If `phi` (x) is a differentiable real valued function satisfying `phi (x) +2phi le 1`, then it can be adjucted as `e^(2x)phi(x)+2e^(2x)phi(x)lee^(2x) or (d)/(dx)(e^(2)phi(x)-(e^(2x))/(2))le or (d)/(dx)e^(2x)(phi(x)-(1)/(2))le0`
Here `e^(2x)` is called integrating factor which helps in creating single differential coefficeint as shown above. Answer the following question:
If p(1)=0 and `dP(x)/(dx)ltP(x)` for all `xge1` then

Options

A. `P(x) gt 0 forall x gt 1`
B. P(x) is a constant function
C. `P(x) lt 0 forall x gt 1`
D. none of these

Explanation

To solve the given problem, we will follow these steps:

### Step 1: Understand the given conditions
We are given that \( p(1) = 0 \) and \( \frac{dP(x)}{dx} < P(x) \) for all \( x \geq 1 \). This means that the derivative of \( P(x) \) is less than the function itself for \( x \geq 1 \).

**Hint:** Identify the implications of the inequality involving the derivative and the function.

### Step 2: Rewrite the inequality
From the condition \( \frac{dP(x)}{dx} < P(x) \), we can rearrange it as:
\[ \frac{dP(x)}{dx} - P(x) < 0 \]
This suggests that the function \( P(x) \) is decreasing.

**Hint:** Consider how the derivative relates to the behavior of the function.

### Step 3: Introduce an integrating factor
To analyze the behavior of \( P(x) \), we can multiply both sides of the inequality by \( e^{-x} \):
\[ e^{-x} \frac{dP(x)}{dx} - e^{-x} P(x) < 0 \]
This can be rewritten as:
\[ \frac{d}{dx}(P(x)e^{-x}) < 0 \]
This indicates that the function \( P(x)e^{-x} \) is decreasing.

**Hint:** Think about the implications of a decreasing function.

### Step 4: Evaluate at a specific point
Since \( P(1) = 0 \), we can evaluate \( P(x)e^{-x} \) at \( x = 1 \):
\[ P(1)e^{-1} = 0 \cdot e^{-1} = 0 \]
Since \( P(x)e^{-x} \) is decreasing and equals 0 at \( x = 1 \), it must be greater than or equal to 0 for all \( x \geq 1 \).

**Hint:** Use the properties of decreasing functions to determine the values at other points.

### Step 5: Conclude the behavior of \( P(x) \)
Since \( P(x)e^{-x} \geq 0 \) for \( x \geq 1 \), we can conclude that:
\[ P(x) \geq 0 \text{ for all } x \geq 1 \]
Thus, \( P(x) \) is non-negative for all \( x \geq 1 \).

**Hint:** Summarize the findings to reach the final conclusion.

### Final Conclusion
We have shown that \( P(x) \geq 0 \) for all \( x \geq 1 \).