Consider two sets $\mathrm{A}=\{\mathrm{x} \in \mathrm{Z}:|(|\mathrm{x}-3|-3)| \leq 1\}$ and $B=\left\{x \in \mathbb{R}-\{1,2\}: \frac{(x-2)(x-4)}{x-1} \log _{e}(|x-2|)=0\right\}$. Then the number of onto functions $f: \mathrm{A} \rightarrow \mathrm{B}$ is equal to :

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Sets Relations and Functions

Consider two sets

A = {x \in Z : ∣ (∣ x - 3∣ - 3) ∣ \leq 1}

and

B = {x \in R - {1, 2} : \frac{( x - 2 ) ( x - 4 )}{x - 1} lo g_{e} (∣ x - 2∣) = 0}

.

Then the number of onto functions

f : A \to B

is equal to :

Answer: (A)

For

A

∣ x - 3∣ - 3 \leq 1

means

2 \leq ∣ x - 3∣ \leq 4

, so

- 1 \leq x \leq 1

5 \leq x \leq 7

; the integer members are

A = {- 1, 0, 1, 5, 6, 7}

with

∣ A ∣ = 6

. For

B

\frac{( x - 2 ) ( x - 4 )}{x - 1} lo g_{e} ∣ x - 2∣ = 0

with

x \neq = 1, 2

holds when

x = 4

∣ x - 2∣ = 1

; the latter gives

x = 3

(as

x = 1

is excluded), so

B = {3, 4}

with

∣ B ∣ = 2

. The number of onto maps from a

6

-element set to a

2

-element set is

2^{6} - 2 = 62

JEE Main · 2026 · 23 Jan 2026 · Shift 2 · Single correct option