# Let 'f' be an even periodic function with period '4' such that f(x) = 2^x-1, 0<=x<=2. The number of solutions of the equation f(x) = 1 in [-10,20] are?

May 13, 2018

The number of solutions is $15$.

#### Explanation:

We know that, by the definition of a periodic function, if $f$ is periodic with period $\rho$, then

$f \left(x\right) = f \left(x + n \rho\right)$ for any integer $n$

Now, in order to find all the solutions of $f \left(x\right) = 1$ for our function, let us apply the same property:

$f \left(x + 4 n\right) = 1$, $\forall n \in \mathbb{Z}$.

Let $n = 0$.

$f \left(x\right) = 1$

We only have a formula for $f \left(x\right)$ within the interval $\left[0 , 2\right]$. Does there exist a value of $x$ in this interval such that $f \left(x\right) = 1$?

Yes. We see that $x = 1$ is a solution.

Thus, we must have

$f \left(1 + 4 n\right) = 1$, $\forall n \in \mathbb{Z}$.

In order to find all the solutions of this equation in the $\left[- 10 , 20\right]$, we must find all the values of $n$ such that

$- 10 \le 1 + 4 n \le 20$

$\therefore - 11 \le 4 n \le 19$

$\therefore - \frac{11}{4} \le n \le \frac{19}{4} \implies - 2.75 \le n \le 4.75$

Since $n$ is an integer, we have the following:

$n \in \left\{- 2 , - 1 , 0 , 1 , 2 , 3 , 4\right\}$

And so, some solutions to the equation $f \left(x\right) = 1$ in $\left[- 10 , 20\right]$ are

$x \in \left\{- 7 , - 3 , 1 , 5 , 9 , 13 , 17\right\}$

The rest of the solutions

Note that the function $f \left(x\right)$ is defined to be ${2}^{x} - 1$ in the interval $\left[0 , 2\right]$ - which is not one complete period. It is defined in the interval $\left[- 2 , 0\right)$ by the fact that it is an even function. Since this means that $f \left(- x\right) = f \left(x\right)$, we have

$f \left(- 1\right) = f \left(1\right) = 1$

as well. Thus $x = - 1$ is a solution.
Using the periodicity of the function we see that in addition to those found above, the following are solutions to the equation in the interval $\left[- 10 , 20\right]$ too :

$\left\{- 9 , - 5 , - 1 , 3 , 7 , 11 , 15 , 19\right\}$

So, the complete set of solutions to $f \left(x\right) = 1$ in the interval $\left[10 , 20\right]$ are

$\left\{- 9 , - 7 , - 5 , - 3 , - 1 , 1 , 3 , 5 , 7 , 9 , 11 , 13 , 15 , 17 , 19\right\}$