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વિધેય $f$ એ ગણ $A=\left\{x \in N: x^{2}-10 x+9 \leq 0\right\}$ થી ગણ $B=\left\{n^{2}: n \in N\right\}$ કે જેથી દરેક $x \in A$ માટે $f(x) \leq(x-3)^{2}+1$ તેવા વિધેય $f$ ની સંખ્યા મેળવો.
$1440$
$1450$
$1460$
$1470$
Solution
$\left( x ^{2}-10 x +9\right) \leq 0 \Rightarrow( x -1)( x -9) \leq 0$
$x \in[1,9] \Rightarrow A =\{1,2,3,4,5,6,7,8,9\}$
$f ( x ) \leq( x -3)^{2}+1$
$x =1: f (1) \leq 5 \Rightarrow 1^{2}, 2^{2}$
$x =2: f (2) \leq 2 \Rightarrow 1^{2}$
$x =3: f (3) \leq 1 \Rightarrow 1^{2}$
$x =4: f (4) \leq 2 \Rightarrow 1^{2}$
$x =5: f (5) \leq 5 \Rightarrow 1^{2}, 2^{2}$
$x =6: f (6) \leq 10 \Rightarrow 1^{2}, 2^{2}, 3^{2}$
$x =7: f (7) \leq 17 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}$
$x =8: f (8) \leq 26 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}$
$x =9: f (9) \leq 37 \Rightarrow 1^{2}, 2^{2}, 3^{2}, 4^{2}, 5^{2}, 6^{2}$
Total number of such function
$=2(6 !)=2(720)=1440$
