Let `f(x)=e^((P+1)x)-e^(x)` for real number `Pgt0,` then
Let `g(t)=int_(t)^(t+1)f(x)e^(t-x)dx.` The value of `t=t_(P),` for which g(t) is minimum, is

Options

A. 0
B. `(1)/(2)`
C. 1
D. non-existent

Answer

To solve the problem step by step, we need to find the value of \( t = t_P \) for which the function \( g(t) \) is minimized. The function \( g(t) \) is defined as:

\[ g(t) = \int_{t}^{t+1} f(x) e^{t-x} \, dx \]

where \( f(x) = e^{(P+1)x} - e^x \).

### Step 1: Differentiate \( g(t) \)

Using the Leibniz rule for differentiation under the integral sign, we can differentiate \( g(t) \):

\[ g'(t) = f(t+1)e^{t-(t+1)} + \int_{t}^{t+1} f(x)(-e^{t-x}) \, dx \]

This simplifies to:

\[ g'(t) = f(t+1)e^{-1} - \int_{t}^{t+1} f(x)e^{t-x} \, dx \]

### Step 2: Substitute \( f(t+1) \) and \( f(t) \)

Now we substitute \( f(t+1) \) and \( f(t) \):

\[ f(t+1) = e^{(P+1)(t+1)} - e^{t+1} = e^{(P+1)t + (P+1)} - e^{t+1} \]
\[ f(t) = e^{(P+1)t} - e^t \]

### Step 3: Rewrite \( g'(t) \)

So we can rewrite \( g'(t) \):

\[ g'(t) = \left( e^{(P+1)t + (P+1)} - e^{t+1} \right)e^{-1} - \int_{t}^{t+1} (e^{(P+1)x} - e^x)e^{t-x} \, dx \]

### Step 4: Set \( g'(t) = 0 \)

To find the critical points, we set \( g'(t) = 0 \):

\[ \left( e^{(P+1)t + (P+1)} - e^{t+1} \right)e^{-1} = \int_{t}^{t+1} (e^{(P+1)x} - e^x)e^{t-x} \, dx \]

### Step 5: Analyze the behavior of \( g'(t) \)

We need to analyze the sign of \( g'(t) \) to determine where \( g(t) \) is minimized.

1. **For \( t < 0 \)**: The function \( g'(t) \) will be positive.
2. **At \( t = 0 \)**: We check the value of \( g'(0) \).
3. **For \( t > 0 \)**: The function \( g'(t) \) will be negative.

### Step 6: Conclusion

From the analysis, we find that \( g(t) \) is decreasing for \( t < 0 \) and increasing for \( t > 0 \). Therefore, the minimum occurs at:

\[ t = 0 \]

Thus, the value of \( t = t_P \) for which \( g(t) \) is minimized is:

\[ \boxed{0} \]