Let a function $f:(0, \infty) \rightarrow (0, \infty)$ be defined by $f(x)=\left|1-\frac{1}{x}\right|$. Then $f$ is:

(C) neither injective nor surjective

(A) not injective but it is surjective

(D) both injective as well as surjective

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Sets Relations and Functions

Let a function

f : (0, \infty) \to (0, \infty)

be defined by

f (x) = 1 - \frac{1}{x}

. Then

f

is:

Answer: (C)

Write

f (x) = 1 - \frac{1}{x}

, which is

1 - \frac{1}{x}

for

x > 1

and

\frac{1}{x} - 1

for

0 < x < 1

. It is not one-one:

f (2) = \frac{1}{2}

and

f (\frac{2}{3}) = \frac{3}{2} - 1 = \frac{1}{2}

, yet

2 \neq = \frac{2}{3}

. It is not onto either:

f (1) = 0

and

f (x) > 0

elsewhere, so the range is

[0, \infty)

, which does not fill the codomain

(0, \infty)

. Hence

f

is neither injective nor surjective.

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option