Key Concepts and Formulas

• Determinant Properties: Row and column operations (adding a multiple of one row/column to another) do not change the value of the determinant. Expanding along a row or column simplifies the calculation.
• Trigonometric Identity: $\sin^2 x + \cos^2 x = 1$.
• Maximum Value of Trigonometric Expression: The maximum value of $a \cos \theta + b \sin \theta$ is $\sqrt{a^2 + b^2}$.

Step-by-Step Solution

Step 1: Initial Setup and Goal We are given the function $f(x)$ as a determinant: f(x) = \left| {\matrix{ {{{\sin }^2}x} & {1 + {{\cos }^2}x} & {\cos 2x} \cr {1 + {{\sin }^2}x} & {{{\cos }^2}x} & {\cos 2x} \cr {{{\sin }^2}x} & {{{\cos }^2}x} & {\sin 2x} \cr } } \right| Our first goal is to simplify this determinant expression for $f(x)$ into a simpler trigonometric function of $x$.

Step 2: Simplifying the Determinant using Column Operation To simplify the entries and potentially create constants or zeros, we can use column operations.

• Operation: Apply the column operation $C_1 \to C_1 + C_2$.
  • Why this step? Observe that $C_1$ contains $\sin^2 x$ terms and $C_2$ contains $\cos^2 x$ terms (or $1+\cos^2 x$). Adding them will utilize the fundamental identity $\sin^2 x + \cos^2 x = 1$. This will introduce constant values into the first column, which are generally easier to work with.
  • Let's see how the entries of $C_1$ change:
    • Row 1, Column 1: $\sin^2 x + (1 + \cos^2 x) = (\sin^2 x + \cos^2 x) + 1 = 1 + 1 = 2$.
    • Row 2, Column 1: $(1 + \sin^2 x) + \cos^2 x = 1 + (\sin^2 x + \cos^2 x) = 1 + 1 = 2$.
    • Row 3, Column 1: $\sin^2 x + \cos^2 x = 1$.
• Resulting Determinant: f(x) = \left| {\matrix{ 2 & {1 + {{\cos }^2}x} & {\cos 2x} \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|

Step 3: Further Simplifying the Determinant using Row Operation Now that we have common entries (specifically '2') in the first column, we can use row operations to create a zero, making expansion easier.

• Operation: Apply the row operation $R_1 \to R_1 - R_2$.
  • Why this step? Subtracting $R_2$ from $R_1$ will make the first entry of $R_1$ zero ($2-2=0$). This is a strategic move to simplify the determinant expansion, as a zero entry eliminates the need to calculate its corresponding minor.
  • Let's see how the entries of $R_1$ change:
    • Row 1, Column 1: $2 - 2 = 0$.
    • Row 1, Column 2: $(1 + \cos^2 x) - \cos^2 x = 1$.
    • Row 1, Column 3: $\cos 2x - \cos 2x = 0$.
• Resulting Determinant: f(x) = \left| {\matrix{ 0 & 1 & 0 \cr 2 & {{{\cos }^2}x} & {\cos 2x} \cr 1 & {{{\cos }^2}x} & {\sin 2x} \cr } } \right|

Step 4: Expanding the Determinant We now have a determinant with two zeros in the first row. This makes expansion along the first row very efficient.

• Method: Expand along $R_1$. The formula for a $3 \times 3$ determinant $\left| {\begin{matrix} a & b & c \\ d & e & f \\ g & h & i \end{matrix}} \right|$ expanded along the first row is $a(ei - fh) - b(di - fg) + c(dh - eg)$.
  • Since $a=0$ and $c=0$, only the term involving $b$ (which is $1$) will remain. Remember the sign pattern for cofactors: $\begin{pmatrix} + & - & + \\ - & + & - \\ + & - & + \end{pmatrix}$.
• Calculation: f(x) = 0 \cdot \left| {\matrix{ {{{\cos }^2}x} & {\cos 2x} \cr {{{\cos }^2}x} & {\sin 2x} \cr } } \right| - 1 \cdot \left| {\matrix{ 2 & {\cos 2x} \cr 1 & {\sin 2x} \cr } } \right| + 0 \cdot \left| {\matrix{ 2 & {{{\cos }^2}x} \cr 1 & {{{\cos }^2}x} \cr } } \right| $f(x) = -1 \cdot (2 \cdot \sin 2x - 1 \cdot \cos 2x)$ $f(x) = -(2 \sin 2x - \cos 2x)$ $f(x) = \cos 2x - 2 \sin 2x$ So, we have simplified the determinant expression for $f(x)$ to a standard trigonometric form.

Step 5: Finding the Maximum Value of $f(x)$ We need to find the maximum value of $f(x) = \cos 2x - 2 \sin 2x$.

• Key Formula: For any expression of the form $A \cos \theta + B \sin \theta$, its maximum value is $\sqrt{A^2 + B^2}$ and its minimum value is $-\sqrt{A^2 + B^2}$.
• Applying the Formula: In our expression $f(x) = \cos 2x - 2 \sin 2x$:
  • $A = 1$ (coefficient of $\cos 2x$)
  • $B = -2$ (coefficient of $\sin 2x$)
  • The angle is $\theta = 2x$.
• Calculation: Maximum value of $f(x) = \sqrt{A^2 + B^2} = \sqrt{(1)^2 + (-2)^2}$ $= \sqrt{1 + 4}$ $= \sqrt{5}$

Step 6: Conclusion The maximum value of the given function $f(x)$ is $\sqrt{5}$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with signs when expanding determinants, especially when using cofactors.
• Trigonometric Identities: Remember fundamental trigonometric identities like $\sin^2 x + \cos^2 x = 1$. They are often crucial for simplifying expressions.
• Formula Application: Correctly identify coefficients $A$ and $B$ when using the $A \cos \theta + B \sin \theta$ formula.

Summary

We simplified the determinant using column and row operations to obtain $f(x) = \cos 2x - 2 \sin 2x$. Then, using the formula for the maximum value of $a \cos \theta + b \sin \theta$, we found the maximum value of $f(x)$ to be $\sqrt{5}$.

The final answer is $\boxed{\sqrt 5}$, which corresponds to option (A).