Question:
If \(P(x)\) is a polynomial such that
\[
P(x^2 + 1) = \{P(x)\}^2 + 1
\]
then \(P'(0)\) is equal to
If \(P(x)\) is a polynomial such that
\[
P(x^2 + 1) = \{P(x)\}^2 + 1
\]
then \(P'(0)\) is equal to
Show Hint
The intensity ratio of fine structure doublets follows statistical weights of initial states.
Updated On: Dec 11, 2025
- 1
- 0
- -1
- None of these
Show Solution
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The Correct Option is A
Solution and Explanation
Try \(P(x) = ax + b\), check if it satisfies the functional equation. You’ll find \(P(x) = x\) works. Then \(P'(x) = 1\), so \(P'(0) = 1\)
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