Let A B=\ x (x-a)(x-b) > 0 , where a < b\ . What are A and B equal to?

Correct answer: B. A=\ x x b\ A B=\ x:(x-a) ·(x-b) > 0 ; a eleminating options (a) x > a has two signs in range (a, ∞) x > a is not possible (b) x b has only one sign in range (b, ∞ ) x > b also satisfies. A=\ x: x b\

NDA 2019

Exam: NDA
Year: 2019
Subject: Mathematics
Chapter: Basics of Mathematics

NDA 2019 · Mathematics · Basics of Mathematics

Let A B=\ x (x-a)(x-b) > 0 , where a < b\ . What are A and B equal...

Let

\mathrm{A} \cup \mathrm{B}=\{x \mid(x-a)(x-b) > 0

, where

a < b\}

. What are

A

and

B

equal to?

Options

A $\mathrm{A}=\{x \mid x > a\}$ and $\mathrm{B}=\{x \mid x > b\}$
B $\mathrm{A}=\{x \mid x < a\}$ and $\mathrm{B}=\{x \mid x > b\}$
C $\mathrm{A}=\{x \mid x < a\}$ and $\mathrm{B}=\{x \mid x < b\}$
D $\mathrm{A}=\{x \mid x > a\}$ and $\mathrm{B}=\{x \mid x < b\}$

Correct answer

B. $\mathrm{A}=\{x \mid x < a\}$ and $\mathrm{B}=\{x \mid x > b\}$

Step-by-step solution

A \cup B=\{x:(x-a) \cdot(x-b) > 0 ; a < b\}

eleminating options (a)

x > a

has two signs in range

(a, \infty)\therefore x > a

is not possible (b)

x < a

satisfies and

x > b

has only one sign in range (b, \infty

) \therefore x > b

also satisfies.

\therefore A=\{x: x < a\}

and

B=\{x: x > b\}

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