The no. of functions f from the set A={x∈N:x2−10x+9≤0} to set B={n2:n∈N} such that f(x)≤(x−3)2+1, for every x∈A is

(x2−10x+9)≤0⇒(x−1)(x−9)≤0x≥1 & x≤9x∈[1,9] then, A{1,2,3,4,5,6,7,8,9}f(x)≤(x−3)2+1Put x=1; f(1)≤5⇒12,22Put x=2; f(2)≤2⇒12 Put x=3; f(3)≤1⇒12Put x=4; f(4)≤2⇒12Put x=5; f(5)≤5⇒12,22Put x=6; f(6)≤10⇒12,22,32Put x=7; f(7)≤17⇒12,22,32,42Put x=8; f(8)≤26⇒12,22,32,42,52Put x=9; f(9)≤37⇒12,22,32,42,52,62Total no.of function =2(66)⇒2(720)=1440

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The no. of functions f from the set $A = {x \in N : x^{2} - 10 x + 9 \leq 0}$ to set $B = {n^{2} : n \in N}$ such that $f (x) \leq {(x - 3)}^{2} + 1$ , for every $x \in A$ is

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answer is 1440.

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Detailed Solution

$\begin{array}{l} (x^{2} - 10 x + 9) \leq 0 \Rightarrow (x - 1) (x - 9) \leq 0 \\ x \geq 1 & x \leq 9 \\ x \in [1, 9] t h e n, A {1, 2, 3, 4, 5, 6, 7, 8, 9} \\ f (x) \leq {(x - 3)}^{2} + 1 \\ P u t x = 1; f (1) \leq 5 \Rightarrow 1^{2}, 2^{2} \\ P u t x = 2; f (2) \leq 2 \Rightarrow 1^{2} \end{array}$
$\begin{array}{l} P u t x = 3; f (3) \leq 1 \Rightarrow 1^{2} \\ P u t x = 4; f (4) \leq 2 \Rightarrow 1^{2} \\ P u t x = 5; f (5) \leq 5 \Rightarrow 1^{2}, 2^{2} \\ P u t x = 6; f (6) \leq 10 \Rightarrow 1^{2}, 2^{2}, 3^{2} \\ P u t x = 7; f (7) \leq 17 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2} \\ P u t x = 8; f (8) \leq 26 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2} \\ P u t x = 9; f (9) \leq 37 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2} \end{array}$
Total no.of function $= 2 (66) \Rightarrow 2 (720) = 1440$