Question

If $\alpha $ and $\beta $ are the real roots of the equation $x^{2}-\left(\right.k-2\left.\right)x+\left(k^{2} + 3 k + 5\right)=0\left(\right.k\in R\left.\right)$ Find the maximum and minimum values of $\left(\left(\alpha \right)^{2} + \left(\beta \right)^{2}\right)$

• A. $18,50/9$
• B. $18,25/9$
• C. $27,50/9$
• D. None of these

Answer 1

Answer: (A)

For real roots $D \geq 0(k-2)^2-4$
$k^2+3 k+5 \geq 0$
$k^2+4-4 k-4 k^2-12 k-20 \geq 0 $
$-3 k^2-16 k-16 \geq 0 ; 3 k^2+16 k+16 \leq 0 $
$k+\frac{4}{3}(k+4) \leq 0$
Now $E=\alpha^2+\beta^2 $
$E=(\alpha+\beta)^2-2 \alpha \beta $
$E=(k-2)^2-2 k^2+3 k+5=-k^2-10 k $
$-6$
$E=-k^2+10 k+6=-(k+5)^2-19$
$=19-(k+5)^2$
$\therefore E_{\min } \text { Occurs when } k=-4 / 3 $
$\therefore E_{\min }=19-\frac{121}{9}=\frac{171-121}{9}=\frac{50}{9} $
$E_{\max } \text { Occurs when } k=-4 $
$E_{\max }=19-1=18$