Question

$<b><big>Good afternoon$

$<b>Answer the above question$

$<b>Chapter : Logarithm$

$<b>No Spam$

Answer 1

Your question needs a correction,

4 option : $\bigg( - \infty ,\dfrac{1}{4} \bigg) \bigcup \bigg( \dfrac{5}{4} , \infty \bigg)$

Given inequality : | 3 - 4x | > 2  

$\overbrace{3-4x>2 \:\: , \:\: 3-4x<-2}^{\text{|3-4x|>2}}$

Case I : when 3 - 4 x > 2  

⇒ 3 - 4x > 2  

⇒ 3 - 2 > 4x  

⇒ 1 > 4x  

⇒ $\dfrac{1}{4}$ > x  

Case II : when 3 - 4x < - 2  

⇒ 3 - 4x < - 2  

⇒ 3 + 2 < 4x

⇒ 5 < 4x  

⇒ $\dfrac{5}{4}$ < x  

$\underline{- \infty}$_0 _1/4 _5/4___$\underline{\infty}$___

In case I, x < $\dfrac{1}{4}$

So, value of x is less 1 /4 , it can be negative too.

$\therefore solution\:\:set\:\:is\:\:\bigg( - \infty , \dfrac{1}{4} \bigg)$  

In case II, x > $\dfrac{5}{4}$

So, value of x more than 5 / 4 ( a positive number ),  it can't be any negative number.

$\therefore solution\:\:set\:\:is\:\:\bigg(\dfrac{5}{4}, \infty \bigg)$  

Therefore the solution of the given inequality is   $\bigg( - \infty ,\dfrac{1}{4} \bigg) \bigcup \bigg( \dfrac{5}{4} , \infty \bigg)$

Option D is correct.