Question

State whether the given statement is true or false

If $|2x-1|+|3-x|>5$ or $\left|a\right|+\left|b\right|>5$ the range of the functions is $\therefore xϵ[-\infty ,-\frac{1}{3}]\cup [3,\infty ]$.

• A. True
• B. False

Answer 1

The correct option is A True

$\left|a\right|+\left|b\right|>5$
where a=2x-1, b=3-xa+ive
$\Rightarrow x>\frac{1}{2}$, a-ive
$\Rightarrow x<\frac{1}{2}$ b+ive $\Rightarrow x<3,b-ive\Rightarrow x>3$
Case I, a+ive, b+ive $\therefore x>\frac{1}{2},x<3$ and $2x-1+3-x>5$
$\Rightarrow x>3.$
Contradictory results.Case II. a+ive, b-ive $\therefore x>\frac{1}{2},x>3\therefore x>3$
$2x-1-(3-x)>5$ or $3x>9$
$\therefore x>3\therefore xϵ[3,\infty ]$ ...(1)
Case III. a-ive, b+ive$x<\frac{1}{2},x<3\therefore x<\frac{1}{2}$ and $-(2x-1)+3-x>5\Rightarrow x<-\frac{1}{3}$
$\therefore x<\frac{1}{2},x<-\frac{1}{3}\Rightarrow x<-\frac{1}{3}$
$\therefore xϵ[-\infty ,-\frac{1}{3}]$ ...(2)
Case IV. a-ive, b-ive. $x<\frac{1}{2},x>3$ and $-(2x-1)-(3-x)>5$ or $-x>7$ or $x<-7$. Contradictory results.
$\therefore xϵ[-\infty ,-\frac{1}{3}]\cup [3,\infty ]$ from (1) and (2).