Let P(x) be a polynomial of degree n with real coefficients and is non negative for all real x then `P(x) + P^(1)(x)+ ...+P^(n)(x)` is

Options

A. non negative for all x
B. non positive for all x
C. may be positive or negative
D. can't say

Answer

To solve the problem, we need to analyze the polynomial \( P(x) \) of degree \( n \) with real coefficients that is non-negative for all real \( x \). We want to determine the nature of the expression \( P(x) + P'(x) + P''(x) + \ldots + P^{(n)}(x) \).

### Step-by-Step Solution:

1. **Understanding the Polynomial**:
Let \( P(x) \) be expressed as:
\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]
where \( a_n, a_{n-1}, \ldots, a_0 \) are real coefficients.

2. **Non-Negativity Condition**:
Since \( P(x) \) is non-negative for all \( x \), it implies that the polynomial does not cross the x-axis and may touch it at certain points. Thus, all coefficients \( a_i \) must be such that they do not allow \( P(x) \) to take negative values.

3. **Finding Derivatives**:
The derivatives of \( P(x) \) are:
\[ P'(x) = n a_n x^{n-1} + (n-1) a_{n-1} x^{n-2} + \ldots + a_1 \]
\[ P''(x) = n(n-1) a_n x^{n-2} + (n-1)(n-2) a_{n-1} x^{n-3} + \ldots \]
Continuing this process, we find the \( n^{th} \) derivative \( P^{(n)}(x) \).

4. **Analyzing the Sum**:
We need to analyze the sum:
\[ S(x) = P(x) + P'(x) + P''(x) + \ldots + P^{(n)}(x) \]
Each term \( P^{(k)}(x) \) for \( k = 0, 1, \ldots, n \) is a polynomial of degree \( n-k \).

5. **Degree of the Sum**:
The highest degree term in \( S(x) \) will be from \( P(x) \) which is of degree \( n \). The lower degree derivatives will contribute lower degree terms, but since \( P(x) \) is non-negative, the entire sum \( S(x) \) will also be non-negative.

6. **Conclusion**:
Since all derivatives \( P^{(k)}(x) \) for \( k = 0, 1, \ldots, n \) are polynomials and \( P(x) \) is non-negative, the sum \( S(x) \) must also be non-negative for all \( x \).

Thus, we conclude that:
\[ P(x) + P'(x) + P''(x) + \ldots + P^{(n)}(x) \text{ is non-negative for all } x. \]