1stupid 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!!!!
1See
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)
:)
36Thanks 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....
1
ninepointcircle
·2011-03-26 00:14:12
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
36
rahul
·2011-03-26 00:42:07
@Ninepointcircle - and that's why u r the most wanted member on targetIIT :P
49@Rahul: is baar bach gaya coz it was a square term...but if u cross-multiply like that in inequalities...nature le na le..maths toh tujhse zaroor badla lega! :P
621/y<2
is same as y>1/2 or y -ve
here y is x-1
so x>3/2 or x<1
:)