JEE Main 2022 — Mathematics Algebra

Finding the cases for $f(a,b)=f(b,a)\geq a$f(a,b)=f(b,a)≥a we have,

$(1,1),(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4)$(1,1),(1,4),(4,1),(2,4),(4,2),(3,4),(4,3),(4,4)- all have one choice for image.

$(2,1),(1,2),(2,2)-$(2,1),(1,2),(2,2)− all have three choices for image.

$(3,2),(2,3),(3,1),(1,3),(3,3)-$(3,2),(2,3),(3,1),(1,3),(3,3)−all have two choices for image.

So the total functions $=3\times 3\times 2\times 2\times 2=72$=3×3×2×2×2=72

Case 1: None of the pre-images have $3$3 as image.

Total functions $=2\times 2\times 1\times 1\times 1=4$=2×2×1×1×1=4

Case 2: None of the pre-images have $2$2 as image

Total functions $=2\times 2\times 2\times 2\times 2=32$=2×2×2×2×2=32

Case 3: None of the pre-images have either $3$3 or $2$2 as image

Total functions $=1\times 1\times 1\times 1\times 1=1$=1×1×1×1×1=1

$\therefore$∴ Total onto functions $=72-4-32+1=37$=72−4−32+1=37