Determine all possible value(s) of a real constant C such that the following system of equations has precisely one real solution in (X, Y).

2^|X| + |X| = Y + X² + C

and, X² + Y² = 1

Note: |X| denotes the absolute value of X.

2^|x|+|x|=y+x^2+c
y=2^|x|+|x|+x^2+c      (1)
x^2+y^2=1              (2)
let f(x)=2^|x|+|x|-x^2

now since x^2+y^2=1 both x and y are restricted to [-1,1]

now examining the graph of f(x) with x=-1 to x=1
it is easy to see that f(x) has a turning point at x=0
f(0)=1

since we want (1) and (2) to have a single point of 
intersection then clearly the turning point at x=0 for
f(x) needs to touch the circle (2).  Thus c=0,  if c>0
then f(x) would be completely above the circle, if c<0
then because of symmetry there would be 2 solutions or 0
solutions.


thus c=0 is the only value that gives a single solution
for (x,y)

Posted by Daniel on 2008-10-17 13:45:18