The domain of the real function $f(x)=\frac{1}{\sqrt{4-x^2}}$ is

the set of all positive real numbers

JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Understanding +4 | -1

The domain of the real function

f(x)=\frac{1}{\sqrt{4-x^2}}

View answer and explanation

Correct option: C

(-2,2)

Option C ((-2,2)) is correct because the function is defined only when the expression inside the square root is positive, i.e.,

4 - x^2 > 0

, which gives

-2 < x < 2

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}