Question

Let $f\left(x\right)=1-(1-x{)}^2,x\le 1$. The number of solutions of the equation $f\left\{f\right(x\left)\right\}=x$ is

• A. one
• B. two
• C. zero
• D. none of these

Answer 1

Answer: (B)

two

Since$\left(x\right)=1-\left(1-x^2\right)=x\left(2-x\right)$, we have $f\left(f\left(x\right)\right)=x\left(2-x\right)\left(2-x\left(2-x\right)\right)$.
Hence $x=0$ is one of the solution for $f\left(f\left(x\right)\right)=x$.
Now if $x\ne 0$, then $f\left(f\left(x\right)\right)=x$ can be written as
$\left(2-x\right)\left(2-x\left(2-x\right)\right)=1$
Simplifying we get, $x^3-4x^2+6x-3=0$.
Note that $x=1$ is solution of this equation and the other two roots are complex numbers.
Hence the for the given function, $f\left(f\left(x\right)\right)=x$ has two real roots for $x\le 1$