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 $x=0$. ■

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