Let A be the set of all prime factors of 2310 and `f : A rightarrow B` be the function given by `f(x) = log_2(x^2 + [x^3/3])`, where B is range of `f(x), [t]` is G.I.F. The number of one-one functions are

Options

A. 20
B. 36
C. 120
D. 720

Answer

To solve the problem, we need to follow these steps:

### Step 1: Identify the prime factors of 2310
We start by finding the prime factorization of 2310.

- 2310 is even, so we divide by 2:
\[ 2310 \div 2 = 1155 \]
- Next, we check for divisibility by 3:
\[ 1155 \div 3 = 385 \]
- Now, we check for divisibility by 5:
\[ 385 \div 5 = 77 \]
- Next, we divide by 7:
\[ 77 \div 7 = 11 \]
- Finally, 11 is a prime number.

Thus, the prime factorization of 2310 is:
\[ 2310 = 2 \times 3 \times 5 \times 7 \times 11 \]

### Step 2: Define the set A
The set \( A \) of all prime factors of 2310 is:
\[ A = \{2, 3, 5, 7, 11\} \]

### Step 3: Analyze the function \( f(x) \)
The function given is:
\[ f(x) = \log_2\left(x^2 + \left\lfloor \frac{x^3}{3} \right\rfloor\right) \]
where \( \lfloor t \rfloor \) denotes the greatest integer function (G.I.F).

### Step 4: Determine the nature of the function
We need to check if \( f(x) \) is a one-to-one function.

- Since \( x \) takes values from the set \( A \), we will evaluate \( f(x) \) for each prime factor:
- For \( x = 2 \):
\[ f(2) = \log_2\left(2^2 + \left\lfloor \frac{2^3}{3} \right\rfloor\right) = \log_2\left(4 + \left\lfloor \frac{8}{3} \right\rfloor\right) = \log_2(4 + 2) = \log_2(6) \]
- For \( x = 3 \):
\[ f(3) = \log_2\left(3^2 + \left\lfloor \frac{3^3}{3} \right\rfloor\right) = \log_2\left(9 + \left\lfloor 9 \right\rfloor\right) = \log_2(18) \]
- For \( x = 5 \):
\[ f(5) = \log_2\left(5^2 + \left\lfloor \frac{5^3}{3} \right\rfloor\right) = \log_2\left(25 + \left\lfloor \frac{125}{3} \right\rfloor\right) = \log_2(25 + 41) = \log_2(66) \]
- For \( x = 7 \):
\[ f(7) = \log_2\left(7^2 + \left\lfloor \frac{7^3}{3} \right\rfloor\right) = \log_2\left(49 + \left\lfloor \frac{343}{3} \right\rfloor\right) = \log_2(49 + 114) = \log_2(163) \]
- For \( x = 11 \):
\[ f(11) = \log_2\left(11^2 + \left\lfloor \frac{11^3}{3} \right\rfloor\right) = \log_2\left(121 + \left\lfloor \frac{1331}{3} \right\rfloor\right) = \log_2(121 + 443) = \log_2(564) \]

Since \( f(x) \) is evaluated at distinct prime numbers and yields distinct outputs, \( f(x) \) is a one-to-one function.

### Step 5: Count the number of one-to-one functions
The number of one-to-one functions from set \( A \) to set \( B \) (where \( B \) is the range of \( f(x) \)) is given by \( m! \) where \( m \) is the number of elements in set \( A \).

- The set \( A \) has 5 elements:
\[ |A| = 5 \]
Thus, the number of one-to-one functions is:
\[ 5! = 120 \]

### Final Answer
The number of one-one functions is \( \boxed{120} \).