Difference between revisions of "MIE 2016/Problem 2"
m |
m (→Step 1) |
||
Line 22: | Line 22: | ||
We now simplify the inequality: | We now simplify the inequality: | ||
+ | |||
Case 0: <math>x=0</math> | Case 0: <math>x=0</math> | ||
+ | |||
This has no solutions since <math>x=0</math> will make the function undefined. | This has no solutions since <math>x=0</math> will make the function undefined. | ||
Revision as of 16:56, 12 September 2020
Problem 2
The following system has integer solutions. We can say that:
(a)
(b)
(c)
(d)
(e)
Solution 1
Objective:
We can solve this problem in two steps: First, we solve for the range of , then combine it with the range of
to get a compound inequality which we can use to find all possible integer solutions.
Step 1
We first find the range of the inequality .
We now simplify the inequality:
Case 0:
This has no solutions since will make the function undefined.
Case 1:
Factoring, we get
Now,
can be greater than
or less than
.
But in this case,
, and this further restricts our solutions.
So, for the case where
, our solutions are
Case 2:
We have in this case that
, but the case statement further restricts our solutions.
For this case, the solutions are
Step 2
Now, we know the solutions for : in the first case, where
, the integer solutions are
In the second case, where , the only integer solution is
The union of these two cases gives .
There are solutions and
, giving
~Windigo