Let \(f: \mathbb{R} \rightarrow \mathbb{R}\) be defined as \(f(x)=x-1\) and \(\mathrm{g}: \mathbb{R}-\{1,-1\} \rightarrow \mathbb{R}\) be defined as \(\mathrm{g}(\mathrm{x})=\frac{\mathrm{x}^{2}}{\mathrm{x}^{2}-1}\). Then the function fog is :

(D) neither one-one nor onto function

(A) one-one but not onto function

(B) onto but not one-one function

(C) both one-one and onto function

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Sets Relations and Functions

Let

f : R \to R

be defined as

f (x) = x - 1

and

g : R - {1, - 1} \to R

be defined as

g (x) = \frac{x ^{2}}{x ^{2} - 1}

.

Then the function fog is :

Answer: (D)

First compute

f (g (x)) = g (x) - 1 = \frac{x ^{2}}{x ^{2} - 1} - 1 = \frac{x ^{2} - ( x ^{2} - 1 )}{x ^{2} - 1} = \frac{1}{x ^{2} - 1}

for

x \neq = \pm 1

.

Since

f (g (x)) = \frac{1}{x ^{2} - 1}

is an even function (

f (g (- x)) = f (g (x))

), it is many-one (not one-one).

For the range: Let

y = \frac{1}{x ^{2} - 1}

. Then

y (x^{2} - 1) = 1

, so

x^{2} = \frac{1 + y}{y}

. For real

x

, we need

\frac{1 + y}{y} \geq 0

, which gives

y \in (- \infty, - 1] \cup (0, \infty)

. Since the range

\neq =

codomain

R

, the function is not onto.

Therefore,

f \circ g

is neither one-one nor onto.

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option