JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Application +4 | -1

f:[1, \infty) \rightarrow[2, \infty)

is given by

f(x)=x+\frac{1}{x}

, then

f^{-1}(x)

equals

View answer and explanation

Correct option: A

\frac{x+\sqrt{x^2-4}}{2}

Option A (

\frac{{x+\sqrt{{x^2-4}}}}{{2}}

) is correct because it represents the inverse function of

f(x) = x + \frac{{1}}{{x}}

with the given domain and range.

More Questions from FUNCTIONS

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\}