Statement I Range of `f(x) = x((e^(2x)-e^(-2x))/(e^(2x)+e^(-2x))) + x^(2) + x^(4)` is not R.
Statement II Range of a continuous evern function cannot be R.

Options

A. (a)Statement I is correct, Statement II is also correct, Statement II is the correct explanation of Statement I
B. (b)Statement I is correct, Statement II is also correct, Statement II is not the correct explanation of Statement I
C. (c)Statement I is correct, Statement II is incorrect
D. (d)Statement I is incorrect, Statement II is correct.

Answer

To solve the problem, we need to analyze the function given in Statement I and determine its range. The function is:

\[ f(x) = x \left( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \right) + x^2 + x^4 \]

### Step 1: Simplify the function
The term \( \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} \) can be recognized as the hyperbolic tangent function:

\[ \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}} = \tanh(2x) \]

Thus, we can rewrite the function as:

\[ f(x) = x \tanh(2x) + x^2 + x^4 \]

### Step 2: Analyze the components of the function
The function \( x^2 + x^4 \) is always non-negative for all \( x \) since both \( x^2 \) and \( x^4 \) are even functions and are always greater than or equal to zero.

### Step 3: Analyze \( x \tanh(2x) \)
The function \( \tanh(2x) \) is bounded between -1 and 1 for all real \( x \). Therefore, the term \( x \tanh(2x) \) will also be bounded, but its behavior depends on the sign of \( x \):
- For \( x > 0 \), \( x \tanh(2x) \) is positive and increases as \( x \) increases.
- For \( x < 0 \), \( x \tanh(2x) \) is negative and decreases as \( x \) decreases.

### Step 4: Determine the range of \( f(x) \)
As \( x \) approaches infinity, \( f(x) \) will tend towards infinity due to the \( x^4 \) term dominating. As \( x \) approaches negative infinity, the \( x^4 \) term will still dominate and approach infinity, but the \( x \tanh(2x) \) term will contribute negatively.

However, since \( x^2 + x^4 \) is always non-negative, the overall function \( f(x) \) cannot take on all real values, particularly it cannot take negative values.

### Conclusion for Statement I
Thus, the range of \( f(x) \) is not all real numbers \( \mathbb{R} \).

### Step 5: Analyze Statement II
Statement II claims that the range of a continuous even function cannot be \( \mathbb{R} \). This is true because an even function is symmetric about the y-axis, and if it were to take on all real values, it would have to take on both positive and negative values, which contradicts the symmetry unless it is constant.

### Final Answer
Both statements are true:
- Statement I: The range of \( f(x) \) is not \( \mathbb{R} \).
- Statement II: The range of a continuous even function cannot be \( \mathbb{R} \).