Question

Let S be the set of all integral values of \( \alpha \) for which the sum of squares of two real roots of the quadratic equation \( 3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0 \) is minimum. Then S :

• A. is an empty set
• B. is a singleton
• C. contains exactly two elements
• D. contains more than two elements

Hint:

Sum of squares of roots \(=(\alpha_1+\alpha_2)^2-2\alpha_1\alpha_2\) as a function of alpha, but alpha must also keep the roots real (discriminant condition).

Answer 1

The Correct Option is A

Step 1: Let the roots be p and q (real, distinct). \(p^2+q^2=(p+q)^2-2pq\) with \(p+q=\frac{6-\alpha}{3}\), \(pq=\frac{\alpha+3}{3}\), giving \(f(\alpha)=\dfrac{(\alpha-9)^2-63}{9}\).
Step 2: f is a parabola in alpha with vertex at 9, but the roots are real and distinct only if \(\alpha(\alpha-24)>0\).
Step 3: The admissible set \((-\infty,0)\cup(24,\infty)\) excludes the vertex; \(f\) takes its infimum 2 as \(\alpha\to0^-\) (value at \(\alpha=0\) is 2, with a double root x = 1 which is not two distinct roots), and this value is not attained.
Step 4: So no integral alpha makes the sum of squares minimum; S is the empty set.
Answer: (A) is an empty set