(C) Given the domain $x \in (4, 6)$.For $x \in (4, 6)$,we have $\frac{x}{2} \in (2, 3)$.The greatest integer function $[\frac{x}{2}]$ for any value in

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(C) Given the domain $x \in (4, 6)$.For $x \in (4, 6)$,we have $\frac{x}{2} \in (2, 3)$.The greatest integer function $[\frac{x}{2}]$ for any value in $(2, 3)$ is equal to $2$.Therefore,the function becomes $f(x) = x + 2$.To find the inverse,let $y = f(x) = x + 2$.Solving for $x$,we get $x = y - 2$.Thus,$f^{-1}(x) = x - 2$.

Class 12 Mathematics Relation and Function Inverse Function

Let $f : (4, 6) \to (6, 8)$ be a function defined by $f(x) = x + [\frac{x}{2}]$ (where $[.]$ denotes the greatest integer function),then $f^{-1}(x)$ is equal to

A
$x - [\frac{x}{2}]$
B
$-x - 2$
C
$x - 2$
D
$\frac{1}{x + [\frac{x}{2}]}$

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Relation and Function

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Inverse Function

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MediumIIT 1995

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Let $f: R - \{-\frac{4}{3}\} \rightarrow R$ be a function defined as $f(x) = \frac{4x}{3x+4}$. The inverse of $f$ is the map $g: \text{Range } f \rightarrow R - \{-\frac{4}{3}\}$ given by

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Let $f : (4, 6) \to (6, 8)$ be a function defined by $f(x) = x + [\frac{x}{2}]$ (where $[.]$ denotes the greatest integer function),then $f^{-1}(x)$ is equal to

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Let $f(x)=(x+1)^2-1$ for $x \geq -1$.
Statement-$1$: $S=\{x:f(x)=f^{-1}(x)\}=\{0, -1\}$
Statement-$2$: $f$ is a bijection.