Given:
Step 1: Express \( P(x) \) in terms of \( \sin^2{x} \): \[ P(x) = \cos^2{x} + \sin^4{x} = (1 - \sin^2{x}) + \sin^4{x} = 1 - \sin^2{x} + \sin^4{x} \]
Step 2: Let \( y = \sin^2{x} \) (note \( 0 \leq y \leq 1 \)): \[ P(x) = 1 - y + y^2 \]
Step 3: Analyze the quadratic function \( f(y) = y^2 - y + 1 \):
Step 4: Determine the range of \( P(x) \): \[ \frac{3}{4} \leq P(x) \leq 1 \]
The correct option is (E) \( \frac{3}{4} \leq P(x) \leq 1 \).
Let \( P(x) = \cos^2(x) + \sin^4(x) \). We want to find the range of \( P(x) \) for all \( x \in \mathbb{R} \).
We can rewrite \( P(x) \) as:
\[ P(x) = \cos^2(x) + \sin^4(x) = \cos^2(x) + \sin^2(x)\sin^2(x) = \cos^2(x) + (1 - \cos^2(x))(1 - \cos^2(x)) \] \[ = \cos^2(x) + 1 - 2\cos^2(x) + \cos^4(x) = \cos^4(x) - \cos^2(x) + 1 \]
Let \( u = \cos^2(x) \). Since \( 0 \leq \cos^2(x) \leq 1 \), we have \( 0 \leq u \leq 1 \). Then:
\[ P(x) = u^2 - u + 1 \]
This is a quadratic function in \( u \). To find its minimum and maximum values on the interval \([0, 1]\), we can complete the square:
\[ P(x) = (u - \frac{1}{2})^2 - \frac{1}{4} + 1 = (u - \frac{1}{2})^2 + \frac{3}{4} \]
Therefore, the range of \( P(x) \) is \([ \frac{3}{4}, 1 ]\).
So the correct option is: \( \frac{3}{4} \leq P(x) \leq 1 \).
Let \( M \) and \( m \) respectively be the maximum and the minimum values of \( f(x) = \begin{vmatrix} 1 + \sin^2x & \cos^2x & 4\sin4x \\ \sin^2x & 1 + \cos^2x & 4\sin4x \\ \sin^2x & \cos^2x & 1 + 4\sin4x \end{vmatrix}, \quad x \in \mathbb{R} \) for \( x \in \mathbb{R} \). Then \( M^4 - m^4 \) is equal to:
If \( \alpha>\beta>\gamma>0 \), then the expression \[ \cot^{-1} \beta + \left( \frac{1 + \beta^2}{\alpha - \beta} \right) + \cot^{-1} \gamma + \left( \frac{1 + \gamma^2}{\beta - \gamma} \right) + \cot^{-1} \alpha + \left( \frac{1 + \alpha^2}{\gamma - \alpha} \right) \] is equal to: