lambda -3{)}^2+4={\lambda }^2-6\lambda +9+4$

$={\lambda }^2-6\lambda +13$

$\therefore f\left(\lambda \right)={\lambda }^2-6\lambda +13$

$f\left(\lambda \right)$ is minimum when $f^{\prime }\left(\lambda \right)=0$

$\Rightarrow f^{\prime }\left(\lambda \right)=2\lambda -6=0$

$\Rightarrow \lambda -3=0$

$\Rightarrow \lambda =3$

$f"\left(\lambda \right)=\frac{d}{d\lambda }(2\lambda -6)=2>0$

$\therefore$ Function is concave upwards, so function is minimum at $\lambda =3$