JEE MAIN - Mathematics (2022 - 27th July Evening Shift - No. 13)
The number of functions $$f$$, from the set $$\mathrm{A}=\left\{x \in \mathbf{N}: x^{2}-10 x+9 \leq 0\right\}$$ to the set $$\mathrm{B}=\left\{\mathrm{n}^{2}: \mathrm{n} \in \mathbf{N}\right\}$$ such that $$f(x) \leq(x-3)^{2}+1$$, for every $$x \in \mathrm{A}$$, is ___________.
Answer
1440
Explanation
$$A = \left\{ {\matrix{ {x \in N,} & {{x^2} - 10x + 9 \le 0} \cr } } \right\}$$
$$ = \{ 1,2,3,\,....,\,9\} $$
$$B = \{ 1,4,9,16,\,.....\} $$
$$f(x) \le {(x - 3)^2} + 1$$
$$f(1) \le 5,\,f(2) \le 2,\,\,..........\,f(9) \le 37$$
$$x = 1$$ has 2 choices
$$x = 2$$ has 1 choice
$$x = 3$$ has 1 choice
$$x = 4$$ has 1 choice
$$x = 5$$ has 2 choices
$$x = 6$$ has 3 choices
$$x = 7$$ has 4 choices
$$x = 8$$ has 5 choices
$$x = 9$$ has 6 choices
$$\therefore$$ Total functions = $$2\times1\times1\times1\times2\times3\times4\times5\times6=1440$$
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