CrackGMAC wrote:I have two general queries in Modlus questions
1. When we say X < Y does that mean x-y < 0 or
x/y < 1. In my view both are same. If not please let me know what the difference is and where we follow these different rules.
2. What is best approach to solve the questions when we have mod on both the sides,
say for example x^2 + |x+2| < x^4 + |x-10
x < y
x + a < y + a
x - a < y -a
x/k < y/k when k is +ve
x/k > y/k when k is negative
All the above same.
x < y
subtract y from both sides, the inequality doesnt change.
so, x -y < 0
x/y < 1
here, we don't know whether y is posiitve or not.
When y is positive, x < y
when y is negative, x > y.
My approach here is to "not bring verbal reasoning": just manipulate algebraically.
x/y < 1
xy/y^2 < 1
xy - y^2 < 0 (we can shuffle around y^2, since it is +ve)
y(x-y) < 0
The above inequality tells us two cases.
(1) y is +ve and x-y is +ve
(2) y is -ve and x-y is -ve.
At this stage, you can translate algebra into words.
x^2 + |x+2| < x^4 + |x-10|
In questions like this, there is a default method.
we got two critical points to look at: -2 and 10
So, solve for each range.
(-inf, -2): x^2 -x -2 < x^4-x +10
x^4 -x^2 + 12 > 0
True for all x's in this range
(-2, 10): x^2 + x+2 < x^4 -x + 10
x^4 -2x^2 +8 > 0
True for all x's in this range
(10, +inf): x^2 + x +2 < x^4 + x -10
x^4 -x^2 -12 > 0
(x^2 - 4)(x^2+3) > 0
x^2 - 4 > 0
or true for all x's.
You can also use conceptual approach to get rid of odd choices
