Question

$f\left(x\right)=x^2,x\in R$
$S\in [0,4]$
$g\left(A\right)=\{x:x\in R,f(x)\in A\}$ where $A\subset R$
which one is incorrect (where $P\subset Q$ means $P$ is subset of $Q$)

• A. $f\left(g\right(s\left)\right)\ne f\left(s\right)$
• B. $f\left(g\right(s\left)\right)=f\left(s\right)$
• C. $g\left(f\right(s\left)\right)=g\left(s\right)$
• D. $g\left(f\right(s\left)\right)\ne s$

Answer 1

The correct option is C $g\left(f\right(s\left)\right)=g\left(s\right)$

$f\left(s\right)=s^2$ $0\le f\left(s\right)\le 16$ ...(i)
$g\left(s\right)=\{x:x\in R,x^2\in S\}$
$=\{x:x^2\in [0,4\left]\right\}$
$\Rightarrow -2\le g\left(s\right)\le 2$ ...(ii)
from (i) $0\le f\left(s\right)\le 16$
$g\left(f\right(s\left)\right)=\{x:f(x)\in f(s\left)\right\}$
$=\{x:x^2\in [0,16\left]\right\}$
$=\{x:-4\le x\le 4]\}$ $-4\le g\left(f\right(s\left)\right)\le 4$ ...(iii)
from (ii) $-2\le g\left(s\right)\le 2\Rightarrow 0\le \left(g\right(Ss){)}^2\le 4$
$f\left(g\right(s\left)\right)=g(s{)}^2$
$0\le f\left(g\right(s\left)\right)\le 4$ ...(iv)
from (iv) and (i) , (1) is true
from (iv) and $S\in [0,4]$, (2) is true
from (iii) and (ii), (3) is False
from (iii) and $S\in [0,4]$, (4) is true
so (3) option is correct