1ans of first ques
x belongs to ( -∞ , -2/3 ] U [ 1/2 , ∞ )
Solve for x
(a) |x2 + 3x| + x2 - 2 > 0
(b) |x + 3| > |2x - 1|
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7 Answers
21a) basically you have to solve
|x2+3x| > 2-x2
which means x2+3x>2-x2 or 2x2+3x-2 >0 (this will be true when x is outside the roots of 2x2+3x-2= 0
and x2+3x<x2-2
which means x <-2/3
u will get same as mohit's
21b)|x+3|>|2x-1|
|x+3||2x-1| >1
|x+32x-1|>1
you may try squaring it and applying quadratic
or do this
|x+32x-1|>1 means
x+32x-1>1
or x+32x-1<1
1@RAHUL
i give u a PERFECT ALWAYS WORKING RULE FOR "MODULUS" or in that case for any "PIECEWISE"...defined inequality
break the inequality according to domain of its definiton and solve each piece.if solutions satisy the domain they are accepted if not they are rejected!
that's it!
21I was completely correct except that i made a typing mistake(which is easily seen)
In the last line it will be x+32x-1<-1
breaking the inequality of course gives the result but takes a lot of time. Better u give stress on
properties of modulus and inequalities.