Question

The value of $p$ for which the sum of the squares of the roots of $2x^2-2(p-2)x-p-1=0$ is least is

• A. $1$
• B. $\frac{3}{2}$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $\frac{3}{2}$

Given equation is $2x^2-2(a-2)x-a-1=0$
Let $\alpha ,\beta$ be roots of the equation.
$\Rightarrow \alpha +\beta =a-2$ and $\alpha \beta =\frac{-(a+1)}{2}$
${\alpha }^2+{\beta }^2=(\alpha +\beta {)}^2-2\alpha \beta$
$=(a-2{)}^2+(a+1)=a^2-3a+5$
$={(a-\frac{3}{2})}^2+\frac{11}{4}$$\therefore$ ${\alpha }^2+{\beta }^2$ value is least at $a=\frac{3}{2}$
Hence, option B.