JEE MAIN - Mathematics (2019 - 10th April Morning Slot - No. 21)
Let f(x) = x2
, x $$ \in $$ R. For any A $$ \subseteq $$ R, define g (A) = { x $$ \in $$ R : f(x) $$ \in $$ A}. If S = [0,4], then which one of the
following statements is not true ?
g(f(S)) $$ \ne $$ S
f(g(S)) = S
f(g(S)) $$ \ne $$ f(S)
g(f(S)) = g(S)
Explanation
f(x) = x2 x $$ \in $$ R
g(A) = {x $$ \in $$ R : f(x) $$ \in $$ A} S $$ \equiv $$ [0, 4]
g(S) = {x $$ \in $$ R : f(x) $$ \in $$ S}
= {x $$ \in $$ R : 0 $$ \le $$ x2 $$ \le $$ 4}
= {x $$ \in $$ R : –2 $$ \le $$ x $$ \le $$ 2}
$$ \therefore $$ g(S) $$ \ne $$ S
$$ \therefore $$ f(g(S)) $$ \ne $$ f(S)
g(f(S)) = {x $$ \in $$ R : f(x) $$ \in $$ f(S)}
= {x $$ \in $$ R : x2 $$ \in $$ S2}
= {x $$ \in $$ R : 0 $$ \le $$ x2 $$ \le $$ 16}
= {x $$ \in $$ R : –4 $$ \le $$ x $$ \le $$ 4}
$$ \therefore $$ g(f(S)) $$ \ne $$ g(S)
$$ \therefore $$ g(f(S)) = g(S) is incorrect
g(A) = {x $$ \in $$ R : f(x) $$ \in $$ A} S $$ \equiv $$ [0, 4]
g(S) = {x $$ \in $$ R : f(x) $$ \in $$ S}
= {x $$ \in $$ R : 0 $$ \le $$ x2 $$ \le $$ 4}
= {x $$ \in $$ R : –2 $$ \le $$ x $$ \le $$ 2}
$$ \therefore $$ g(S) $$ \ne $$ S
$$ \therefore $$ f(g(S)) $$ \ne $$ f(S)
g(f(S)) = {x $$ \in $$ R : f(x) $$ \in $$ f(S)}
= {x $$ \in $$ R : x2 $$ \in $$ S2}
= {x $$ \in $$ R : 0 $$ \le $$ x2 $$ \le $$ 16}
= {x $$ \in $$ R : –4 $$ \le $$ x $$ \le $$ 4}
$$ \therefore $$ g(f(S)) $$ \ne $$ g(S)
$$ \therefore $$ g(f(S)) = g(S) is incorrect
Comments (0)
