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Mathematics · Functions

JEE Main 2026 — 28 January, Evening Shift — Question 8

Given below are two statements : one is labelled as Statement I and the other is labelled as Statement II :

Statement I : The function f:R→R\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R} defined by f(x)=X1+∣x∣\mathrm{f}(\mathrm{x})=\frac{\mathrm{X}}{1+|\mathrm{x}|} is one-one.

Statement II : The function f:R→R\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R} defined by f(x)=x2+4x−30x2−8x+18f(x)=\frac{x^{2}+4 x-30}{x^{2}-8 x+18} is many-one.

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

  1. Option A:

    Both Statement I and Statement II are false.

  2. Option B:

    Both Statement I and Statement II are true.

    Correct
  3. Option C:

    Statement I is false but Statement II is true .

  4. Option D:

    Statement I is true but Statement II is false.

Answer: B

Step-by-step solution

Statement 1: f(x)=x1+∣x∣\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}

f(x)={x1+xifx≥0x1−xifx<0\mathrm{f}(\mathrm{x})= \begin{cases}\frac{\mathrm{x}}{1+\mathrm{x}} & if \mathrm{x} \geq 0 \\ \frac{\mathrm{x}}{1-\mathrm{x}} &if \mathrm{x}<0\end{cases}

f(x)\mathrm{f}(\mathrm{x}) is one-one

Statement 2: f(x)=x2+4x−30x2−8x+18,f(x)=\frac{x^{2}+4 x-30}{x^{2}-8 x+18},

f(0)=−3018=−53f(0)=\frac{-30}{18}=\frac{-5}{3}

−53=x2+4x−30x2−8x+18\frac{-5}{3}=\frac{\mathrm{x}^{2}+4 \mathrm{x}-30}{\mathrm{x}^{2}-8 \mathrm{x}+18}

On solving x=0,−1\mathrm{x}=0,-1

⇒f(0)=f(−1)=−53\Rightarrow \mathrm{f}(0)=\mathrm{f}(-1)=\frac{-5}{3}

∴f(x)\therefore \mathrm{f}(\mathrm{x}) is many-one

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
28 January, Evening Shift
Subject
Mathematics
Chapter
Functions
Topic
One-One, many-one, onto, into, bijective functions

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