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

Options

A. `a gt 3`
B. `a lt 3`
C. `a gt 9`
D. f(x) can't be an odd function for any real value of a.

Answer

To determine the conditions under which the function \( f(x) = \frac{x^2}{a} \tan(ax) + \sec(ax) \) is an odd function, we need to follow these steps:

### Step 1: Understand the condition for odd functions
For a function \( f(x) \) to be classified as an odd function, it must satisfy the condition:
\[ f(-x) = -f(x) \]

### Step 2: Calculate \( f(-x) \)
We need to find \( f(-x) \):
\[ f(-x) = \frac{(-x)^2}{a} \tan(a(-x)) + \sec(a(-x)) \]
This simplifies to:
\[ f(-x) = \frac{x^2}{a} \tan(-ax) + \sec(-ax) \]
Using the properties of the tangent and secant functions, we know that:
\[ \tan(-\theta) = -\tan(\theta) \quad \text{and} \quad \sec(-\theta) = \sec(\theta) \]
Thus, we can rewrite \( f(-x) \):
\[ f(-x) = \frac{x^2}{a} (-\tan(ax)) + \sec(ax) = -\frac{x^2}{a} \tan(ax) + \sec(ax) \]

### Step 3: Set up the equation
Now we set \( f(-x) \) equal to \( -f(x) \):
\[ -\frac{x^2}{a} \tan(ax) + \sec(ax) = -\left(\frac{x^2}{a} \tan(ax) + \sec(ax)\right) \]
This simplifies to:
\[ -\frac{x^2}{a} \tan(ax) + \sec(ax) = -\frac{x^2}{a} \tan(ax) - \sec(ax) \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ \sec(ax) + \sec(ax) = 0 \]
This simplifies to:
\[ 2\sec(ax) = 0 \]

### Step 5: Analyze the implications
The equation \( 2\sec(ax) = 0 \) implies that:
\[ \sec(ax) = 0 \]
However, the secant function, defined as \( \sec(ax) = \frac{1}{\cos(ax)} \), can never be zero since the cosine function ranges from -1 to 1 and is never zero. Therefore, this condition cannot be satisfied.

### Conclusion
Since there are no real values of \( a \) for which \( f(x) \) can be an odd function, we conclude that:
\[ f(x) \text{ cannot be an odd function for any real value of } a. \]