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(B) First,find the prime factorization of $2310$: $2310 = 231 	imes 10 = 3 	imes 7 	imes 11 	imes 2 	imes 5$. Thus,the set $A = \{2, 3, 5, 7, 11\

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(B) First,find the prime factorization of $2310$: $2310 = 231 	imes 10 = 3 	imes 7 	imes 11 	imes 2 	imes 5$. Thus,the set $A = \{2, 3, 5, 7, 11\}$. Now,calculate the values of $f(x)$ for each $x \in A$: $f(2) = [\log_2(2^2 + [2^3/5])] = [\log_2(4 + [1.6])] = [\log_2(5)] = 2$. $f(3) = [\log_2(3^2 + [3^3/5])] = [\log_2(9 + [5.4])] = [\log_2(14)] = 3$. $f(5) = [\log_2(5^2 + [5^3/5])] = [\log_2(25 + 25)] = [\log_2(50)] = 5$. $f(7) = [\log_2(7^2 + [7^3/5])] = [\log_2(49 + [68.6])] = [\log_2(117)] = 6$. $f(11) = [\log_2(11^2 + [11^3/5])] = [\log_2(121 + [266.2])] = [\log_2(387)] = 8$. The range of $f$ is $B = \{2, 3, 5, 6, 8\}$. Since the number of elements in $A$ is $5$ and the number of elements in $B$ is $5$,the number of one-to-one functions from $A$ to $B$ is given by $5! = 120$.

Let $[t]$ be the greatest integer less than or equal to $t$. Let $A$ be the set of all prime factors of $2310$ and $f: A \rightarrow Z$ be the function $f(x) = \left[\log_2\left(x^2 + \left[\frac{x^3}{5}\right]\right)\right]$. The number of one-to-one functions from $A$ to the range of $f$ is:

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