Q. If $f:R \rightarrow R$ and $g:R \rightarrow R$ are defined by $f(x) = |x|$ and $g(x) = \lbrack x - 3\rbrack$ for $x \in R$, then $\left\{ g\left( f(x) \right): - \frac{8}{5} < x < \frac{8}{5} \right\}$ is equal to
A) {0, 1}
B) {1, 2}
C) {-3, -2}
D) {2, 3}
Explanation
(c)
Given that, $f(x) = |x|$ and $g(x) = \lbrack x - 3\rbrack$
For $\ \ \ - \frac{8}{3} < x < \frac{8}{5},\ 0 \leq f(x) < \frac{8}{5}$
Now, for $0 \leq f(x) < 1$,
$\ \ \ \ g\left( f(x) \right) = \left\lbrack f(x) - 3 \right\rbrack = - 3$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because\ - 3 \leq f(x) - 3 < - 2\rbrack$
Again, for $1 \leq f(x) < 16$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ g\left( f(x) \right) = - 2$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \lbrack\because\ - 2 \leq f(x) - 3 < - 14\rbrack$
Hence, required set is $\{ - 3,\ - 2\}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.