Consider two sets A=\ x ∈ Z:|(|x-3|-3)| ≤ 1\ and B=\ x ∈ ℝ-\ 1,2\ : ((x-2)(x-4))/(x-1) _ e (|x-2|)=0\ . Then the number of onto functions f: A arrow B is equal to

Correct answer: C. 62 For set A : |(|x-3|-3)| ≤ 1 2 ≤ |x-3| ≤ 4 . If x ≥ 3 : 5 ≤ x ≤ 7 x ∈ \ 5,6,7\ . If x So A = \ -1,0,1,5,6,7\ , |A|=6 . For set B : ((x-2)(x-4))/(x-1) _e(|x-2|) = 0 with x ∈ ℝ-\ 1,2\ . Either ((x-2)(x-4))/(x-1)=0 x=4 (since x ≠ 1,2 ), or _e|x-2|=0 |x-2|=1 x=3 (since x ≠ 1 ). So B = \ 3,4\ , |B|=2 . Number of onto functions from A to B = 2^6 - 2 = 62 .

JEE Main 202623 JanuaryEvening Shift

Exam: JEE Main
Year: 2026
Date: 23 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2026 · 23 January · Evening Shift · Mathematics · Functions

Consider two sets A=\ x ∈ Z:|(|x-3|-3)| ≤ 1\ and B=\ x ∈ ℝ-\ 1,2\ :...

Consider two sets

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

and

\mathrm{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

Options

Correct answer

C. 62

Step-by-step solution

For set

A

|(|x-3|-3)| \le 1 \Rightarrow 2 \le |x-3| \le 4

.
If

x \ge 3

5 \le x \le 7 \Rightarrow x \in \{5,6,7\}

.
If

x < 3

-1 \le x \le 1 \Rightarrow x \in \{-1,0,1\}

.
So

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

|A|=6

.
For set

B

\frac{(x-2)(x-4)}{x-1}\log_e(|x-2|) = 0

with

x \in \mathbb{R}-\{1,2\}

.
Either

\frac{(x-2)(x-4)}{x-1}=0 \Rightarrow x=4

(since

x \neq 1,2

), or

\log_e|x-2|=0 \Rightarrow |x-2|=1 \Rightarrow x=3

(since

x \neq 1

).
So

B = \{3,4\}

|B|=2

.
Number of onto functions from

A

B

= 2^6 - 2 = 62

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