Question

The value of ${}^{\prime }{a}^{\prime }$ for which the sum of the squares of the roots of $2x^2-2(a-2)x-a-1=0$ is least is-

• A. $1$
• B. $3/2$
• C. $2$
• D. $-1$

Answer 1

The correct option is B $3/2$

$2x^2-2(a-2)x-a-1=0$

$\Rightarrow 2x^2-2(a-2)x-(a+1)=0$

let $\alpha ,\beta$ be the roots of equation. Then we can write

$\Rightarrow \alpha +\beta =-\frac{-2(a-2)}{2}=a-2$

$\Rightarrow \alpha \beta =\frac{-(a+1)}{2}$

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

$\Rightarrow {\alpha }^2+{\beta }^2=(a-2{)}^2-(2\times \frac{-(a+1)}{2})$

$\Rightarrow {\alpha }^2+{\beta }^2=(a-2{)}^2+a+1$

$\Rightarrow {\alpha }^2+{\beta }^2=a^2+4-4a+a+1$

$\Rightarrow {\alpha }^2+{\beta }^2=a^2-3a+5$

Let ${\alpha }^2+{\beta }^2=S$

Given that $S$ is minimum

Therefore $\Rightarrow \frac{dS}{da}=0$

$\Rightarrow \frac{d}{da}(a^2-3a+5)=0$

$\Rightarrow 2a-3=0$

$\Rightarrow a=\frac{3}{2}$