JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

The domain of definition of the real function

f\left( x \right) = \sqrt {{{\log }_{12}}{x^2}}

of the real variable

x

, is

View answer and explanation

Correct option: B

\left| x \right| \ge 1

Option B (

\left| x \right| \ge 1

) is correct because the logarithm must be non-negative, meaning log_{16}(x²) >= 0, which implies x² >= 1, and thus |x| >= 1.

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