First slide

Introduction to sets

Question

$Let A = \{x ∣ x^{2} - 4 x + 3 < 0, x \in R\}; B = \{x ∣ 2^{1 - x} + p \leq 0; x^{2} - 2 (p + 7) x + 5 \leq 0\}$ $If B \subseteq A, then p \in$

Moderate

A

$[- 4, - 1]$
B

$[- 4, \infty)$
C

$(- \infty, 1]$
D

$[0, 1]$

Solution

$\begin{array}{l} A = (1, 3) \\ For B \subseteq A \begin{matrix} 2^{1 - 1} + p \leq 0 \Rightarrow p \leq - 1 \\ 2^{1 - 3} + p \leq 0 \Rightarrow p \leq - 1 / 4 \end{matrix}\} p \leq - 1 \\ f (x) = x^{2} - 2 (p + 7) x + 5 \\ \begin{matrix} f (1) \leq 0 \Rightarrow p \geq - 4 \\ f (3) \leq 0 \Rightarrow P \geq - 1 \end{matrix}\} P \geq - 4 \\ So p \in [- 4, - 1] \end{array}$

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