If `g:[-2,2]vecR` , where `f(x)=x^3+tanx+[(x^2+1)/P]` is an odd function, then the value of parametric P, where [.] denotes the greatest integer function, is

Options

A. `-5 lt P lt 5`
B. `P lt 5`
C. `P gt 5`
D. None of these

Explanation

To solve the problem, we need to determine the value of the parameter \( P \) such that the function

\[ f(x) = x^3 + \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor \]

is an odd function. An odd function satisfies the property that \( f(-x) = -f(x) \).

### Step-by-Step Solution:

1. **Understanding the Function**:
The function is given as \( f(x) = x^3 + \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor \). We need to check if this function is odd.

2. **Finding \( f(-x) \)**:
We calculate \( f(-x) \):
\[ f(-x) = (-x)^3 + \tan(-x) + \left\lfloor \frac{(-x)^2 + 1}{P} \right\rfloor \]
Simplifying this, we have:
\[ f(-x) = -x^3 - \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor \]

3. **Setting Up the Odd Function Condition**:
For \( f(x) \) to be odd, we need:
\[ f(-x) = -f(x) \]
Substituting the expressions we found:
\[ -x^3 - \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor = -\left( x^3 + \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor \right) \]
This simplifies to:
\[ -x^3 - \tan x + \left\lfloor \frac{x^2 + 1}{P} \right\rfloor = -x^3 - \tan x - \left\lfloor \frac{x^2 + 1}{P} \right\rfloor \]

4. **Equating Terms**:
From the above equation, we can see that:
\[ \left\lfloor \frac{x^2 + 1}{P} \right\rfloor = -\left\lfloor \frac{x^2 + 1}{P} \right\rfloor \]
This implies:
\[ 2\left\lfloor \frac{x^2 + 1}{P} \right\rfloor = 0 \]
Therefore:
\[ \left\lfloor \frac{x^2 + 1}{P} \right\rfloor = 0 \]

5. **Analyzing the Floor Function**:
The condition \( \left\lfloor \frac{x^2 + 1}{P} \right\rfloor = 0 \) means:
\[ 0 \leq \frac{x^2 + 1}{P} < 1 \]
This leads to:
\[ 0 \leq x^2 + 1 < P \]

6. **Finding the Range of \( x^2 + 1 \)**:
Since \( x \) is in the interval \([-2, 2]\), the minimum value of \( x^2 + 1 \) is \( 1 \) (when \( x = 0 \)) and the maximum value is \( 5 \) (when \( x = -2 \) or \( x = 2 \)). Thus:
\[ 1 \leq x^2 + 1 \leq 5 \]

7. **Setting the Condition for \( P \)**:
From the inequality \( x^2 + 1 < P \), we conclude:
\[ 5 < P \]

### Conclusion:
The value of the parameter \( P \) must be greater than \( 5 \).