Question

If $|9-x|<2-3x$, then find the range of $xϵR$

• A. $(-\infty ,0)$
• B. $(-\infty ,\frac{-7}{2})$
• C. $(-\infty ,-3)$
• D. $(-16,8)$

Answer 1

Answer: (B)

$(-\infty ,\frac{-7}{2})$

Given, $|9-x|<2-3x$
Case 1: When $9-x>0\Rightarrow |9-x|=9-x$
Then, $9-x<2-3x$
$\Rightarrow 2x<-7$
$\Rightarrow x<-\frac{7}{2}$ -- (1)
Case 2: When $9-x<0\Rightarrow |9-x|=-(9-x)$
Then, $-9+x<2-3x$
$\Rightarrow 4x<11$
$\Rightarrow x<\frac{11}{2}$ --- (2)
From, solutions $1,2$, we get
$\Rightarrow x<-\frac{7}{2}$ or $(-\infty ,-\frac{7}{2})$