If `P(x)` is a polynomial such that
`P(x^(2)+1)={P(x)}^(2)+1` and P(0)=0, then P'(0) is equal to

Options

A. 1
B. 0
C. -1
D. none of these

Explanation

To solve the problem, we need to find the value of \( P'(0) \) given the polynomial \( P(x) \) satisfies the equation:

\[ P(x^2 + 1) = P(x)^2 + 1 \]

and the condition \( P(0) = 0 \).

### Step 1: Substitute \( x = 0 \) into the equation

First, we substitute \( x = 0 \) into the equation:

\[ P(0^2 + 1) = P(0)^2 + 1 \]

This simplifies to:

\[ P(1) = 0^2 + 1 \]

Thus, we have:

\[ P(1) = 1 \]

### Step 2: Substitute \( x = 1 \) into the equation

Next, we substitute \( x = 1 \):

\[ P(1^2 + 1) = P(1)^2 + 1 \]

This simplifies to:

\[ P(2) = 1^2 + 1 \]

Thus, we have:

\[ P(2) = 2 \]

### Step 3: Substitute \( x = 2 \) into the equation

Now, we substitute \( x = 2 \):

\[ P(2^2 + 1) = P(2)^2 + 1 \]

This simplifies to:

\[ P(5) = 2^2 + 1 \]

Thus, we have:

\[ P(5) = 4 + 1 = 5 \]

### Step 4: Identify the polynomial

We have found:

- \( P(0) = 0 \)
- \( P(1) = 1 \)
- \( P(2) = 2 \)
- \( P(5) = 5 \)

From these results, we can conjecture that \( P(x) = x \) is a possible solution.

### Step 5: Verify if \( P(x) = x \) satisfies the original equation

Let's verify if \( P(x) = x \) satisfies the original equation:

\[ P(x^2 + 1) = x^2 + 1 \]
\[ P(x)^2 + 1 = x^2 + 1 \]

Both sides are equal, confirming that \( P(x) = x \) is indeed a solution.

### Step 6: Find \( P'(x) \)

Now, we differentiate \( P(x) \):

\[ P'(x) = 1 \]

### Step 7: Evaluate \( P'(0) \)

Finally, we evaluate \( P'(0) \):

\[ P'(0) = 1 \]

### Conclusion

Thus, the value of \( P'(0) \) is:

\[ \boxed{1} \]

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