JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Sets Relations and Functions

Let

f (x) = x^{2}, 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?

Answer: (D)

Solution

Take

f (x) = x^{2}

and

g (A) = {x \in R : f (x) \in A}

with

S = [0, 4]

. Since

x^{2} \in [0, 4]

needs

x \in [- 2, 2]

, we have

g (S) = [- 2, 2]

, and then

f (g (S)) = f ([- 2, 2]) = [0, 4] = S

. Also

f (S) = f ([0, 4]) = [0, 16]

, so

g (f (S)) = {x : x^{2} \in [0, 16]} = [- 4, 4]

. Comparing,

f (g (S)) = S

holds,

f (S) \neq = f (g (S))

holds, and

g (f (S)) = [- 4, 4] \neq = S

holds; but

g (f (S)) = [- 4, 4]

is not equal to

g (S) = [- 2, 2]

, so the claim that these last two are equal is the one that fails.

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option