Let `A={x:x^(2)-4x+3 lt 0,x in R }` `B={x: 2^(1-x)+p le 0 , x^(2)-2(p+7)x+5 le0}` If `B sube A`, then `p in `

Options

A. `[-4,-1]`
B. `[-4,oo)`
C. `(-oo,1)`
D. `[0,1]`

Explanation

To solve the problem, we need to analyze the sets \( A \) and \( B \) and find the range of \( p \) such that \( B \subseteq A \).

### Step 1: Determine the set \( A \)

We start with the inequality defining set \( A \):
\[ A = \{ x : x^2 - 4x + 3 < 0, \, x \in \mathbb{R} \} \]

We can factor the quadratic expression:
\[ x^2 - 4x + 3 = (x - 1)(x - 3) \]

Now, we need to find where this product is less than zero:
\[ (x - 1)(x - 3) < 0 \]

### Step 2: Analyze the sign of the product

The critical points are \( x = 1 \) and \( x = 3 \). We can test the intervals:
- For \( x < 1 \) (e.g., \( x = 0 \)): \( (0 - 1)(0 - 3) = 3 > 0 \)
- For \( 1 < x < 3 \) (e.g., \( x = 2 \)): \( (2 - 1)(2 - 3) = -1 < 0 \)
- For \( x > 3 \) (e.g., \( x = 4 \)): \( (4 - 1)(4 - 3) = 3 > 0 \)

Thus, the solution to the inequality is:
\[ A = (1, 3) \]

### Step 3: Determine the set \( B \)

Now we analyze set \( B \):
\[ B = \{ x : 2^{1-x} + p \leq 0, \, x^2 - 2(p+7)x + 5 \leq 0 \} \]

#### Step 3.1: Solve the first inequality in \( B \)

From the first inequality:
\[ 2^{1-x} + p \leq 0 \implies p \leq -2^{1-x} \]

As \( x \) varies, \( 2^{1-x} \) decreases. The maximum value occurs at \( x = 1 \):
\[ p \leq -2^{1-1} = -2^0 = -1 \]

#### Step 3.2: Solve the second inequality in \( B \)

Now we analyze the second inequality:
\[ x^2 - 2(p+7)x + 5 \leq 0 \]

This is a quadratic in \( x \). The roots can be found using the quadratic formula:
\[ x = \frac{2(p+7) \pm \sqrt{(2(p+7))^2 - 4 \cdot 1 \cdot 5}}{2} \]

The discriminant must be non-negative for real roots:
\[ (2(p+7))^2 - 20 \geq 0 \]
\[ 4(p+7)^2 - 20 \geq 0 \]
\[ (p+7)^2 \geq 5 \]

Taking square roots:
\[ |p + 7| \geq \sqrt{5} \]

This leads to two cases:
1. \( p + 7 \geq \sqrt{5} \implies p \geq \sqrt{5} - 7 \)
2. \( p + 7 \leq -\sqrt{5} \implies p \leq -\sqrt{5} - 7 \)

### Step 4: Combine conditions for \( B \subseteq A \)

For \( B \subseteq A \), we need:
1. \( p \leq -1 \)
2. \( p \geq \sqrt{5} - 7 \) or \( p \leq -\sqrt{5} - 7 \)

### Step 5: Determine the range of \( p \)

From \( p \leq -1 \) and \( p \leq -\sqrt{5} - 7 \):
- Since \( -\sqrt{5} \approx -2.236 \), we have \( -\sqrt{5} - 7 \approx -9.236 \).

Thus, the valid range for \( p \) is:
\[ p \in (-\infty, -\sqrt{5} - 7] \cup [\sqrt{5} - 7, -1] \]

### Final Answer

The range of \( p \) such that \( B \subseteq A \) is:
\[ p \in (-\infty, -9.236] \text{ or } p \in [-1, -1] \]