JEE Main 30 June 2022 Shift 1 Solved Paper

Given quadratic equation,
3x2+(α−6)x+(α+3)=0
Let, a and b are the roots of the equation,
∴a+b=−‌

α−6

3

and ab=‌

α+3

3

For real roots,
D≥0
⇒(α−6)2−4⋅3⋅(α+9)≥0
⇒α2−12α+36−12α−36≥0
⇒α2−24α≥0
⇒α(α−24)≥0∴ Real roots of the equation possible for α>24 or α<0.
Now, sum of square of roots
=a2+b2
=(a+b)2−2ab
=‌

(α−6)2

9

−2⋅‌

(α+3)

3

=‌

α2−12α+36−6α−18

9

=‌

α2−18α+18

9

=f(x)
∴ Sum of square of roots are minimum when a2+b2= minimum.
∴f(α)min=‌

α2−18α+18

9

Value of quadratic equation α2−18α+18 is minimum at α=−‌

b

2a

=−‌

(−18)

2.1

=9
But for real roots α should be less than 0 or greater than 24 .
So, there is no value of α in the range α>24∪α<0 where sum of squares of two real roots is minimum.
∴S is an empty set.