QJEEM22Ma031c: JEE Main 2022 Mathematics - Class 11 JEEM22Ma031c Solution: Let S be the set of all integral values of for which the sum of …

Given quadratic equation,

$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$

Let, a and b are the roots of the equation,

$\therefore$ $a + b = - {{\alpha - 6} \over 3}$

and $ab = {{\alpha + 3} \over 3}$

For real roots,

$D \ge 0$

$\Rightarrow {(\alpha - 6)^2} - 4\,.\,3\,.\,(\alpha + 9) \ge 0$

$\Rightarrow {\alpha ^2} - 12\alpha + 36 - 12\alpha - 36 \ge 0$

$\Rightarrow {\alpha ^2} - 24\alpha \ge 0$

$\Rightarrow \alpha (\alpha - 24) \ge 0$

$\therefore$ \alpha > 24 or \alpha < 0

$\therefore$ Real roots of the equation possible for \alpha > 24 or \alpha < 0 .

Now, sum of square of roots

$= {a^2} + {b^2}$

$= {(a + b)^2} - 2ab$

$= {{{{(\alpha - 6)}^2}} \over 9} - 2\,.\,{{(\alpha + 3)} \over 3}$

$= {{{\alpha ^2} - 12\alpha + 36 - 6\alpha - 18} \over 9}$

$= {{{\alpha ^2} - 18\alpha + 18} \over 9} = f(x)$

$\therefore$ Sum of square of roots are minimum when ${a^2} + {b^2} =$ minimum.

$\therefore$ $f{(\alpha )_{\min }} = {{{\alpha ^2} - 18\alpha + 18} \over 9}$

Value of quadratic equation ${\alpha ^2} - 18\alpha + 18$ is minimum at $\alpha = - {b \over {2a}} = - {{( - 18)} \over {2\,.\,1}} = 9$

But for real roots $\alpha$ should be less than 0 or greater than 24.

So, there is no value of $\alpha$ in the range \alpha > 24 \cup \alpha < 0 where sum of squares of two real roots is minimum.

$\therefore$ S is an empty set.