Let `f(x) = e^(x)- x and g(x) = x^(2) - x , AA x in R`. Then, the set of all `x in R`, when the function `h(x)= (fog)(x)` is increasing, in

Options

A. (a) `[0, infty)`
B. (b) `[-1,-1/2] cup [1/2,infty)`
C. (c) `[0,1/2] cup [1,infty)`
D. (d) `[-1/2,0] cup [1, infty)`

Explanation

To solve the problem, we need to determine the set of all \( x \in \mathbb{R} \) for which the function \( h(x) = (f \circ g)(x) \) is increasing. Here, \( f(x) = e^x - x \) and \( g(x) = x^2 - x \).

### Step 1: Define the composite function
We start by defining the composite function:
\[ h(x) = f(g(x)) = f(x^2 - x) \]
Substituting \( g(x) \) into \( f(x) \):
\[ h(x) = e^{(x^2 - x)} - (x^2 - x) \]

### Step 2: Differentiate \( h(x) \)
To find when \( h(x) \) is increasing, we need to compute the derivative \( h'(x) \):
\[ h'(x) = f'(g(x)) \cdot g'(x) \]
Now, we need to find \( f'(x) \) and \( g'(x) \).

### Step 3: Compute \( f'(x) \) and \( g'(x) \)
The derivative of \( f(x) \):
\[ f'(x) = e^x - 1 \]
The derivative of \( g(x) \):
\[ g'(x) = 2x - 1 \]

### Step 4: Substitute into the derivative of \( h \)
Now substituting \( g(x) \) into \( f' \):
\[ h'(x) = f'(g(x)) \cdot g'(x) = (e^{(x^2 - x)} - 1)(2x - 1) \]

### Step 5: Set \( h'(x) \geq 0 \)
For \( h(x) \) to be increasing, we need:
\[ (e^{(x^2 - x)} - 1)(2x - 1) \geq 0 \]

### Step 6: Analyze the two factors
We analyze the two factors separately:

1. **Factor 1: \( e^{(x^2 - x)} - 1 \geq 0 \)**
- This implies \( e^{(x^2 - x)} \geq 1 \), which occurs when \( x^2 - x \geq 0 \).
- Factoring gives \( x(x - 1) \geq 0 \), which is satisfied for \( x \leq 0 \) or \( x \geq 1 \).

2. **Factor 2: \( 2x - 1 \geq 0 \)**
- This simplifies to \( x \geq \frac{1}{2} \).

### Step 7: Combine the conditions
Now we combine the conditions from both factors:
- From the first factor, we have \( x \leq 0 \) or \( x \geq 1 \).
- From the second factor, we have \( x \geq \frac{1}{2} \).

Thus, the intervals where both conditions are satisfied are:
- For \( x \geq 1 \): This satisfies both conditions.
- For \( x \leq 0 \): This does not satisfy \( x \geq \frac{1}{2} \).

### Step 8: Final intervals
The solution set for \( h(x) \) being increasing is:
\[ x \in \left[0, \frac{1}{2}\right) \cup [1, \infty) \]

### Conclusion
Thus, the set of all \( x \in \mathbb{R} \) for which \( h(x) \) is increasing is:
\[ \left[0, \frac{1}{2}\right) \cup [1, \infty) \]