Let $f: R \rightarrow R$ be a function such that $f(x)=\frac{x^{2}+2 x+1}{x^{2}+1}$. Then

(C) $f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$

(A) $f(x)$ is many-one in $(-\infty,-1)$

(B) $f(x)$ is many-one in $(1, \infty)$

(D) $f(x)$ is one-one in $(-\infty, \infty)$

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Sets Relations and Functions

Let

f : R \to R

be a function such that

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

. Then

(A) $f (x)$ is many-one in $(- \infty, - 1)$
(B) $f (x)$ is many-one in $(1, \infty)$
(C) $f (x)$ is one-one in $[1, \infty)$ but not in $(- \infty, \infty)$ ✓ Correct
(D) $f (x)$ is one-one in $(- \infty, \infty)$

Answer: (C)

Write

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

, so

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

. Thus

f^{'} (x) < 0

for

∣ x ∣ > 1

and

f^{'} (x) > 0

for

∣ x ∣ < 1

: the function is strictly decreasing on

(- \infty, - 1)

and on

(1, \infty)

, and strictly increasing on

(- 1, 1)

. On

[1, \infty)

it is strictly monotonic and hence one-one, while over the whole real line it first rises and then falls, taking equal values on opposite sides of

x = 1

. The function is therefore one-one on

[1, \infty)

but not on

(- \infty, \infty)

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option