inequality doubt

stupid question from a stupid guy [3][3] answered by another stupid and discussed by another stupid in STUPID chtbx makes this site stupidiit!

great tongue twister!!!!

See
1x-2 ≤ 2
≡ \frac{1}{x-2}\leq 2 \Rightarrow \frac{1}{x-2}- 2 \leq 0 \Rightarrow \frac{1-2(x-1)}{x-1} \Rightarrow 0 \Rightarrow \frac{3-2x}{x-1} \leq 0 \Rightarrow (3-2x)(x-1) \leq 0 \Rightarrow\frac{1}{x-2}\leq 2 \Rightarrow \frac{1}{x-2}- 2 \leq 0 \Rightarrow \frac{1-2(x-1)}{x-1} \Rightarrow 0 \Rightarrow \frac{3-2x}{x-1} \leq 0 \Rightarrow (3-2x)(x-1) \leq 0

So either (x-1) < 0 and (3-2x) > 0 or (x-1) > 0 and (3-2x) < 0

If (x-1) < 0 and (3-2x) > 0 then x\subseteq ( 32, + infinty)

If (x-1) > 0 and (3-2x) < 0 then x\subseteq ( - infinty, 1 )

therefore solution is \left(-infinty, 1 \right)\bigcup{} \left(\frac{3}{2}, +infinty \right)

:)

Thanks but why that big...

1x - 1 < 2

(x - 1)(x - 1)² < 2

=> (x - 1) < 2 (x² - 2x + 1)

=> (x - 1) < 2x² - 4x + 2

=> 2x² - 5x + 3 > 0

=> 2x² - 3x - 2x + 3 > 0

=> 2x (x - 1) - 3 (x - 1) > 0

=> (x - 1)(2x - 3) > 0

The critical points are 1, 3/2

Thus on applying sign scheme we get

x = (-∞ , 1) U [3/2 , ∞)

on neglecting 1 as it dosen't satisfy the inequality.

Dosen't this makes the question easier. Well ur solution is jhakaas....

its the same only you have reduced some steps but whenever i write a solution i write with all steps so that everyone can understand :0

@Ninepointcircle - and that's why u r the most wanted member on targetIIT :P