The true set values of 'a' for which the inequality `int _(a)^(0)(3^(-2x)-2.3^(-x)) dx ge 0` is true, is

Options

A. `[0,1]`
B. `(-infty,-1]`
C. `[0,infty)`
D. `(-infty,-1]cup[0,infty)`

Answer

To solve the inequality

\[ \int_{a}^{0} (3^{-2x} - 2 \cdot 3^{-x}) \, dx \geq 0, \]

we will follow these steps:

### Step 1: Rewrite the Integral
We can rewrite the integral as:

\[ \int_{a}^{0} 3^{-2x} \, dx - 2 \int_{a}^{0} 3^{-x} \, dx \geq 0. \]

### Step 2: Compute the First Integral
The integral of \(3^{-2x}\) is:

\[ \int 3^{-2x} \, dx = \frac{1}{-2 \ln(3)} \cdot 3^{-2x} + C. \]

Now, we evaluate this from \(a\) to \(0\):

\[ \left[ \frac{1}{-2 \ln(3)} \cdot 3^{-2x} \right]_{a}^{0} = \frac{1}{-2 \ln(3)} \left(3^{0} - 3^{-2a}\right) = \frac{1}{-2 \ln(3)} \left(1 - \frac{1}{3^{2a}}\right). \]

### Step 3: Compute the Second Integral
The integral of \(3^{-x}\) is:

\[ \int 3^{-x} \, dx = \frac{1}{-\ln(3)} \cdot 3^{-x} + C. \]

Now, we evaluate this from \(a\) to \(0\):

\[ \left[ \frac{1}{-\ln(3)} \cdot 3^{-x} \right]_{a}^{0} = \frac{1}{-\ln(3)} \left(3^{0} - 3^{-a}\right) = \frac{1}{-\ln(3)} \left(1 - \frac{1}{3^{a}}\right). \]

### Step 4: Substitute Back into the Inequality
Substituting both integrals back into the inequality gives:

\[ \frac{1}{-2 \ln(3)} \left(1 - \frac{1}{3^{2a}}\right) - 2 \cdot \frac{1}{-\ln(3)} \left(1 - \frac{1}{3^{a}}\right) \geq 0. \]

### Step 5: Simplify the Inequality
Combining terms, we have:

\[ \frac{1}{-2 \ln(3)} \left(1 - \frac{1}{3^{2a}}\right) + \frac{2}{\ln(3)} \left(1 - \frac{1}{3^{a}}\right) \geq 0. \]

Multiplying through by \(-2 \ln(3)\) (noting that this reverses the inequality):

\[ 1 - \frac{1}{3^{2a}} - 4 \left(1 - \frac{1}{3^{a}}\right) \leq 0. \]

### Step 6: Expand and Rearrange
Expanding gives:

\[ 1 - \frac{1}{3^{2a}} - 4 + \frac{4}{3^{a}} \leq 0, \]

which simplifies to:

\[ -\frac{1}{3^{2a}} + \frac{4}{3^{a}} - 3 \leq 0. \]

### Step 7: Let \( t = 3^{a} \)
Let \(t = 3^{a}\), then the inequality becomes:

\[ -t^{-2} + \frac{4}{t} - 3 \leq 0. \]

Multiplying through by \(t^2\) (assuming \(t > 0\)) gives:

\[ -1 + 4t - 3t^2 \leq 0. \]

### Step 8: Rearranging
Rearranging yields:

\[ 3t^2 - 4t + 1 \geq 0. \]

### Step 9: Find Roots
Using the quadratic formula, the roots are:

\[ t = \frac{4 \pm \sqrt{(4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}. \]

Thus, the roots are:

\[ t = 1 \quad \text{and} \quad t = \frac{1}{3}. \]

### Step 10: Analyze the Quadratic
The quadratic \(3t^2 - 4t + 1\) opens upwards (since the coefficient of \(t^2\) is positive). The intervals where it is non-negative are:

\[ (-\infty, \frac{1}{3}] \cup [1, \infty). \]

### Step 11: Convert Back to 'a'
Since \(t = 3^a\), we have:

1. \(3^a \leq \frac{1}{3} \Rightarrow a \leq -1\)
2. \(3^a \geq 1 \Rightarrow a \geq 0\)

### Final Result
Thus, the true set of values for \(a\) is:

\[ (-\infty, -1] \cup [0, \infty). \]