Given below are two statements : Statement I: The function \(\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}\) defined by \(\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{1+|\mathrm{x}|}\) is one-one. Statement II : The function \(\mathrm{f}: \mathbf{R} \rightarrow \mathbf{R}\) 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 :

(B) Both Statement I and Statement II are true.

(A) Both Statement I and Statement II are false.

(C) Statement I is false but Statement II is true .

(D) Statement I is true but Statement II is false.

Sets Relations and Functions — JEE Main · 2026 · 28 Jan 2026 · Shift 2 Question with Solution

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 :

Answer: (B)

For Statement I,

f (x) = \frac{x}{1 + ∣ x ∣}

equals

\frac{x}{1 + x}

for

x \geq 0

and

\frac{x}{1 - x}

for

x < 0

. On

x \geq 0

f^{'} (x) = \frac{1}{( 1 + x ) ^{2}} > 0

; on

x < 0

f^{'} (x) = \frac{1}{( 1 - x ) ^{2}} > 0

; and the two branches meet continuously at

x = 0

, so

f

increases strictly on all of

R

and is therefore one-one.

For Statement II, the denominator

x^{2} - 8 x + 18 = (x - 4)^{2} + 2 > 0

, so

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

is defined for every real

x

. Solving

f (x) = - \frac{5}{3}

gives

3 (x^{2} + 4 x - 30) = - 5 (x^{2} - 8 x + 18)

, i.e.

8 x^{2} - 28 x = 0

, so

x = 0

x = \frac{7}{2}

. Two different inputs give the same value

- \frac{5}{3}

, so

f

is many-one.

Both Statement I and Statement II are true.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option