JEE Main 2026 — Jan 28, Shift 2 — Mathematics solution

JEE Main2026Jan 28, Shift 2Mathematics

Given below are two statements :

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

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

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

A
Both Statement I and Statement II are false.
B
Both Statement I and Statement II are true.
C
Statement I is false but Statement II is true.
D
Statement I is true but Statement II is false.

Solution

Statement I: Write $f (x) = \frac{x}{1 + ∣ x ∣}$ piecewise. For $x \geq 0$ , $f (x) = \frac{x}{1 + x}$ , strictly increasing from $0$ to $1$ (as $x \to \infty$ ). For $x < 0$ , $f (x) = \frac{x}{1 - x}$ , strictly increasing from $- 1$ (as $x \to - \infty$ ) to $0$ . The function is monotonic across all of $R$ , hence one-one.

Statement II: Check whether $f (x) = \frac{x ^{2} + 4 x - 30}{x ^{2} - 8 x + 18}$ takes any value twice. Try $f (0) = \frac{- 30}{18} = - \frac{5}{3}$ . Set $\frac{x ^{2} + 4 x - 30}{x ^{2} - 8 x + 18} = - \frac{5}{3}$ , cross-multiply: $3 (x^{2} + 4 x - 30) = - 5 (x^{2} - 8 x + 18) \Rightarrow 8 x^{2} - 28 x - 180 = 0 \Rightarrow 2 x^{2} - 7 x - 45 = 0$ . Discriminant $> 0$ , so there are two real roots ( $x = 0$ being one and $x = - 1$ another by check). Hence $f (0) = f (- 1)$ , making $f$ many-one.

$∴$ Both statements are true.

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