Question

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?

Answer 1

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(x)=f(x+n rho)` for any integer `n`

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

`f(x+4n) = 1`, `forall n in ZZ`.

Let `n = 0`.

`f(x) = 1`

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

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

Thus, we must have

`f(1+4n) = 1`, `forall n in ZZ`.

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

`-10<= 1+4n <= 20`

`:.-11 <= 4n <= 19`

`:.-11/4 <= n <= 19/4 =>-2.75<=n<=4.75`

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

`n in {-2, -1, 0, 1, 2, 3, 4}`

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

`x in {-7,-3,1,5,9,13,17}`

The rest of the solutions

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

`f(-1) = f(1) = 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 `[-10,20]` too :

`{-9,-5,-1,3,7,11,15,19}`

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

`{-9,-7,-5,-3,-1,1,3,5,7,9,11,13,15,17,19}`