JEE Mathematics 2025 Question with Solution

Given: The quadratic equation is

$$(1-a)x^2 + 2(a-3)x + 9 = 0$$

and both its roots are positive.

Find: The value of $2\alpha + \beta + \gamma$ if the set of all such $a$ is $(-\infty,-\alpha] \cup [\beta,\gamma]$.

For a quadratic $Ax^2+Bx+C=0$ to have both roots positive, we need:

1. Real roots:

$$\Delta = B^2 - 4AC \ge 0$$

2. Sum of roots positive:

$$-\frac{B}{A} > 0$$

3. Product of roots positive:

$$\frac{C}{A} > 0$$

Here,

$$A = 1-a, \qquad B = 2(a-3), \qquad C = 9$$

From the product condition,

$$\frac{9}{1-a} > 0$$

Since $9>0$, this gives

$$1-a > 0$$

So,

$$a < 1$$

Now check the sum condition:

$$-\frac{B}{A} = -\frac{2(a-3)}{1-a} = \frac{2(a-3)}{a-1} > 0$$

Because $a<1$, we have $a-1<0$. Hence for the fraction to be positive, we need

$$a-3 < 0$$

which means

$$a<3$$

This is already satisfied whenever $a<1$. So no new restriction is added.

Now apply the discriminant condition:

$$\Delta = [2(a-3)]^2 - 4(1-a)(9) \ge 0$$ $$\Delta = 4(a-3)^2 - 36(1-a) \ge 0$$ $$\Delta = 4\big[(a-3)^2 - 9(1-a)\big] \ge 0$$ $$\Delta = 4(a^2 - 6a + 9 - 9 + 9a) \ge 0$$ $$\Delta = 4(a^2 + 3a) \ge 0$$ $$a(a+3) \ge 0$$

Therefore,

a \in (-\infty,-3] \cup [0,\infty) $$](streamdown:incomplete-link)

Combining with $a<1$, we get

$$a \in (-\infty,-3] \cup [0,1)$$

Thus,

\alpha = 3, \qquad \beta = 0, \qquad \gamma = 1 $$](streamdown:incomplete-link)

Now compute:

$$2\alpha + \beta + \gamma = 2\cdot 3 + 0 + 1 = 7$$

Therefore, the correct option is A.

Given: The equation is

$$(1-a)x^2 + 2(a-3)x + 9 = 0$$

with both roots positive.

Find: The value of $2\alpha+\beta+\gamma$.

If roots are positive, then their product must be positive:

$$\frac{9}{1-a} > 0$$

which immediately gives

$$a<1$$

The sum of roots is

$$-\frac{2(a-3)}{1-a} = \frac{2(a-3)}{a-1}$$

For $a<1$, the denominator is negative. So positivity of the sum requires the numerator to be negative too, which gives

$$a-3<0$$

This is automatically true under $a<1$.

Now require real roots:

$$[2(a-3)]^2 - 4(1-a)(9) \ge 0$$ $$4(a-3)^2 - 36(1-a) \ge 0$$ $$4a(a+3) \ge 0$$ $$a(a+3) \ge 0$$

Hence,

$$a \le -3 \quad \text{or} \quad a \ge 0$$

Intersecting with $a<1$ gives

$$a \in (-\infty,-3] \cup [0,1)$$

So the interval form matches

$$(-\infty,-\alpha] \cup [\beta,\gamma]$$

with the endpoint $1$ excluded already by $a \in \mathbb{R}\setminus\{1\}$. Hence,

$$\alpha=3, \quad \beta=0, \quad \gamma=1$$

Therefore,

$$2\alpha+\beta+\gamma=7$$

So the correct option is A.