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Let [t] be the greatest integer less than or equal to t . Let A be the set of all prime factors of 2310 and f: A Z be the function f(x)= [ ₂ (x^2+ [ x^3 5 ] ) ] . The number of one-to-one functions from A to the range of f is

Options

  1. A25
  2. B24
  3. C20
  4. D120

Correct answer

D. 120

Step-by-step solution

N =2310=231 10=3 11 7 2 5A= 2,3,5,7,11 aligned & f(x)= [ ₂ (x^2+ [ x^3 5 ] ) ] & f(2)= [ ₂(5) ]=2 & f(3)= [ ₂(14) ]=3 & f(5)= [ ₂(25+25) ]=5 & f(7)= [ ₂(117) ]=6 & f(11)= [ ₂ 387 ]=8 aligned Range of f: B= 2,3,5,6,8 No. of one-one functions =5 !=120

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