JEE Main 2022 (Online) 27th July Evening ShiftQuestion 1
The number of functions $f$, from the set $A=\{x\in\mathbf{N}:x^2-10x+9\le0\}$ to the set $B=\{n^2:n\in\mathbf{N}\}$ such that $f(x)\le(x-3)^2+1$, for every $x\in A$, is ________.
Sets, Relations and Function
Answer$1440$
Solution
$x^2-10x+9\le0\Rightarrow 1\le x\le 9$, so $A=\{1,\ldots,9\}$; $B=\{1,4,9,16,\ldots\}$. Count of squares $\le (x-3)^2+1$: $x=1$:bound 5 gives 2; $x=2$:2 gives 1; $x=3$:1 gives 1; $x=4$:2 gives 1; $x=5$:5 gives 2; $x=6$:10 gives 3; $x=7$:17 gives 4; $x=8$:26 gives 5; $x=9$:37 gives 6. Number of functions $=2\cdot1\cdot1\cdot1\cdot2\cdot3\cdot4\cdot5\cdot6=1440$.