JEE Main 2026 — 23 January, Evening Shift — Question 5

$$A=\{x\in \mathbb{Z}:\ |(|x-3|-3)|\le 1\}$$

Let $y=|x-3|$. Then

$$|y-3|\le 1 \;\Rightarrow\; 2\le y\le 4$$

So,

$$|x-3|=2,3,4$$

Hence

$$x=3\pm2,\;3\pm3,\;3\pm4$$ $$A=\{-1,0,1,5,6,7\}$$

Thus,

$$|A|=6$$ $$B=\left\{x\in\mathbb{R}\setminus\{1,2\}:\frac{(x-2)(x-4)}{x-1}\ln|x-2|=0\right\}$$

The product equals zero when any factor is zero (keeping domain restrictions in mind).

1. $\ln|x-2|=0\Rightarrow |x-2|=1\Rightarrow x=1,3$ But $x=1$ is excluded, so $x=3$.

2. $x-4=0\Rightarrow x=4$

3. $x-2=0\Rightarrow x=2$ (excluded)

Hence,

$$B=\{3,4\}$$

So,$|B|=2$

Number of onto functions $f:A\to B$

Number of surjections from a set of size $6$ onto a set of size $2$ is

$$2^6 - 2$$

(Subtract the two constant functions.)

$$=64-2=62$$ $$\boxed{62}$$