JEE Main 2026 — Jan 23, Shift 2 — Mathematics solution

JEE Main2026Jan 23, Shift 2Mathematics

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 :

Solution

Set A: $∣∣ x - 3∣ - 3∣ \leq 1 \Rightarrow - 1 \leq ∣ x - 3∣ - 3 \leq 1 \Rightarrow 2 \leq ∣ x - 3∣ \leq 4$ .

Splitting: $2 \leq x - 3 \leq 4$ gives $5 \leq x \leq 7$ , and $- 4 \leq x - 3 \leq - 2$ gives $- 1 \leq x \leq 1$ .

Integer values: $A = {- 1, 0, 1, 5, 6, 7}$ , so $∣ A ∣ = 6$ .

Set B: The product is zero when $(x - 2) (x - 4) = 0$ or $lo g_{e} ∣ x - 2∣ = 0$ .

From $(x - 2) (x - 4) = 0$ : $x = 2$ (excluded) or $x = 4$ .

From $lo g_{e} ∣ x - 2∣ = 0$ : $∣ x - 2∣ = 1 \Rightarrow x = 3$ or $x = 1$ (excluded).

So $B = {3, 4}$ , $∣ B ∣ = 2$ .

Onto functions from $A$ (6 elements) to $B$ (2 elements):

Total functions $= 2^{6} = 64$ ; subtract the 2 constant functions: $64 - 2 = 62$ .

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