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If $f: R \rightarrow R$ and $g: R \rightarrow R$ are defined by $f(x)=|x|$ and $g(x)=[x-3]$ for $x \in R$, then $\left\{g(f(x)):-\frac{8}{5} < x < \frac{8}{5}\right\}$ is equal to
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Verified Answer
The correct answer is:
$\{-3,-2\}$
Given that, $f(x)=|x|$ and $g(x)=[x-3]$
For $\quad-\frac{8}{5} < x < \frac{8}{5}, 0 \leq f(x) < \frac{8}{5}$
Now, for $\quad 0 \leq f(x) < 1$,
$$
\begin{aligned}
g(f(x)) & =[f(x)-3] \\
& =-3[\because-3 \leq f(x)-3 < -2]
\end{aligned}
$$
Again, for $1 \leq f(x) < 1.6$
$$
g(f(x))=-2
$$
Hence, required set is $\{-3,-2\}$.
For $\quad-\frac{8}{5} < x < \frac{8}{5}, 0 \leq f(x) < \frac{8}{5}$
Now, for $\quad 0 \leq f(x) < 1$,
$$
\begin{aligned}
g(f(x)) & =[f(x)-3] \\
& =-3[\because-3 \leq f(x)-3 < -2]
\end{aligned}
$$
Again, for $1 \leq f(x) < 1.6$
$$
g(f(x))=-2
$$
Hence, required set is $\{-3,-2\}$.
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