JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

f\left( x \right) = {x^n} + 4

, then the value of

f\left( 1 \right) + \frac{{f'\left( 1 \right)}}{{1!}} + \frac{{f''\left( 1 \right)}}{{2!}} + ... + \frac{{{f^n}\left( 1 \right)}}{{n!}}

View answer and explanation

Correct option: B

{2^n} + 4

Option B (

{2^n} + 4

) is correct because it is the result of applying Taylor's theorem and the binomial theorem.

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