Solution set of inequality ${\log _e}\frac{{x - 2}}{{x - 3}}$ is

$\left( { - \infty ,\;\infty } \right)$

JEE MATHEMATICS MCQ: FUNCTIONS

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Analyzing +4 | -1

Solution set of inequality

{\log _e}\frac{{x - 2}}{{x - 3}}

View answer and explanation

Correct option: B

\left( { - \infty ,\;2} \right)

Option B (

\left( { - \infty ,\;2} \right)

) is correct because the inequality is satisfied when the argument of the logarithm is between 0 and 1.

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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\}