Question

pl tell the answer for the attached photo​

Answer 1

Step-by-step explanation:

We have,

$\left| \dfrac{2}{x - 4} \right| > 1$

$\implies \dfrac{2}{ | x - 4 | } > 1$

$\implies \dfrac{ | x - 4 | }{2} < 1$

$\implies | x - 4 | < 2$

$\implies | x - 4 |^{2} < (2) ^{2}$

$\implies x^{2} - 8x + 16 < 4$

$\implies x^{2} - 8x + 12 < 0$

$\implies x^{2} - 6x - 2x + 12 < 0$

$\implies x(x- 6)- 2(x - 6) < 0$

$\implies (x - 2)(x- 6) < 0$

$\implies x \in ( 2 ,6)$

Answer 2

$\large\underline{\sf{Given \:Question - }}$

Solve :-

$\rm :\longmapsto\:\bigg |\dfrac{2}{x - 4} \bigg| > 1 \: and \: x \: \ne \: 4$

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

Given inequality is

$\rm :\longmapsto\:\bigg |\dfrac{2}{x - 4} \bigg| > 1 \: and \: x \: \ne \: 4$

We know,

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

So,

$\rm :\longmapsto\:\dfrac{2}{ |x - 4| } > 1$

$\rm :\longmapsto\:\dfrac{ |x - 4| }{2} < 1$

$\rm :\longmapsto\: |x - 4| < 2$

$\rm :\longmapsto\: - 2 < x - 4 < 2$

On adding 4 in each term, we get

$\rm :\longmapsto\: - 2 + 4 < x - 4 + 4 < 2 + 4$

$\rm :\longmapsto\:2 < x < 6 \: \: and \: \: x \: \ne \: 4$

$\rm \implies\:x \: \in \: (2,4) \: \cup \: (4,6)$

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More to Know :-

$\boxed{\tt{ |x| < y \: \: \rm \implies\: - y < x < y \: }}$

$\boxed{\tt{ |x| \leqslant y \: \: \rm \implies\: - y \leqslant x \leqslant y \: }}$

$\boxed{\tt{ |x| > y \: \: \rm \implies\:x < - y \: \: or \: \: x > y}}$

$\boxed{\tt{ |x| \geqslant y \: \: \rm \implies\:x \leqslant - y \: \: or \: \: x \geqslant y}}$

$\boxed{\tt{ |x - z| < y \: \: \rm \implies\: z- y < x <z + y \: }}$

$\boxed{\tt{ |x - z| \leqslant y \: \: \rm \implies\: z- y \leqslant x \leqslant z + y \: }}$