Question

solve 1/|2x-1|<6 and express it using interval notation
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Answer 1

We have to solve inequality 1/|2x - 1| < 6 using interval notation.

Solution : we see both side of inequality sign are positive so if we reciprocate, sign of inequality will chnage.

I.e., |2x - 1 | > 1/6

now break the modulus function,

2x - 1 = 0 ⇒x = 1/2

case 1 : x > 1/2

(2x - 1) > 1/6

⇒2x - 1 - 1/6 > 0

⇒2x - 7/6 > 0

⇒2x > 7/6

⇒x > 7/12

now put 1/2 and 7/12 in number line and apply their conditions , you get, x > 7/12

Case 2 : x < 1/2

-(2x - 1) > 1/6

⇒-2x + 1 - 1/6 > 0

⇒-2x + 5/6 > 0

⇒5/6 > 2x

⇒x < 5/12

Similarly put 1/2 and 5/12 in number line and apply their conditions, you get x < 5/12

Therefore the value of x ∈ (-∞, 5/12) U (7/12, ∞)

Answer 2

SOLUTION

TO EXPRESS

$\displaystyle \sf{ \frac{1}{ |2x - 1|} < 6\: }$

in interval notation

CONCEPT TO BE IMPLEMENTED

We are aware of the formula on inequality that

$\sf{ |x - a| > b \: implies \: x < a - b \: or \: x > a + b \: }$

EVALUATION

Here the given inequality is

$\displaystyle \sf{ \frac{1}{ |2x - 1|} < 6\: }$

$\displaystyle \implies \sf{ |2x - 1| > \frac{1}{6} \: }$

$\displaystyle \implies \sf{ 2x - 1 > \frac{1}{6} \: \: or \: \: 2 x- 1 < - \frac{1}{6} }$

Now

$\displaystyle \sf{ 2x - 1 > \frac{1}{6} } \: \: \: \: gives$

$\displaystyle \sf{ 2x > 1 + \frac{1}{6} } \: \: \: \:$

$\implies \displaystyle \sf{ 2x > \frac{7}{6} } \: \: \: \:$

$\implies \displaystyle \sf{ x > \frac{7}{12} } \: \: \: \:$

$\implies \displaystyle \sf{ x \in \: \bigg(\frac{7}{12} } \: , \infty \bigg) \: \:$

Again

$\displaystyle \sf{ 2x - 1 < - \frac{1}{6} } \: \: \: gives$

$\displaystyle \sf{ 2x < 1 - \frac{1}{6} }$

$\implies \displaystyle \sf{ 2x < \frac{5}{6} }$

$\implies \displaystyle \sf{ x < \frac{5}{12} }$

$\implies \displaystyle \sf{ x \in \: \bigg( - \infty , \frac{5}{12} \: \bigg) \: \:}$

Hence the required solution of the given inequality is

$\displaystyle \sf{ x \in \bigg( \frac{7}{12} , \infty \bigg) \cup\bigg( - \infty , \frac{5}{12} \: \bigg) }$

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