SonyGmat wrote:
I have a question regarding inequalities using extreme values.
1st Example
Lets say that I know that \(y>4\) and I want to find the values that x can take. Moreover, I have 2 more inequalities:
\(x-y>-2\)
\(x-2*y<-6\)
I know that by multiplying the first one by -2 (which changes the sign) and then adding the two aforementioned inequalities I can find that \(x>2\).
How can we solve it using extreme values?
I know that y must be between 4.00000001 and a value like 10000000000.
If I use the smallest possible value to the first equation: x must be a bit bigger than 2
If I use the largest possible value to the first equation: x must be a bit bigger than a number really really big.
How do we continue our reasoning??
2nd Example.
Lets say that we have 2 inequalities \(a^2*b>1\) and \(b<2\) and I want to find the values that a can take.
From this we can understand that b must be between 0 and 2 because if b was negative then the first equation would be wrong. (I can't figure out how to solve this algebraically and I would really like to see how, if possible).
Attempt using extreme values
Assume that b is a bit bigger than 0, lets say 1/10000000 we can conclude that \(a^2\) will be bigger than a very big number. Squaring both sides we can see that a is larger that a big value and smaller than a very small.
How do we continue our reasoning...
I dislike 'making equations and solving' but when it comes to solving some inequalities simultaneously, I generally prefer to tame my imagination and follow algebra because it is risk free. Let me discuss how to use algebra in both cases and then I will talk about 'guessing the range'
Golden Rule: Given two inequalities, you can always add them up if the inequality signs point in the same direction.
\(x-y>-2\)
\(x-2*y<-6\)
Here, the signs point in opposite directions.
To get the range of x, we multiply the first inequality by -2 and add.
\(-2x+2y<4\)
\(x-2y<-6\)
-x < -2 or x > 2
To get the range of y, we multiply the first inequality by -1 and add.
\(-x+y<2\)
\(x-2y<-6\)
-y < -4 or y > 4
We get the range of x and y.
\(a^2*b>1\)
\(b<2\)
To get the range of a, multiply the second inequality by -a^2 (which we know will be negative since a^2 is positive) and add. (just solving them blindly)
\(a^2*b>1\)
\(-a^2*b>-2a^2\)
We get \(0 > 1-2a^2\)
\(a^2 > 1/2\)
\(a^2 - (1/\sqrt{2})^2 > 0\)
\(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)
Now, if you want to use the extreme value to guess the range, here is the logic for that:
Given:\(y>4\)
Find the range of x if \(x-y>-2\)
\(x > y-2\)
If y is a little more than 4, x is more than 2.
If y is 5, x is more than 3.
and so on...
The range that includes all these ranges is 'x is a little more than 2.'
Given: \(y>4\)
Find the range of x if \(x-2y<-6\)
\(x < 2y -6\)
If y is a little more than 4, x is less than something more than 2.
If y is 5, x is less than 4.
If y is 6, x is less than 6.
and so on...
The range that includes all these ranges is 'x is less than infinity.' which doesn't add much information. Algebraically, if you try to solve these two inequalities, you see you cannot get rid of y and hence don't get the range for x.
The diagram explains the above given two discussions.
Attachment:
Ques4.jpg [ 7.55 KiB | Viewed 4109 times ]
Given: \(a^2*b>1\)
\(b< 2\)
If b is a little less than 2, a^2 is greater than 1/2 which means \(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)
If b is 1, a^2 is greater than 1 which means \(a > 1\)or \(a < -1\)
b cannot be 0 or negative.
The range that includes all these ranges will be \(a > 1/\sqrt{2}\) or a \(< -1/\sqrt{2}\)
Attachment:
Ques5.jpg [ 4.73 KiB | Viewed 4107 times ]
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15 Nov 2011, 05:20
