$ ext{If } A = \cos^2 heta + \sin^4 heta ext{, then for all values of } heta ext{ is :}$

Q: $\text{If } A = \cos^2 \theta + \sin^4 \theta \text{, then for all values of } \theta \text{ is :}$

$\text{If } A = \cos^2 \theta + \sin^4 \theta \text{, then for all values of } \theta \text{ is :}$

Question

$	ext{If } A = \cos^2 	heta + \sin^4 	heta 	ext{, then for all values of } 	heta 	ext{ is :}$

Accepted Answer

To determine why Option 1 is the correct answer for the expression $ A = \cos^2 \theta + \sin^4 \theta $, we need to analyze the behavior of this expression for all values of $\theta$.

### Step-by-Step Explanation:

1. **Expression Analysis:**
   $$
   A = \cos^2 \theta + \sin^4 \theta
   $$
   We know that $\cos^2 \theta = 1 - \sin^2 \theta$. Substituting this into the expression for $A$, we get:
   $$
   A = (1 - \sin^2 \theta) + \sin^4 \theta = 1 - \sin^2 \theta + \sin^4 \theta
   $$
   Let $ x = \sin^2 \theta $. Then, $ 0 \leq x \leq 1 $.

2. **Rewriting the Expression:**
   $$
   A = 1 - x + x^2
   $$
   This is a quadratic expression in terms of $x$.

3. **Finding the Range of $A$:**
   - The expression $ A = 1 - x + x^2 $ is a quadratic function of $x$.
   - The vertex form of a quadratic $ ax^2 + bx + c $ is given by $ a(x - h)^2 + k $, where $ h = -\frac{b}{2a} $.
   - Here, $ a = 1 $, $ b = -1 $, and $ c = 1 $.
   - The vertex $ x $ value is $ h = -\frac{-1}{2 \times 1} = \frac{1}{2} $.

4. **Calculating $A$ at Critical Points:**
   - Evaluate $ A $ at $ x = 0 $, $ x = 1 $, and $ x = \frac{1}{2} $:
     - $ A(0) = 1 - 0 + 0^2 = 1 $
     - $ A(1) = 1 - 1 + 1^2 = 1 $
     - $ A\left(\frac{1}{2}\right) = 1 - \frac{1}{2} + \left(\frac{1}{2}\right)^2 = 1 - \frac{1}{2} + \frac{1}{4} = \frac{3}{4} $

5. **Range of $A$:**
   - The minimum value of $A$ is $\frac{3}{4}$ and the maximum value is $1$.
   - Therefore, the range of $A$ is $\frac{3}{4} \leq A \leq 1$.

### Justification for Other Options:

- **Option 2: $\frac{3}{4} \le A \le \frac{13}{16}$**
  - This option is incorrect because the maximum value of $A$ is $1$, not $\frac{13}{16}$.

- **Option 3: $1 \le A \le 2$**
  - This option is incorrect because the minimum value of $A$ is $\frac{3}{4}$, not $1$.

- **Option 4: $\frac{13}{16} \le A \le 1$**
  - This option is incorrect because the minimum value of $A$ is $\frac{3}{4}$, not $\frac{13}{16}$.

Thus, the correct answer is indeed Option 1: $\frac{3}{4} \le A \le 1$.