Question

$\Large{\bf{\green{\mathfrak{\dag{\underline{\underline{Question:-}}}}}}}$

| 2x - 1 | > - 2

Solve for x.

$\orange{\textsf{Class:-}}$ 11th

$\orange{\textsf{Topic:-}}$ Moduless function

I want complete explanation of the answer with suitable steps.

@Takenname,@Amansharma and mods help me.

$\red{\textsf{Don't Spam.}}$​

Answer 1

Question :

• | 2x - 1 | > - 2 Solve for x.

Concept :

• We have to solve x on this question.

Let's Start :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies}\sf |2x - 1| > - 2 \\ \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: Since \: |2x - 1| \: is \: always +ve \: and -2 \: is -ve \\ \\ \\ \\ \sf|2x-1| is \: always \: great \: than -2 \\ \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \: So \: the \: inequality \: is \: always \: true \: for \: any \: value \: of \: x \\ \\ \\ \\ : \implies \sf All \: Real \: numbers$

The Result can be shown in multiple forms.

All real numbers Interval Notation : ( -∞, ∞ )

Other information :

• Simplify = Reducing the lowest term .

• Inequality = A mathematical expression which shows that two quantities are not equal.

• variable = A letter used to represent a number value in an expression or an equation.

• Interval = A set of values between two endpoints.

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\: |2x - 1| > - 2$

As we know,

Modulus function is defined as

$\begin{gathered}\begin{gathered}\bf\:\rm :\longmapsto\: |x| = \begin{cases} &\sf{x \: \: \: if \: x \geqslant 0} \\ &\sf{ - x \: \: \: \: if \: x \: < \: 0} \end{cases}\end{gathered}\end{gathered}$

$\bf\implies \: |x| \: can \: never \: be \: negative \: for \: x \in \: R$

Now, It is given that

$\rm :\longmapsto\: |2x - 1| > - 2$

As

$\rm :\longmapsto\: |2x - 1| \: \: \cancel{ < } \: \: 0$

$\bf\implies \: |2x - 1| \geqslant 0$

$\bf\implies \:x \: \in \: R$

Additional Information :-

$\rm :\longmapsto\: |x| < y \implies \: - y < x < y$

$\rm :\longmapsto\: |x| \leqslant y \implies \: - y \leqslant x \leqslant y$

$\rm :\longmapsto\: |x| > y \implies \: x < - y \: or \: x > y$

$\rm :\longmapsto\: |x| \geqslant y \implies \: x \leqslant - y \: or \: x \geqslant y$

$\rm :\longmapsto\: |x - a| < b \implies \: a - b < x < a + b$

$\rm :\longmapsto\: |x - a| \leqslant b \implies \: a - b \leqslant x \leqslant a + b$