The solution of the inequation `4^(-x+0.5) - 7.2^(-x) lt 4, x in R` is

Options

A. `(-2, oo)`
B. `(2, oo)`
C. `(2, (7)/(2))`
D. None of these

Answer

To solve the inequation \( 4^{-x + 0.5} - 7 \cdot 2^{-x} < 4 \), we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting \( 4^{-x + 0.5} \) in terms of base 2:
\[ 4^{-x + 0.5} = (2^2)^{-x + 0.5} = 2^{-2x + 1} \]
Thus, we can rewrite the inequation as:
\[ 2^{-2x + 1} - 7 \cdot 2^{-x} < 4 \]

### Step 2: Substitute \( t = 2^{-x} \)
Let \( t = 2^{-x} \). Then \( 2^{-2x} = t^2 \) and the inequation becomes:
\[ 2t^2 - 7t < 4 \]

### Step 3: Rearrange the inequation
Rearranging gives us:
\[ 2t^2 - 7t - 4 < 0 \]

### Step 4: Solve the quadratic inequality
We will find the roots of the quadratic equation \( 2t^2 - 7t - 4 = 0 \) using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -7, c = -4 \):
\[ t = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot (-4)}}{2 \cdot 2} \]
\[ t = \frac{7 \pm \sqrt{49 + 32}}{4} = \frac{7 \pm \sqrt{81}}{4} = \frac{7 \pm 9}{4} \]
Calculating the two roots:
\[ t_1 = \frac{16}{4} = 4, \quad t_2 = \frac{-2}{4} = -\frac{1}{2} \]

### Step 5: Determine the intervals
The quadratic \( 2t^2 - 7t - 4 < 0 \) is a parabola that opens upwards. The solution will be between the roots:
\[ -\frac{1}{2} < t < 4 \]

### Step 6: Substitute back for \( t \)
Since \( t = 2^{-x} \) and \( t > 0 \), we only consider the interval:
\[ 0 < t < 4 \]

### Step 7: Convert back to \( x \)
Now substituting back for \( t \):
\[ 0 < 2^{-x} < 4 \]
Taking logarithm base 2:
\[ 0 < -x < 2 \]
This simplifies to:
\[ -2 < x < 0 \]

### Final Solution
Thus, the solution for the inequation is:
\[ x \in (-2, 0) \]