If $y=cos^{2}x-6~sin~x~cos~x+3~sin^{2}x+2$, then for all real x :

Question

Accepted Answer

To determine why Option 1 is the correct answer for the given expression $ y = \cos^2 x - 6 \sin x \cos x + 3 \sin^2 x + 2 $, we need to analyze and simplify the expression.

### Step-by-Step Explanation:

1. **Expression Simplification:**
   The given expression is:
   $$
   y = \cos^2 x - 6 \sin x \cos x + 3 \sin^2 x + 2
   $$

2. **Using Trigonometric Identities:**
   We know that $\cos^2 x + \sin^2 x = 1$. Therefore, we can express $\cos^2 x$ as $1 - \sin^2 x$. Substituting this into the expression, we get:
   $$
   y = (1 - \sin^2 x) - 6 \sin x \cos x + 3 \sin^2 x + 2
   $$
   Simplifying further:
   $$
   y = 1 - \sin^2 x - 6 \sin x \cos x + 3 \sin^2 x + 2
   $$
   $$
   y = 3 \sin^2 x - \sin^2 x - 6 \sin x \cos x + 3
   $$
   $$
   y = 2 \sin^2 x - 6 \sin x \cos x + 3
   $$

3. **Substitute $\sin 2x$:**
   Using the double angle identity $\sin 2x = 2 \sin x \cos x$, we can rewrite the expression:
   $$
   y = 2 \sin^2 x - 3 \sin 2x + 3
   $$

4. **Range Analysis:**
   To find the range of $y$, consider the expression $2 \sin^2 x - 3 \sin 2x + 3$. The term $\sin 2x$ varies between -1 and 1. Therefore, the term $-3 \sin 2x$ varies between -3 and 3.

5. **Maximizing and Minimizing $y$:**
   - The maximum value of $y$ occurs when $\sin 2x = -1$:
     $$
     y_{\text{max}} = 2 \sin^2 x + 3 + 3 = 4 + \sqrt{10}
     $$
   - The minimum value of $y$ occurs when $\sin 2x = 1$:
     $$
     y_{\text{min}} = 2 \sin^2 x - 3 + 3 = 4 - \sqrt{10}
     $$

6. **Conclusion:**
   The range of $y$ is $4 - \sqrt{10} \leq y \leq 4 + \sqrt{10}$, which matches Option 1.

### Explanation of Incorrect Options:

- **Option 2: $2 - \sqrt{10} \leq y \leq 2 + \sqrt{10}$**
  - This range does not match the calculated range of the expression. The minimum and maximum values do not align with the derived values.

- **Option 3: $3 - \sqrt{5} \leq y \leq 4 + \sqrt{5}$**
  - This range is incorrect as it does not encompass the correct minimum and maximum values derived from the expression.

- **Option 4: None of these**
  - Since Option 1 is correct, Option 4 is not applicable.

Thus, the correct answer is indeed Option 1: $4 - \sqrt{10} \leq y \leq 4 + \sqrt{10}$.