Key Concepts: Properties of Determinants and Trigonometric Identities

This problem requires us to evaluate a determinant whose entries involve trigonometric functions. The primary strategy is to simplify the determinant using elementary row or column operations before expanding it. This process often reduces the complexity of the entries, introduces zeros, and makes the expansion much easier. After simplifying the determinant to an expression involving trigonometric functions, we will use standard trigonometric identities to further simplify it and then determine its maximum value over the given interval.

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Step-by-Step Solution

1. Simplify the Determinant using Row Operations

The given function is: f(x) = \left| {\matrix{ {{{\sin }^2}x} & { - 2 + {{\cos }^2}x} & {\cos 2x} \cr {2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr } } \right|

To simplify, we aim to create zeros in rows or columns. Let's perform the row operation $R_1 \to R_1 - R_2$.

• For the first element: $\sin^2 x - (2 + \sin^2 x) = -2$
• For the second element: $(-2 + \cos^2 x) - \cos^2 x = -2$
• For the third element: $\cos 2x - \cos 2x = 0$

Applying this operation, the determinant becomes: f(x) = \left| {\matrix{ {-2} & {-2} & {0} \cr {2 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {1 + \cos 2x} \cr } } \right| Reasoning: This operation does not change the value of the determinant but simplifies the first row significantly by introducing a zero, which will make the expansion easier.

2. Expand the Determinant

Now, we can expand the determinant along the first row ($R_1$) because it contains a zero, reducing the number of $2 \times 2$ determinants to calculate. f(x) = (-2) \cdot \left| {\matrix{ {{{\cos }^2}x} & {\cos 2x} \\ {{{\cos }^2}x} & {1 + \cos 2x} \\ } } \right| - (-2) \cdot \left| {\matrix{ {2 + {{\sin }^2}x} & {\cos 2x} \\ {{{\sin }^2}x} & {1 + \cos 2x} \\ } } \right| + (0) \cdot (\text{minor}) Reasoning: Expanding along a row/column with zeros simplifies calculations. The formula for expanding a $3 \times 3$ determinant along the first row is $a(ei-fh) - b(di-fg) + c(dh-eg)$.

Let's calculate the $2 \times 2$ determinants:

• First minor: $(\cos^2 x)(1 + \cos 2x) - (\cos 2x)(\cos^2 x)$ $= \cos^2 x + \cos^2 x \cos 2x - \cos^2 x \cos 2x$ $= \cos^2 x$

• Second minor: $(2 + \sin^2 x)(1 + \cos 2x) - (\cos 2x)(\sin^2 x)$ $= (2 + \sin^2 x + 2\cos 2x + \sin^2 x \cos 2x) - \sin^2 x \cos 2x$ $= 2 + \sin^2 x + 2\cos 2x$

Substitute these back into the expansion: $f(x) = -2 (\cos^2 x) + 2 (2 + \sin^2 x + 2\cos 2x)$ $f(x) = -2\cos^2 x + 4 + 2\sin^2 x + 4\cos 2x$

3. Simplify the Trigonometric Expression

Rearrange the terms and use the identity $\sin^2 x - \cos^2 x = -\cos 2x$: $f(x) = 4 + 4\cos 2x + (2\sin^2 x - 2\cos^2 x)$ $f(x) = 4 + 4\cos 2x + 2(\sin^2 x - \cos^2 x)$ $f(x) = 4 + 4\cos 2x + 2(-\cos 2x)$ $f(x) = 4 + 4\cos 2x - 2\cos 2x$ $f(x) = 4 + 2\cos 2x$ Reasoning: Combining like terms and using the fundamental trigonometric identity $\sin^2 x - \cos^2 x = -\cos 2x$ simplifies the expression into a more manageable form involving only $\cos 2x$.

4. Find the Maximum Value of f(x)

We have $f(x) = 4 + 2\cos 2x$. The domain for $x$ is given as $[0, \pi]$. For $x \in [0, \pi]$, the range of $2x$ is $[0, 2\pi]$. In the interval $[0, 2\pi]$, the cosine function $\cos(2x)$ takes all values from its minimum to its maximum, i.e., $-1 \le \cos(2x) \le 1$.

To find the maximum value of $f(x)$, we substitute the maximum possible value of $\cos(2x)$ into the expression: Maximum value of $\cos(2x) = 1$. $f_{max} = 4 + 2(1) = 6$

To find the minimum value of $f(x)$, we substitute the minimum possible value of $\cos(2x)$ into the expression: Minimum value of $\cos(2x) = -1$. $f_{min} = 4 + 2(-1) = 4 - 2 = 2$

Thus, the range of $f(x)$ for $x \in [0, \pi]$ is $[2, 6]$. The question asks for the maximum value of $f(x)$.

The maximum value of $f(x)$ is $6$.

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Tips and Common Mistakes:

• Systematic Row/Column Operations: Always aim to create zeros in a row or column using elementary operations ($R_i \to R_i + k R_j$ or $C_i \to C_i + k C_j$) as these operations do not change the determinant's value. This is much more efficient than direct expansion.
• Careful with Signs: When expanding a determinant, pay close attention to the alternating signs for cofactors (e.g., $+,-,+$ for a $3 \times 3$ matrix expansion along the first row).
• Trigonometric Identities: Keep common identities like $\sin^2 x + \cos^2 x = 1$ and $\cos 2x = \cos^2 x - \sin^2 x$ handy. They are crucial for simplifying expressions.
• Range of Trigonometric Functions: Remember the range of $\sin \theta$ and $\cos \theta$ is $[-1, 1]$. Also, be mindful of the interval given for $x$ as it might restrict the range of $2x$ or other arguments.

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Summary: We successfully simplified the given determinant using row operations, reducing it to a simple trigonometric expression $f(x) = 4 + 2\cos 2x$. By understanding the range of the cosine function, we determined that the maximum value of $f(x)$ over the interval $x \in [0, \pi]$ is $6$.