JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

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

Choose the correct answer:

Explanation

Function Analysis and Derivation

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

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

Now, $f'(x) = 0 + \frac{2(x^2 + 1) - 2x(2x)}{(x^2 + 1)^2}$

$\Rightarrow f'(x) = \frac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2}$

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

Critical Points and Values

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

$f(-1) = \frac{1 - 2 + 1}{1 + 1} = 0$

Interval Analysis

For $x < -1$ , $f'(x)$ is negative ( $\ominus$ ), so $f$ is decreasing.
For $-1 < x < 1$ , $f'(x)$ is positive ( $\oplus$ ), so $f$ is increasing.
For $x > 1$ , $f'(x)$ is negative ( $\ominus$ ), so $f$ is decreasing.

Conclusion

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