QJEEM23Ma2350: JEE Main 2023 Mathematics - Class 11 JEEM23Ma2350 Solution: Let be a function such that . Then

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

$\Rightarrow f'(x) = {{({x^2} + 1)(2x + 2) - ({x^2} + 2x + 1)(2x)} \over {{{({x^2} + 1)}^2}}}$

$\Rightarrow f'(x) = {{2 - 2{x^2}} \over {{{({x^2} + 1)}^2}}}$

$= {{2\left( {1 + x} \right)\left( {1 - x} \right)} \over {{{\left( {{x^2} + 1} \right)}^2}}}$

$\therefore$ $f'(x) = 0$ at x = 1 and x = -1. So, x = 1 and x = -1 are point of maxima/minima.

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

As x = -1 is a point of maxima/minima. So it is a boundary point in the range $( - \infty , - 1)$ . So, $f(x)$ is one-one in $( - \infty , - 1)$ .

$\therefore$ Option A is incorrect.

Option B : $f(x)$ is one-one in $( - \infty ,\infty )$ .

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$ . So, $f(x)$ is many-one function in $( - \infty ,\infty )$ .

$\therefore$ Option B is incorrect.

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

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$ . So, $f(x)$ is one-one in $[1,\infty )$ .

As, x = 1 and x = -1 are point of maxima/minima which is present inside the range $( - \infty ,\infty )$ . So, $f(x)$ is many-one function in $( - \infty ,\infty )$ .

$\therefore$ Option C is correct.

Option D : $f(x)$ is many-one in $(1,\infty )$ .

As x = 1 is a point of maxima/minima. So it is a boundary point in the range $[1,\infty )$ . So, $f(x)$ is one-one in $[1,\infty )$ .

$\therefore$ Option D is incorrect.

Note :

Methods to Check One-One Function :

(i) If a function is one-one, any line parallel to $x$ -axis cuts the graph of the function maximum at one point.

(ii) Any continuous function which is entirely increasing or decreasing (no maxima/minima is present) in whole domain will always be one-one function.