Let `f(x) = sin (sin^-1 2x) + cosec (cosec^-1 2x) + tan(tan^-1 2x)`, then which one of the following statements is/are incorrect ?

Options

A. f(x) is odd function
B. f(x) is injective
C. Range of f (x) contains only two integers.
D. The value of `f' (1/2) ` is equal to 6.

Answer

To solve the problem, we need to analyze the function \( f(x) = \sin(\sin^{-1}(2x)) + \csc(\csc^{-1}(2x)) + \tan(\tan^{-1}(2x)) \) and determine which statements about it are incorrect.

### Step 1: Determine the domain of \( f(x) \)

1. **For \( \sin^{-1}(2x) \)**: The argument \( 2x \) must lie in the interval \([-1, 1]\). Therefore, we have:
\[ -1 \leq 2x \leq 1 \implies -\frac{1}{2} \leq x \leq \frac{1}{2} \]

2. **For \( \csc^{-1}(2x) \)**: The argument \( 2x \) must be either \( \geq 1 \) or \( \leq -1 \). Thus:
\[ 2x \geq 1 \quad \text{or} \quad 2x \leq -1 \implies x \geq \frac{1}{2} \quad \text{or} \quad x \leq -\frac{1}{2} \]

3. **For \( \tan^{-1}(2x) \)**: There are no restrictions on \( 2x \) since the domain of \( \tan^{-1} \) is all real numbers.

### Conclusion on Domain
The overall domain of \( f(x) \) is the intersection of the domains from the three parts:
\[ x \in \left[-\frac{1}{2}, \frac{1}{2}\right] \cap \left( (-\infty, -\frac{1}{2}] \cup [\frac{1}{2}, \infty) \right) = \left\{-\frac{1}{2}, \frac{1}{2}\right\} \]
Thus, the domain of \( f(x) \) is \( x = -\frac{1}{2} \) and \( x = \frac{1}{2} \).

### Step 2: Evaluate \( f(x) \) at the endpoints

1. **Evaluate \( f\left(-\frac{1}{2}\right) \)**:
\[ f\left(-\frac{1}{2}\right) = \sin(\sin^{-1}(-1)) + \csc(\csc^{-1}(-1)) + \tan(\tan^{-1}(-1)) \]
\[ = -1 - 1 - 1 = -3 \]

2. **Evaluate \( f\left(\frac{1}{2}\right) \)**:
\[ f\left(\frac{1}{2}\right) = \sin(\sin^{-1}(1)) + \csc(\csc^{-1}(1)) + \tan(\tan^{-1}(1)) \]
\[ = 1 + 1 + 1 = 3 \]

### Step 3: Check if \( f(x) \) is an odd function

A function \( f(x) \) is odd if \( f(-x) = -f(x) \). We check:
\[ f\left(-\frac{1}{2}\right) = -3 \quad \text{and} \quad f\left(\frac{1}{2}\right) = 3 \]
Thus, \( f(-x) = -f(x) \) holds true, so \( f(x) \) is an odd function.

### Step 4: Check if \( f(x) \) is injective

A function is injective if it gives different outputs for different inputs. Since \( f(-\frac{1}{2}) = -3 \) and \( f(\frac{1}{2}) = 3 \) are distinct, and there are no other values in the domain, \( f(x) \) is injective.

### Step 5: Determine the range of \( f(x) \)

The outputs we found are \( -3 \) and \( 3 \). Therefore, the range of \( f(x) \) contains only these two values, which are integers.

### Step 6: Calculate the derivative at \( x = \frac{1}{2} \)

Since \( f(x) \) is linear in this case, we can differentiate:
\[ f'(x) = 6 \quad \text{(as derived from the function)} \]
Thus, \( f'(\frac{1}{2}) = 6 \).

### Conclusion on Statements
- **Statement A**: \( f(x) \) is an odd function. (Correct)
- **Statement B**: \( f(x) \) is injective. (Correct)
- **Statement C**: The range of \( f(x) \) contains only two integers. (Incorrect, it only contains -3 and 3)
- **Statement D**: \( f'(\frac{1}{2}) = 6 \). (Correct)

### Final Answer
The incorrect statement is **C**.