Identify the options having correct statement:

Options

A. `f (x) =""^(3) sqrt(x^(2) |x|) -|x|` is no where non-differentiable
B. `lim _(x to oo)((x+5) tan ^(-1)(x+1)) -((x+1) tan ^(-1) (x+1))=2pi`
C. `f (x)=sin (ln (x+ sqrt(x^(2)+1)))` is an odd function
D. `f (x) =(4-x ^(2))/(4x-x ^(3))` is discontinous at exactly one point

Answer

To solve the problem, we need to analyze each option provided in the question and determine whether the statements are correct or incorrect based on the properties of continuity and differentiability.

### Step-by-Step Solution:

**Option 1: \( f(x) = \sqrt[3]{x^2} \cdot |x| - |x| \) is not differentiable.**

1. **Simplify the function:**
- We know that \( |x| = \sqrt{x^2} \).
- Therefore, \( f(x) = \sqrt[3]{x^2} \cdot |x| - |x| = |x|(\sqrt[3]{x^2} - 1) \).

2. **Check differentiability:**
- The function \( |x| \) is differentiable everywhere except at \( x = 0 \).
- At \( x = 0 \), we need to check if the limit of the derivative exists.
- The limit from the left and right can be calculated, and since both limits exist and are equal, \( f(x) \) is differentiable everywhere.

**Conclusion:** The statement is incorrect; \( f(x) \) is differentiable everywhere.

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**Option 2: \( \lim_{x \to \infty} (x + 5 - x - 1) \tan^{-1}(x + 1) = 2\pi \)**

1. **Simplify the limit:**
- The expression simplifies to \( \lim_{x \to \infty} (4) \tan^{-1}(x + 1) \).
- As \( x \to \infty \), \( \tan^{-1}(x + 1) \to \frac{\pi}{2} \).

2. **Calculate the limit:**
- Therefore, \( 4 \cdot \frac{\pi}{2} = 2\pi \).

**Conclusion:** The statement is correct.

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**Option 3: \( f(x) = \sin(x) + \sqrt{1 + x^2} \) is an odd function.**

1. **Check if \( f(-x) = -f(x) \):**
- Calculate \( f(-x) = \sin(-x) + \sqrt{1 + (-x)^2} = -\sin(x) + \sqrt{1 + x^2} \).
- For \( f(x) \) to be odd, we need \( f(-x) = -f(x) \).

2. **Evaluate:**
- \( f(-x) = -\sin(x) + \sqrt{1 + x^2} \) is not equal to \( -(\sin(x) + \sqrt{1 + x^2}) \).

**Conclusion:** The statement is incorrect; \( f(x) \) is not an odd function.

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**Option 4: \( f(x) = \frac{4 - x^2}{4x - x^3} \) is continuous.**

1. **Identify points of discontinuity:**
- The function is undefined when the denominator \( 4x - x^3 = 0 \).
- Factor the denominator: \( x(4 - x^2) = 0 \) gives \( x = 0, \pm 2 \).

2. **Check continuity:**
- The function is discontinuous at \( x = 0, 2, -2 \) because the function is not defined at these points.

**Conclusion:** The statement is incorrect; \( f(x) \) is not continuous at these points.

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### Final Summary of Options:
- Option 1: Incorrect
- Option 2: Correct
- Option 3: Incorrect
- Option 4: Incorrect