Problem:
The graph of y=ex+1+e−x−2 has an axis of symmetry. What is the reflection of the point (−1,21) over this axis?
Answer Choices:
A. (−1,−23)
B. (−1,0)
C. (−1,21)
D. (0,21)
E. (3,21)
Solution:
Let f(x)=ex+1+e−x−2. Because f(x) approaches infinity as ∣x∣ increases without bound, the only possible axis of symmetry is a vertical line. If the axis of symmetry has equation x= c, then f(x)=f(2c−x) for every real x, which is equivalent to e⋅ex+e−x=e2c+1e−x+e−2cex. Multiplying through by ex and simplifying gives (e−e−2c)e2x=e2c+1−1. Because this equation holds for all x, it follows that e−e−2c=0 and e2c+1−1=0. Thus c=−21. The reflection of (−1,21) with respect to the vertical line x=−21 is (D)(0,21).
The problems on this page are the property of the MAA's American Mathematics Competitions