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Maths Question 21 – JEE-MAIN 2025

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If the set of all a∈R−{1}, for which the roots of the equation (1−a)x2+2(a−3)x+9=0 are positive is (−∞,−α]∪[β,γ), then 2α+β+γ is equal to _______.

For a quadratic equation Ax2+Bx+C=0 to have both roots positive, three conditions must be satisfied: the discriminant must be non-negative, the sum of the roots must be positive, and the product of the roots must be positive.

Step 1: Apply conditions for positive roots✦ Active

For the quadratic equation (1−a)x2+2(a−3)x+9=0 to have positive roots, we must satisfy three conditions:

1. Discriminant D≥0:

D=(2(a−3))2−4(1−a)(9)≥0 ⟹4(a2−6a+9)−36(1−a)≥0 ⟹4a2−24a+36−36+36a≥0 ⟹4a2+12a≥0⟹4a(a+3)≥0

This implies a∈(−∞,−3]∪[0,∞).

2. Sum of roots S>0:

S=−2(a−3)1−a>0⟹a−31−a<0⟹a−3a−1>0

This implies a∈(−∞,1)∪(3,∞).

3. Product of roots P>0:

P=91−a>0⟹1−a>0⟹a<1

This implies a∈(−∞,1).

Step 2: Find the intersection of the conditions○ Expand

We need to find the values of a that satisfy all three conditions:

a∈((−∞,−3]∪[0,∞))∩((−∞,1)∪(3,∞))∩(−∞,1)

First, intersect the sum and product conditions:

((−∞,1)∪(3,∞))∩(−∞,1)=(−∞,1)

Now, intersect this result with the discriminant condition:

((−∞,−3]∪[0,∞))∩(−∞,1)=(−∞,−3]∪[0,1)

The problem states that a∈R−{1}, which is consistent with our derived interval.

💡 Teacher's Secret Hint

Remember to consider all conditions simultaneously by finding the intersection of their solution sets.

Step 3: Determine α,β,γ and calculate the final expression○ Expand

Comparing the derived interval (−∞,−3]∪[0,1) with the given form (−∞,−α]∪[β,γ), we identify:

−α=−3⟹α=3 β=0 γ=1

The required value is 2α+β+γ:

2(3)+0+1=6+0+1=7
💡 Teacher's Secret Hint

Carefully match the endpoints of the derived interval with the given symbolic representation to find the values of α, β, and γ.

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