Question

If $|3x-4|<11$, then $x\in$

• A. $(\frac{7}{3},5)$
• B. $[-\frac{7}{3},5]$
• C. $R-(-\frac{7}{3},5)$
• D. $(-\frac{7}{3},5)$

Answer 1

The correct option is D $(-\frac{7}{3},5)$

$|3x-4|<11$
$\Rightarrow -11<3x-4<11$
$\Rightarrow -7<3x<15$
$\Rightarrow -\frac{7}{3}<x<5$
$\therefore x\in (-\frac{7}{3},5)$

Alternate:
Case $1:$ $x\ge \frac{4}{3}$
$\Rightarrow 3x-4<11$
$\Rightarrow x<5$
$\Rightarrow x\in [\frac{4}{3},5)$

Case $2:$ $x<\frac{4}{3}$
$\Rightarrow -3x+4<11\Rightarrow x>-\frac{7}{3}\Rightarrow x\in (-\frac{7}{3},\frac{4}{3})$

$\therefore x\in (-\frac{7}{3},\frac{4}{3})\cup [\frac{4}{3},5)\Rightarrow x\in (-\frac{7}{3},5)$