If `varphi(x)` is a polynomial function and `varphi^(prime)(x)>varphi(x)AAxgeq1a n dvarphi(1)=0,` then `varphi(x)geq0AAxgeq1` `varphi(x)

Options

A. `phi(x)ge0forallxge1`
B. `phi(x)lt0forallxge1`
C. `phi(x)=0forallxge1`
D. none of these

Explanation

To solve the problem, we need to analyze the given conditions about the polynomial function \(\varphi(x)\).

### Step-by-Step Solution:

1. **Understanding the Given Conditions**:
We know that \(\varphi'(x) > \varphi(x)\) for all \(x \geq 1\) and that \(\varphi(1) = 0\). This means that the derivative of the function is always greater than the function itself for \(x\) values greater than or equal to 1.

**Hint**: Identify what the implications of the derivative being greater than the function itself are.

2. **Analyzing the Implication of \(\varphi'(x) > \varphi(x)\)**:
Since \(\varphi'(x) - \varphi(x) > 0\), we can rewrite this as:
\[ \varphi'(x) - \varphi(x) > 0 \implies \varphi'(x) > \varphi(x) \]
This indicates that the function \(\varphi(x)\) is increasing faster than its own value.

**Hint**: Consider the behavior of the function when its derivative exceeds its value.

3. **Using the Initial Condition \(\varphi(1) = 0\)**:
At \(x = 1\), we have \(\varphi(1) = 0\). Since \(\varphi(x)\) is increasing for \(x \geq 1\) (as established from the previous step), and it starts at 0, it must be that \(\varphi(x) > 0\) for all \(x > 1\).

**Hint**: Think about how an increasing function behaves when it starts at zero.

4. **Conclusion**:
From the analysis, we conclude that:
\[ \varphi(x) \geq 0 \quad \text{for all } x \geq 1 \]
However, since \(\varphi(1) = 0\) and the function is increasing, we can say:
\[ \varphi(x) > 0 \quad \text{for all } x > 1 \]

Therefore, the correct answer is:
- \(\varphi(x) \geq 0 \quad \text{for } x \geq 1\)

### Final Answer:
The answer is: **(a) \(\varphi(x) \geq 0\) for \(x \geq 1\)**.