If f(x) = x100 + x99 + …..+ x + 1, then f’(1) is equal to
As, f(x) = 1 + x + x2 + x3 + ….. + x99 + x100
We know that,
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∴ ![]()
⇒ ![]()
⇒ ![]()
⇒ f’(1) = 1 + 2 + 3 + … + 99 + 100 (total 100 terms)
We know that sum of n terms of an A.P = ![]()
∴ f’(1) = ![]()
Clearly, the above solution suggests that only 1 result is possible which is 5050.
Hence,
The only option (a) is the correct answer.