Let `f : [-3, 3] rarr R` defined by `f(x) = [(x^(2))/(a)] tan ax + sex ax`. Where [] represents greatest integer function
If f(x) is an even function, then

Options

A. `a gt 3`
B. `a lt 3`
C. `a gt 9`
D. `a lt 9`

Answer

To determine the value of \( a \) such that the function \( f(x) = \left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) + \sec(ax) \) is an even function, we need to follow these steps:

### Step 1: Understand the condition for an even function
A function \( f(x) \) is considered even if it satisfies the condition:
\[ f(x) = f(-x) \]
for all \( x \) in its domain.

### Step 2: Calculate \( f(-x) \)
Given:
\[ f(x) = \left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) + \sec(ax) \]
Let's find \( f(-x) \):
\[ f(-x) = \left\lfloor \frac{(-x)^2}{a} \right\rfloor \tan(a(-x)) + \sec(a(-x)) \]
Since \( (-x)^2 = x^2 \), we have:
\[ f(-x) = \left\lfloor \frac{x^2}{a} \right\rfloor \tan(-ax) + \sec(-ax) \]
Using the trigonometric identities \( \tan(-\theta) = -\tan(\theta) \) and \( \sec(-\theta) = \sec(\theta) \), we can simplify this to:
\[ f(-x) = \left\lfloor \frac{x^2}{a} \right\rfloor (-\tan(ax)) + \sec(ax) \]
This simplifies to:
\[ f(-x) = -\left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) + \sec(ax) \]

### Step 3: Set \( f(x) = f(-x) \)
Now we equate \( f(x) \) and \( f(-x) \):
\[ \left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) + \sec(ax) = -\left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) + \sec(ax) \]
Subtracting \( \sec(ax) \) from both sides, we get:
\[ \left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) = -\left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) \]
This implies:
\[ 2\left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) = 0 \]

### Step 4: Analyze the equation
For the equation \( 2\left\lfloor \frac{x^2}{a} \right\rfloor \tan(ax) = 0 \) to hold true, either:
1. \( \left\lfloor \frac{x^2}{a} \right\rfloor = 0 \) or
2. \( \tan(ax) = 0 \)

### Step 5: Solve for \( \left\lfloor \frac{x^2}{a} \right\rfloor = 0 \)
The greatest integer function \( \left\lfloor \frac{x^2}{a} \right\rfloor = 0 \) implies:
\[ 0 \leq \frac{x^2}{a} < 1 \]
This leads to:
\[ 0 \leq x^2 < a \]
Given the domain \( x \in [-3, 3] \), the maximum value of \( x^2 \) is \( 9 \). Thus, we require:
\[ a > 9 \]

### Conclusion
The function \( f(x) \) is even if \( a > 9 \).

### Final Answer
The value of \( a \) must be greater than 9.

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