BITSAT-2015 Problem

Calculus Level 4

If P(x)P(x) is a quadratic polynomial satisfying

P(0)=6P(0) = 6, P(1)=2P(1) = 2, P(2)=0P(2) = 0.

And f(x)=n=0P(n)n!.2nxn\displaystyle f(x) = \sum_{ n = 0}^{\infty } \dfrac{P(n)}{n!.2^n} x^n

Then f(x)f(x) is

Note:- I have changed the options from the original ones.
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