Question

If $f\left(x\right)$ is a real valued function discontinuous at all integral points lying in $[0,n]$ and if $\left(f\right(x){)}^2=1\forall x\in [0,n],$ then number of functions $f\left(x\right)$ are

• A. $2^{n+1}$
• B. $6\times 3^n$
• C. $2\times 3^{n-1}$
• D. $3^{n+1}$

Answer 1

The correct option is C $2\times 3^{n-1}$

There are four possible functions defined in $0\le x\le 1$,
Of them $2$ are continuous and two are discontinuous,
Now for each of the points $(1,2,...,n-1),$
Keep functions fixed from left of the point,
So there are $4$ possible functions defined in between next two consecutive integral points of them only one is continuous and at last for $x=n$,
There is only one possibility of discontinuity of the function.
So total number of functions
$=2\times 3^{n-1}\times 1$