If `f(x)=(sin^(4)x+cos^(2)x)/(sin^(2)x+cos^(4)x)"for "x in R`, then f(2010)

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the function \( f(x) = \frac{\sin^4 x + \cos^2 x}{\sin^2 x + \cos^4 x} \) for \( x = 2010 \), we will first simplify the function \( f(x) \).

### Step 1: Rewrite the function
We start with:
\[ f(x) = \frac{\sin^4 x + \cos^2 x}{\sin^2 x + \cos^4 x} \]

### Step 2: Use trigonometric identities
Recall that \( \sin^2 x + \cos^2 x = 1 \). We can express \( \sin^4 x \) and \( \cos^4 x \) in terms of \( \sin^2 x \) and \( \cos^2 x \):
\[ \sin^4 x = (\sin^2 x)^2 \quad \text{and} \quad \cos^4 x = (\cos^2 x)^2 \]

### Step 3: Substitute \( \sin^2 x \) and \( \cos^2 x \)
Let \( a = \sin^2 x \) and \( b = \cos^2 x \). Then \( a + b = 1 \). We can rewrite the function as:
\[ f(x) = \frac{a^2 + b}{a + b^2} \]

### Step 4: Substitute \( b = 1 - a \)
Substituting \( b \) gives:
\[ f(x) = \frac{a^2 + (1 - a)}{a + (1 - a)^2} \]
Simplifying the numerator:
\[ = \frac{a^2 + 1 - a}{a + (1 - 2a + a^2)} = \frac{a^2 - a + 1}{a^2 - a + 1} \]

### Step 5: Simplify the expression
Since the numerator and the denominator are the same, we have:
\[ f(x) = 1 \]

### Step 6: Evaluate \( f(2010) \)
Since \( f(x) = 1 \) for all \( x \) in \( \mathbb{R} \), we can conclude that:
\[ f(2010) = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]