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(C) Given the function $f:(5,10) \rightarrow (7,12)$ defined by $f(x) = x + 2\left[\frac{x}{5}\right]$.For $x \in (5, 10)$,the value of $\frac{x}{5}$

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$f^{-1}(x)=x-2$

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(C) Given the function $f:(5,10) \rightarrow (7,12)$ defined by $f(x) = x + 2\left[\frac{x}{5}\right]$.For $x \in (5, 10)$,the value of $\frac{x}{5}$ lies in the interval $(1, 2)$.Therefore,the greatest integer function $\left[\frac{x}{5}\right] = 1$ for all $x \in (5, 10)$.Substituting this into the function definition,we get $f(x) = x + 2(1) = x + 2$.Since $f(x) = x + 2$ is a linear function,it is strictly increasing and therefore one-to-one (injective).To check if it is onto (surjective),we find the range: as $x$ varies from $5$ to $10$,$f(x)$ varies from $5+2=7$ to $10+2=12$. Thus,the range is $(7, 12)$,which matches the codomain.Since $f$ is bijective,$f^{-1}(x)$ exists.Let $y = x + 2$,then $x = y - 2$.Thus,$f^{-1}(x) = x - 2$.

If $[\cdot]$ denotes the greatest integer function and if $f:(5,10) \rightarrow(7,12)$ is a function defined by $f(x)=x+2\left[\frac{x}{5}\right]$,then

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