aflaam messih wrote:if (x-2)^4/4 + (y+2)^4/9 = 4,then the maximum possible value of x is between
a) -1 and 1
b) 1 and 3
c) 3 and 5
d) 5 and 7
e) 7 and 9
I don't know how to solve it. Some help will be highly appreciated.
Given (x-2)^4/4 + (y+2)^4/9 = 4, we can note that since the exponents are even (4) for both (x-2)^4/4 and (y+2)^4/9, they are non-negative numbers . Thus, two non-negative numbers add up 4. To get the maximum value of x, we must have the maximum value of (x-2)^4/4. This follows that we must have the minimum value of (y + 2)^4/4.
Since (y + 2)^4/4 is a non-negative number, the minimum value of (y + 2)^4/4 would be 0.
Again, (x-2)^4/4 + (y+2)^4/9 = 4
=> Maximum possible value of (x-2)^4/4 + Minimum possible value of (y+2)^4/9 = 4
=> Maximum possible value of (x-2)^4/4 + 0 = 4
Maximum possible value of (x-2)^4 = 16
x = 0 or 4
Thus, the maximum possible value of x is 4.
The correct answer:
C
Hope this helps!
-Jay
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