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 find the values of \( p \) such that the set \( B \) is a subset of the set \( A \). Let's break this down step by step.

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

The set \( A \) is defined by the inequality:
\[ x^2 - 4x + 3 < 0 \]
To solve this inequality, we first find the roots of the quadratic equation \( x^2 - 4x + 3 = 0 \).

Factoring, we have:
\[ (x - 1)(x - 3) = 0 \]
Thus, the roots are \( x = 1 \) and \( x = 3 \).

Next, we analyze the sign of the quadratic expression \( x^2 - 4x + 3 \):
- For \( x < 1 \), \( (x - 1)(x - 3) > 0 \) (positive).
- For \( 1 < x < 3 \), \( (x - 1)(x - 3) < 0 \) (negative).
- For \( x > 3 \), \( (x - 1)(x - 3) > 0 \) (positive).

Thus, the solution to the inequality \( x^2 - 4x + 3 < 0 \) is:
\[ A = (1, 3) \]

### Step 2: Determine the set \( B \)

The set \( B \) is defined by two inequalities:
1. \( 2^{1-x} + p \leq 0 \)
2. \( x^2 - 2(p + 7)x + 5 \leq 0 \)

#### Analyzing the first inequality:
From the first inequality \( 2^{1-x} + p \leq 0 \), we can rearrange it to find:
\[ p \leq -2^{1-x} \]
To find the values of \( p \) at the endpoints of the interval \( A \):
- At \( x = 1 \):
\[ p \leq -2^{1-1} = -2^0 = -1 \]
- At \( x = 3 \):
\[ p \leq -2^{1-3} = -2^{-2} = -\frac{1}{4} \]

Thus, combining these results, we have:
\[ p \leq -1 \quad \text{and} \quad p \leq -\frac{1}{4} \]
So, the more restrictive condition is:
\[ p \leq -1 \]

#### Analyzing the second inequality:
Now, we analyze the second inequality \( x^2 - 2(p + 7)x + 5 \leq 0 \). This is a quadratic in \( x \) and will be non-positive between its roots.

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

For the quadratic to be non-positive, the roots must be real, which requires:
\[ (p + 7)^2 - 5 \geq 0 \]
Solving this:
\[ (p + 7)^2 \geq 5 \]
Taking square roots:
\[ |p + 7| \geq \sqrt{5} \]
This gives us two inequalities:
1. \( p + 7 \geq \sqrt{5} \) or
2. \( p + 7 \leq -\sqrt{5} \)

From the first inequality:
\[ p \geq -7 + \sqrt{5} \]
From the second inequality:
\[ p \leq -7 - \sqrt{5} \]

### Step 3: Combine the conditions

Now we have two conditions:
1. \( p \leq -1 \)
2. \( p \geq -7 + \sqrt{5} \) or \( p \leq -7 - \sqrt{5} \)

Since \( -7 - \sqrt{5} \) is less than \( -1 \), we focus on the interval:
\[ p \in [-7 + \sqrt{5}, -1] \]

### Conclusion

Thus, the final result for \( p \) such that \( B \subseteq A \) is:
\[ p \in [-7 + \sqrt{5}, -1] \]