Question

Consider the two sets: $A=\{m\in R:$ both the roots of$x^2-(m+1)x+m+4=0$ are real$\}$ and$B=[-3,5)$ Which of the following is not true?

• A. $A-B=(-\infty ,-3)\cup (5,\infty )$
• B. $A\cap B=\{-3\}$
• C. $B-A=(-3,5)$
• D. $A\cup B=R$

Answer 1

The correct option is A

$A-B=(-\infty ,-3)\cup (5,\infty )$

Explanation for correct option:

Step 1:Find the intervals in which $m$ lies.

We have given $A=\{m\in R:$ both the roots of$x^2-(m+1)x+m+4=0$ are real$\}$ and $B=[-3,5)$

As roots of$x^2-(m+1)x+m+4=0$ are real

$\begin{array}{rcl}\therefore D& \ge & 0\\ b^2-4ac& \ge & 0\\ {\{-(m+1\left)\right\}}^2-4\left(1\right)(m+4)& \ge & 0\\ m^2+2m+1-4m-16& \ge & 0\\ m^2-2m-15& \ge & 0\\ (m-5)(m+3)& \ge & 0\\ & & m\in (-\infty ,-3]\cup [5,\infty )\end{array}$

$\begin{array}{rcl}\therefore A& =& (-\infty ,-3]\cup [5,\infty )\\ B& =& [-3,5)\end{array}$

Step 2: Check the different situations.

$\begin{array}{rcl}A-B& =& (-\infty ,-3)\cup [5,\infty )\\ A\cap B& =& \{-3\}\\ B-A& =& (-3,5)\\ A\cup B& =& R\end{array}$

Hence, the correct option is$\left(A\right)$