Question

If $|-2x+1|<1$, what will be one of the possible value of $x$?

• A. $0<x<1$
• B. $-1<x<1$
• C. $-1<x<2$
• D. None of the above

Answer 1

The correct option is A $0<x<1$

Given

$|-2x+1|$ $<$ $1$

We know that

If $\left|x\right|$ $<$ $1$ then $-$ $1$ $<$ $x$ $<$ $1$.

Hence, $|-2x+1|$ $<$ $1$

$\Rightarrow$ $-$ $1$ $<$ $-$ $2x$ $+$ $1$ $<$ $1$

Subtracting ${}^{\prime }{1}^{\prime }$ from all, we get

$\Rightarrow$ $-$ $1$ $-$ $1$ $<$ $-$ $2x$ $+$ $1$ $-$ $1$ $<$ $1$ $-$ $1$

$\Rightarrow$ $-$ $2$ $<$ $-$ $2x$ $<$ $0$

Dividing by ${}^{\prime }{2}^{\prime }$ to all, we get

$\Rightarrow$ $\frac{-2}{2}$ $<$ $\frac{-2x}{2}$ $<$ $\frac{0}{2}$

$\Rightarrow$ $-$ $1$ $<$ $-$ $x$ $<$ $0$

Multiplying with ${}^{\prime }-$ $1^{\prime }$ to all, we get

$\Rightarrow$ $0$ $<$ $x$ $<$ $1$

Therefore, Possible values of ${}^{\prime }{x}^{\prime }$ are $0$ $<$ $x$ $<$ $1$.