Consider the inequation `9^(x) -a3^(x) - a+ 3 le 0`, where a is real parameter.
The given inequality has at least one negative solution for `a in ` (a) `(-oo,2)` (b) `(3,oo)` (c) `(-2,oo)` (d) `(2,3)`

Options

A. (a) `(-oo,2)`
B. (b) `(3,oo)`
C. (c) `(-2,oo)`
D. (d) `(2,3)`

Explanation

To solve the inequality \(9^x - a \cdot 3^x - a + 3 \leq 0\), where \(a\) is a real parameter, we will follow these steps:

### Step 1: Rewrite the inequality
We can express \(9^x\) as \((3^x)^2\). Let \(r = 3^x\). Then the inequality becomes:
\[ r^2 - ar - (a - 3) \leq 0 \]

### Step 2: Identify the quadratic equation
The quadratic equation in terms of \(r\) is:
\[ r^2 - ar - (a - 3) = 0 \]
To find the roots of this equation, we will use the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = -a\), and \(c = -(a - 3)\).

### Step 3: Calculate the discriminant
The discriminant \(D\) of the quadratic equation is given by:
\[ D = b^2 - 4ac = (-a)^2 - 4 \cdot 1 \cdot (-(a - 3)) = a^2 + 4(a - 3) \]
Simplifying this gives:
\[ D = a^2 + 4a - 12 \]

### Step 4: Determine when the quadratic has real roots
For the quadratic to have real roots, the discriminant must be non-negative:
\[ a^2 + 4a - 12 \geq 0 \]
We can solve this inequality by finding the roots of the equation \(a^2 + 4a - 12 = 0\).

### Step 5: Find the roots using the quadratic formula
Using the quadratic formula:
\[ a = \frac{-4 \pm \sqrt{16 + 48}}{2} = \frac{-4 \pm 8}{2} \]
This gives us the roots:
\[ a_1 = 2 \quad \text{and} \quad a_2 = -6 \]

### Step 6: Analyze the intervals
The quadratic \(a^2 + 4a - 12\) opens upwards (as the coefficient of \(a^2\) is positive). Thus, it is non-negative outside the roots:
- The intervals where \(D \geq 0\) are:
\[ (-\infty, -6] \cup [2, \infty) \]

### Step 7: Determine the conditions for negative solutions
Since \(r = 3^x\) and we are looking for negative solutions, we need \(r\) to be in the range \(0 < r < 1\) (which corresponds to \(x < 0\)). The quadratic must be less than or equal to zero between its roots.

### Step 8: Find the range of \(a\) for negative solutions
For the quadratic \(r^2 - ar - (a - 3) \leq 0\) to have at least one negative solution, we need:
- The roots must be real (which we found occurs for \(a \leq -6\) or \(a \geq 2\)).
- The vertex of the parabola must lie in the interval \(0 < r < 1\).

### Conclusion
After analyzing the conditions, we find that the values of \(a\) for which the inequality has at least one negative solution are:
\[ a \in (-\infty, 2) \cup (3, \infty) \]

Thus, the correct answer is:
**(a) \((-∞, 2)\)** and **(b) \((3, ∞)\)**.