`Let f(x)=tan^(-1)x-x +(x^3)/6 `
Statement -1: `f(x) lt g(x) for 0 lt x le 1`
Statement -2:`h(X)= tan^(-1) x-x +(x^3)/(6)` decreases on [-1,1]

Options

A. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is a correct explanation for Statement-2
B. Statement-1 True statement -1 is True,Statement -2 is True statement -2 is not a correct explanation for Statement-2
C. Statement-1 True statement -1 is True,Statement -2 is False
D. Statement-1 is False ,Statement -2 is True

Explanation

To solve the problem, we will analyze the function \( f(x) = \tan^{-1} x - x + \frac{x^3}{6} \) and the statements provided.

### Step 1: Analyze the function \( f(x) \)

We start with the function:
\[ f(x) = \tan^{-1} x - x + \frac{x^3}{6} \]

### Step 2: Find the derivative \( f'(x) \)

To determine if \( f(x) \) is increasing or decreasing, we need to find its derivative:
\[ f'(x) = \frac{d}{dx}(\tan^{-1} x) - \frac{d}{dx}(x) + \frac{d}{dx}\left(\frac{x^3}{6}\right) \]
Using the derivatives:
\[ \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}, \quad \frac{d}{dx}(x) = 1, \quad \frac{d}{dx}\left(\frac{x^3}{6}\right) = \frac{x^2}{2} \]
Thus, we have:
\[ f'(x) = \frac{1}{1+x^2} - 1 + \frac{x^2}{2} \]

### Step 3: Simplify the derivative

Now, we simplify \( f'(x) \):
\[ f'(x) = \frac{1}{1+x^2} - 1 + \frac{x^2}{2} = \frac{1 - (1+x^2) + \frac{x^2(1+x^2)}{2}}{1+x^2} \]
This leads to:
\[ f'(x) = \frac{1 - 1 - x^2 + \frac{x^2 + x^4}{2}}{1+x^2} = \frac{-x^2 + \frac{x^2 + x^4}{2}}{1+x^2} \]
\[ = \frac{-2x^2 + x^2 + x^4}{2(1+x^2)} = \frac{x^4 - x^2}{2(1+x^2)} = \frac{x^2(x^2 - 1)}{2(1+x^2)} \]

### Step 4: Analyze the sign of \( f'(x) \)

The critical points occur when \( f'(x) = 0 \):
\[ x^2(x^2 - 1) = 0 \]
This gives us \( x = 0, 1 \).

Now, we analyze the intervals:
- For \( 0 < x < 1 \), \( x^2 > 0 \) and \( x^2 - 1 < 0 \) implies \( f'(x) < 0 \) (decreasing).
- For \( x > 1 \), both \( x^2 > 0 \) and \( x^2 - 1 > 0 \) implies \( f'(x) > 0 \) (increasing).

### Step 5: Conclusion for Statement 2

Since \( f'(x) < 0 \) for \( x \in (0, 1) \), \( f(x) \) is decreasing on \( [-1, 1] \). Thus, Statement 2 is true.

### Step 6: Analyze Statement 1

To check if \( f(x) < g(x) \) for \( 0 < x \leq 1 \), we need to compare \( f(x) \) with \( g(x) \). Assuming \( g(x) \) is another function that we need to define or compare with \( f(x) \).

Since \( f(x) \) is decreasing in the interval \( (0, 1) \), we can evaluate \( f(0) \) and \( f(1) \):
- \( f(0) = \tan^{-1}(0) - 0 + 0 = 0 \)
- \( f(1) = \tan^{-1}(1) - 1 + \frac{1^3}{6} = \frac{\pi}{4} - 1 + \frac{1}{6} \)

Calculating \( f(1) \):
\[ f(1) = \frac{\pi}{4} - 1 + \frac{1}{6} \approx 0.785 - 1 + 0.1667 \approx -0.0483 \]
Thus, \( f(x) < f(0) = 0 \) for \( 0 < x \leq 1 \).

### Final Conclusion

Both statements are true:
- Statement 1: \( f(x) < g(x) \) for \( 0 < x \leq 1 \) is true.
- Statement 2: \( h(x) \) decreases on \( [-1, 1] \) is true.