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 ) = [ log 2 ( x 2 + [ x 3 5 ] ) ] . The number of one-to-one functions from A to the range of f is

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Solution

Q.
Let [ $t$ ] be the greatest integer less than or equal to $t$ . Let A be the set of all prime factors of 2310 and $f : A \to Z be the function f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])] .$ The number of one-to-one functions from A to the range of $f$ is [2024]

1 25

2 24

3 20

4 120

Ans.
(4)

$2310 = 2 \times 3 \times 5 \times 7 \times 11$

$∴ A = {2, 3, 5, 7, 11}$

$f : A \to Z is a function such that$

$f (x) = [\log_{2} (x^{2} + [\frac{x^{3}}{5}])]$

$f (2) = [\log_{2} (4 + [1.6])] = [\log_{2} (4 + 1)] = [\log_{2} 5] = [\log_{2} (2^{2} + 1)] = 2$

   $f (3) = [\log_{2} (9 + 5)] = [\log_{2} (2^{3} + 6)] = 3$

   $f (5) = [\log_{2} (25 + 25)] = [\log_{2} (2^{5} + 18)] = 5$

   $f (7) = [\log_{2} (49 + 68)] = [\log_{2} (2^{6} + 53)] = 6$

   $f (11) = [\log_{2} (121 + 266)] = [\log_{2} (2^{8} + 131)] = 8$

   $Range of f = {2, 3, 5, 6, 8}$

   $Number of one-one functions = 5! = 120$

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