Question

Consider the two sets \[A = \{ m \in \mathbb{R}:\] both the roots of \[{x^2} - \left( {m + 1} \right)x + m + 4 = 0\] are real} and \[B = \left[ { - 3,5} \right)\]. Which of the following is not true?
A. \[A - B = \left( { - \infty , - 3} \right) \cup \left( {5,\infty } \right)\]
B. \[A \cap B = \left\{ { - 3} \right\}\]
C. \[B - A = \left( { - 3,5} \right)\]
D. \[A \cup B = \mathbb{R}\]

Answer 1

Hint: If both the roots of a quadratic equation are real, then the discriminant of the equation is non-negative. Using this you’ll get an inequation in \[m\]. Solve this inequation and obtain the set \[A\] in roster form. Then check all the given options one by one.

Formula Used:
Discriminant of a quadratic equation \[a{x^2} + bx + c = 0\] is \[D = {b^2} - 4ac\]
If both the roots are real then \[D \ge 0\]

Complete step-by-step answer:
Given that \[A = \{ m \in \mathbb{R}:\]both the roots of \[{x^2} - \left( {m + 1} \right)x + m + 4 = 0\] are real} and \[B = \left[ { - 3,5} \right)\]
The discriminant of the quadratic equation \[{x^2} - \left( {m + 1} \right)x + m + 4 = 0\] is
\[D = {\left\{ { - \left( {m + 1} \right)} \right\}^2} - 4 \times 1 \times \left( {m + 4} \right)\]
\[ = {\left( {m + 1} \right)^2} - 4\left( {m + 4} \right)\]
\[ = {m^2} + 2m + 1 - 4m - 16\]
\[ = {m^2} - 2m - 15\]
\[ = {m^2} - 5m + 3m - 15\]
\[ = m\left( {m - 5} \right) + 3\left( {m - 5} \right)\]
\[ = \left( {m - 5} \right)\left( {m + 3} \right)\]
As both the roots of the equation are real, so \[D \ge 0\]
\[ \Rightarrow \left( {m - 5} \right)\left( {m + 3} \right) \ge 0\]
\[ \Rightarrow m \in \left( { - \infty , - 3} \right] \cup \left[ {5,\infty } \right)\]
\[\therefore A = \left( { - \infty , - 3} \right] \cup \left[ {5,\infty } \right)\]
Checking option A:
\[A - B = \left( { - \infty , - 3} \right] \cup \left[ {5,\infty } \right) - \left[ { - 3,5} \right) = \left( { - \infty , - 3} \right) \cup \left[ {5,\infty } \right)\]
So, option A is not correct.
Checking option B:
\[A \cap B = \left\{ { - 3} \right\}\]
So, option B is correct.
Checking option C:
\[B - A = \left( { - 3,5} \right)\]
So, option C is correct.
Checking option D:
\[A \cup B = \left( { - \infty , - 3} \right] \cup \left[ {5,\infty } \right) \cup \left[ { - 3,5} \right) = \mathbb{R}\]
So, option D is correct.
Hence option A is not true.

Note: In this type of MCQ you should check all the options properly. We can easily solve the problems of set theory if a set is given in roster form. So, at first you should find the roster form of the set \[A\]. Then apply the operations union, intersection and minus.