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JEE Mains · Maths · STD 12 - 1. relation and function

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.

  1. A \(1440\)
  2. B \(1450\)
  3. C \(1460\)
  4. D \(1470\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(1440\)

Step-by-step Solution

Detailed explanation

\(\left( x ^{2}-10 x +9\right) \leq 0 \Rightarrow( x -1)( x -9) \leq 0\) \(x \in[1,9] \Rightarrow A =\{1,2,3,4,5,6,7,8,9\}\) \(f ( x ) \leq( x -3)^{2}+1\) \(x =1: f (1) \leq 5 \Rightarrow 1^{2}, 2^{2}\) \(x =2: f (2) \leq 2 \Rightarrow 1^{2}\)…
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