JEE MATHEMATICS MCQ: FUNCTIONS

1 Easy

1 of 1

Analyzing +4 | -1

Let

f: R^{+} \rightarrow R

be a function which satisfy

f(x) f(y)=f(x y)+2\left(\frac{1}{x}+\frac{1}{y}+1\right) \forall x, y>0

then the value of

f\left(\frac{1}{2}\right)

View answer and explanation

Correct option: C

\frac{-8}{3}

Option C (

\frac{-8}{3}

) is correct because it is the value obtained by substituting specific values into the given functional equation and solving for f(1/2).

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

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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\}