surya167 wrote:
Brunel and all,
Is it a rule to apply one value as zero whenever it is given:
1) Both x and y are non-negative integers
2) we need to find the max value of x-y
What if we are asked to find the min ? how do we solve those questions and also, what would be the approach for min and max value of x+y ? Can u guys pls advise?
Usually, when you are checking for numbers, you do check for 0. It's often a transition point for patterns. Secondly, the question used the term 'non-negative integers' instead of 'positive integers' - this means 0 would probably have a role to play.
There are no such rules but common sense says that we must not ignore 0.
Also, this question doesn't really test max min concepts. It is a direct application of your understanding of exponential relations discussed in the post:
http://www.veritasprep.com/blog/2013/01 ... cognition/Now, when we look at the equation, 2^x + 2^y = x^2 + y^2, some things come to mind:
1. It is not very easy to find values that satisfy this equation.
2. But there must be some values which satisfy since we are looking for a value of |x – y|
3. If x = y = 2, the equation is satisfied since all terms become equal and |x – y| = 0 which is the minimum value of |x – y|.
Usually, the left hand side will be greater than the right hand side (as discussed in the post, 2^n will usually be greater than x^2 except in very few cases). So we must focus on those 'very few cases'. Also, we need to make x and y unequal.
We know (from the post) that 2^4 = 4^2 is one solution so we could put x = 4 while keeping y = 2. The equation will be satisfied and |x – y| = 2
Now, we also know that 2^x < x^2 when x = 3. So is there a solution there as well? The difference between 2^3 and 3^2 is of 1 so can we create a difference of 1 between the other two terms? Sure! If y = 0, then 2^0 = 1 but 0^2 = 0.
So another solution is 2^3 + 2^0 = 3^2 + 0^2.
Here, |x – y| = 3 which is the maximum difference.
The reason we can be sure that there are no other values is that as you go ahead of 4 on the number line, 2^n will be greater than n^2 (again, discussed in the post). So both left hand side terms will be greater than the right hand side terms i.e. 2^x > x^2 and 2^y > y^2. So, for no other values can we satisfy this equation.
Answer (D)
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