Let `f(x)=2^(2x-1)" and "g(x)=-2^(x)+2xlog2.` Then the set of points satisfying `f'(x)gtg'(x)`, is

Options

A. `(0,1)`
B. `[0,1)`
C. `(0,oo)`
D. `[0,oo)`

Answer

To solve the problem, we need to find the set of points satisfying the inequality \( f'(x) > g'(x) \) for the given functions:

1. **Define the functions**:
\[ f(x) = 2^{2x - 1} \]
\[ g(x) = -2^x + 2x \log 2 \]

2. **Find the derivatives \( f'(x) \) and \( g'(x) \)**:

- **Derivative of \( f(x) \)**:
Using the formula for the derivative of \( a^x \), which is \( a^x \log a \), we differentiate \( f(x) \):
\[ f'(x) = 2^{2x - 1} \cdot \log 2 \cdot \frac{d}{dx}(2x - 1) = 2^{2x - 1} \cdot \log 2 \cdot 2 = 2^{2x - 1} \cdot 2 \log 2 = 2^{2x} \log 2 \]

- **Derivative of \( g(x) \)**:
We differentiate \( g(x) \):
\[ g'(x) = -\frac{d}{dx}(2^x) + \frac{d}{dx}(2x \log 2) \]
Using the same derivative formula:
\[ g'(x) = -2^x \log 2 + 2 \log 2 \]

3. **Set up the inequality**:
We need to solve the inequality:
\[ f'(x) > g'(x) \]
Substituting the derivatives we found:
\[ 2^{2x} \log 2 > -2^x \log 2 + 2 \log 2 \]

4. **Simplify the inequality**:
We can factor out \( \log 2 \) (assuming \( \log 2 > 0 \)):
\[ 2^{2x} > -2^x + 2 \]
Rearranging gives:
\[ 2^{2x} + 2^x - 2 > 0 \]

5. **Let \( y = 2^x \)**:
This substitution simplifies our inequality:
\[ y^2 + y - 2 > 0 \]
Factoring the quadratic:
\[ (y - 1)(y + 2) > 0 \]

6. **Find the critical points**:
The critical points are \( y = 1 \) and \( y = -2 \). Since \( y = 2^x \) is always positive, we only consider \( y = 1 \).

7. **Determine the intervals**:
The inequality \( (y - 1)(y + 2) > 0 \) holds when:
- \( y < -2 \) (not applicable since \( y > 0 \))
- \( y > 1 \)

Thus, we have:
\[ y > 1 \implies 2^x > 1 \implies x > 0 \]

8. **Conclusion**:
The set of points satisfying \( f'(x) > g'(x) \) is:
\[ x \in (0, \infty) \]