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If f(x) is a real valued function discontinuous at all integral points lying on [0,n] and if $(f(x))^2$=1 for all x in [0,n], then number of functions f(x) are:?

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According to me, only these two cases are possible since the question says that f(x) can only take the value 1or -1 both at integral and non integral points. But the answer given is $(2/3).3^n$.

What am I missing here??

3 Answers 3

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For any integer, you have the following options:

  1. Make the function have the same value on both adjacent intervals, and a different one at the point.

  2. Make the function have different values on both adjacent intervals, and the point having the same one as the interval to its left.

  3. Make the function have different values on both adjacent intervals, and the point having the same one as the interval to its right.

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    what do you mean by adjacent interval, the interval from one integer to next integer? – 2017-02-25
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    Can you please draw any one case that I have not drawn? – 2017-02-25
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    For example, just modify your first case such that the function takes the value 1 rather than -1 at 1. – 2017-02-25
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A function could have the same value from about an integral point.

For example, $f(x) = 1$ for $x \in (n-1,n) \cup (n,n+1)$, but $f(n) = -1$

EDIT : Consider $f(x) = a$, for $x \in (n-1,n)$ where $a$ equals either 1 or -1

Then there are three possible cases for discontinuity -

1)$f(n) = a$, $f(x) = -a$ for $x\in (n,n+1)$

2)$f(n) = -a$, $f(x) = a$ for $x\in (n,n+1)$

3)$f(n) = -a$, $f(x) = -a$ for $x\in (n,n+1)$

A caveat; for $f(x), x\in(0,1)$, you have two choices only. Thus total possible cases equals $2.3^{n-1} = \dfrac 23.3^n$

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    This way I am getting the answer as $2.2^n$ – 2017-02-26
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    My point was to give an example of other possible cases. I'll edit my answer to provide the full solution. – 2017-02-26
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It is tacitly assumed that $f$ should be continuous (hence constant) on the open intervals between integers.

You have two choices for $f(0)$, and at each of the $n-1$ interior integer points you can choose between a left jump, a right jump, or two jumps. It follows that there are $2\cdot 3^{n-1}$ admissible functions in all.