If `T_(1)` is the period of the function `f(x)=e^(3x-[x]) and T_(2)` is the period of the function `g(x)=e^(3-[x])``([*]` denotes the greatest integer function ), then (a) `T_(1)=T_(2)` (b) `T_(1)=(T_(2))/(3)` (c) `T_(1)=3T_(2)` (d) None of these

Options

A. `T_(1)=T_(2)`
B. `T_(1)=(T_(2))/(3)`
C. `T_(1)=3T_(2)`
D. None of these

Answer

To find the periods \( T_1 \) and \( T_2 \) of the functions \( f(x) = e^{3x - [x]} \) and \( g(x) = e^{3 - [x]} \), we will analyze each function step by step.

### Step 1: Understanding the Greatest Integer Function
The greatest integer function, denoted by \([x]\), gives the largest integer less than or equal to \(x\). We can express \(x\) as:
\[ x = [x] + \{x\} \]
where \(\{x\} = x - [x]\) is the fractional part of \(x\).

### Step 2: Analyzing \( f(x) \)
For the function \( f(x) = e^{3x - [x]} \):
\[ f(x) = e^{3([x] + \{x\}) - [x]} = e^{3[x] + 3\{x\} - [x]} = e^{(3-1)[x] + 3\{x\}} = e^{2[x] + 3\{x\}} \]
The term \(2[x]\) is periodic with a period of 1 because \([x]\) increases by 1 for every integer increase in \(x\). The term \(3\{x\}\) is periodic with a period of 1 as well since \(\{x\}\) resets every time \(x\) increases by 1.

Thus, the overall function \(f(x)\) has a period of:
\[ T_1 = 1 \]

### Step 3: Analyzing \( g(x) \)
For the function \( g(x) = e^{3 - [x]} \):
\[ g(x) = e^{3 - [x]} \]
Here, the term \(-[x]\) is periodic with a period of 1, as it also increases by 1 for every integer increase in \(x\).

Thus, the function \(g(x)\) also has a period of:
\[ T_2 = 1 \]

### Step 4: Comparing the Periods
Now we have:
\[ T_1 = 1 \quad \text{and} \quad T_2 = 1 \]
This implies:
\[ T_1 = T_2 \]

### Conclusion
The correct answer is:
\[ \text{(a) } T_1 = T_2 \]