2020 MEXT Japanese Government Scholarship Undergraduate Students Natural Sciences Qualifying Examination Mathematics (B): Problem 1(2)
Problem 1(2)
Let f(x) = 1 + \frac{1}{x-1}(x\neq 1). The solution of the equation f\left(f\left(x\right)\right) = f\left(x\right) is x = \fbox{A }, \fbox{B }.
Solution
The right-hand side of the equation is f(x), which is already given in the problem to be f\left(x\right) = 1 + \frac{1}{x-1}.
The left-hand side of the equation is \begin{align} f\left(f\left(x\right)\right) &= 1 + \frac{1}{f(x) - 1}\\ &= 1 + \frac{1}{\left(1+\frac{1}{x-1}\right) - 1}\\ &= 1 + \frac{1}{\frac{1}{x-1}}\\ &= 1 + {x-1}\\ &= x. \end{align}
Equating the left-hand side and right-hand side of the equation, we obtain \begin{align} x &= 1 + \frac{1}{x-1}\\ x-1 &= \frac{1}{x-1}\\ (x-1)^2 &= 1\\ x-1 &= \pm 1\\ x &= 1\pm 1 = 2,0. \end{align}
We check our solutions. For x=2, the right-hand side of the equation becomes f(2) = 2 and the left-hand side becomes f\left(f\left(2\right)\right) = f(2) =2. Because both sides are equal (both are equal to 2), x=2 is indeed a solution.
For x=0, the right-hand side of the equation becomes f(0) = 0 and the left-hand side becomes f\left(f\left(0\right)\right) = f(0) =0. Because both sides are equal (both are equal to 0), x=0 is indeed a solution. Thus, the two solutions are x=2 and x=0. ā
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