Range of `y = (sin^4 x + 3 cos^2 x)/(sin^4 x + cos^2 x)` is

Options

A. `[0, 2]`
B. `[1, 3]`
C. `[2, 3]`
D. `[1, 2]`

Answer

To find the range of the function

\[ y = \frac{\sin^4 x + 3 \cos^2 x}{\sin^4 x + \cos^2 x}, \]

we will follow these steps:

### Step 1: Rewrite the function in terms of \( t = \sin^2 x \)

Since \(\cos^2 x = 1 - \sin^2 x\), we can express the function in terms of \( t \):

\[ y = \frac{t^2 + 3(1 - t)}{t^2 + (1 - t)}. \]

### Step 2: Simplify the function

Substituting \(\cos^2 x\) gives:

\[ y = \frac{t^2 + 3 - 3t}{t^2 + 1 - t} = \frac{t^2 - 3t + 3}{t^2 - t + 1}. \]

### Step 3: Determine the limits for \( t \)

Since \( t = \sin^2 x \), the range of \( t \) is from \( 0 \) to \( 1 \) (i.e., \( 0 \leq t \leq 1 \)).

### Step 4: Analyze the function

We need to find the critical points of \( y \) by differentiating it with respect to \( t \) and setting the derivative to zero.

Let

\[ f(t) = t^2 - 3t + 3, \]
\[ g(t) = t^2 - t + 1. \]

Using the quotient rule:

\[ y' = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}. \]

Calculating \( f'(t) \) and \( g'(t) \):

\[ f'(t) = 2t - 3, \]
\[ g'(t) = 2t - 1. \]

Setting the numerator of \( y' \) to zero to find critical points:

\[ (2t - 3)(t^2 - t + 1) - (t^2 - 3t + 3)(2t - 1) = 0. \]

### Step 5: Evaluate at the endpoints

Evaluate \( y \) at the endpoints \( t = 0 \) and \( t = 1 \):

1. For \( t = 0 \):
\[ y(0) = \frac{0^2 - 3(0) + 3}{0^2 - 0 + 1} = \frac{3}{1} = 3. \]

2. For \( t = 1 \):
\[ y(1) = \frac{1^2 - 3(1) + 3}{1^2 - 1 + 1} = \frac{1 - 3 + 3}{1 - 1 + 1} = \frac{1}{1} = 1. \]

### Step 6: Conclusion

The maximum value of \( y \) is \( 3 \) and the minimum value is \( 1 \). Therefore, the range of \( y \) is:

\[ \text{Range of } y = [1, 3]. \]