JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Understanding +4 | -1

f\left( x \right) = \sum\nolimits_{n = 0}^\infty {\frac{{{x^n}}}{{n!}}{{\left( {{{\log }_e}a} \right)}^n}} ,a > 0,a \ne 0

, then at

x = 0,f\left( x \right)

View answer and explanation

Correct option: B

Everywhere differentiable

The correct answer is B (Everywhere differentiable) because the function simplifies to

a^x

, which is differentiable everywhere.

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