Let f: `[-3,3] rarr R` where f(x)=`x^(3)+sin x+[(x^(2)+2)/(a)]` be an odd function then value of a is where [.] represents greatest integer functions

Options

A. less than 11
B. 11
C. greater than 11
D. None of these

Answer

To solve the problem, we need to determine the value of \( a \) such that the function \( f(x) = x^3 + \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor \) is an odd function. An odd function satisfies the property \( f(-x) = -f(x) \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = x^3 + \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor \]

2. **Calculate \( f(-x) \)**:
\[ f(-x) = (-x)^3 + \sin(-x) + \left\lfloor \frac{(-x)^2 + 2}{a} \right\rfloor \]
Simplifying this, we get:
\[ f(-x) = -x^3 - \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor \]

3. **Set up the equation for odd function**:
For \( f(x) \) to be odd, we need:
\[ f(-x) = -f(x) \]
This gives us:
\[ -x^3 - \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor = -\left( x^3 + \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor \right) \]

4. **Simplify the equation**:
Expanding the right side:
\[ -x^3 - \sin x + \left\lfloor \frac{x^2 + 2}{a} \right\rfloor = -x^3 - \sin x - \left\lfloor \frac{x^2 + 2}{a} \right\rfloor \]
Rearranging gives:
\[ 2\left\lfloor \frac{x^2 + 2}{a} \right\rfloor = 0 \]

5. **Condition for the greatest integer function**:
The greatest integer function \( \left\lfloor \frac{x^2 + 2}{a} \right\rfloor = 0 \) when:
\[ 0 \leq \frac{x^2 + 2}{a} < 1 \]

6. **Inequalities from the condition**:
From \( \frac{x^2 + 2}{a} \geq 0 \):
\[ x^2 + 2 \geq 0 \quad \text{(always true since \( x^2 \geq 0 \))} \]
From \( \frac{x^2 + 2}{a} < 1 \):
\[ x^2 + 2 < a \]

7. **Determine the range of \( x^2 + 2 \)**:
Since \( x \) varies from \(-3\) to \(3\):
- The minimum value of \( x^2 \) is \( 0 \) (when \( x = 0 \)).
- The maximum value of \( x^2 \) is \( 9 \) (when \( x = \pm 3 \)).
Thus, \( x^2 + 2 \) varies from \( 2 \) to \( 11 \).

8. **Set up the inequality for \( a \)**:
Therefore, we have:
\[ 2 < a \quad \text{and} \quad a > 11 \]

9. **Conclusion**:
The only condition that satisfies both inequalities is:
\[ a > 11 \]

### Final Answer:
The value of \( a \) must be greater than \( 11 \).