JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

The domain of the real valued function,

f(x)=\sqrt{5-4 x-x^2}+x^2 \log (x+4)

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Correct option: C

-4 < x \leq 1

Option C (-4 < x ≤ 1) is correct because it satisfies both the square root condition (5-4x-x² ≥ 0) and the logarithm condition (x+4 > 0).

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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\}