Aspire Faculty ID #18118 · Topic: JEE Main 2026 (28 January Evening Shift) · Just now
JEE Main 2026 (28 January Evening Shift)

Given below are two statements :

Statement I : The function $ f : \mathbb{R} \rightarrow \mathbb{R} $ defined by
$ f(x) = \frac{x}{1 + |x|} $ is one-one.

Statement II : The function $ f : \mathbb{R} \rightarrow \mathbb{R} $ defined by
$ f(x) = \frac{x^2 + 4x - 30}{x^2 - 8x + 18} $ is many-one.

In the light of the above statements, choose the correct answer from the options given below :

Solution

Statement I:

$ f(x) = \frac{x}{1 + |x|} $

$ f(x) = \begin{cases} \frac{x}{1 + x}, & x \ge 0 \ \frac{x}{1 - x}, & x < 0 \end{cases} $

[Image used in solution — graph]

$ f(x) $ is one-one

Statement II:

$ f(x) = \frac{x^2 + 4x - 30}{x^2 - 8x + 18} $

$ f(0) = -\frac{30}{18} = -\frac{5}{3} $

$ f(-1) = -\frac{5}{3} $

$ \Rightarrow f(0) = f(-1) $

$ \Rightarrow f(x) $ is many-one

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