if f(x) is real valued function discontinuous at all integral points lying in [0,n] and if (f(x))² =1 for x ε [0,n] then number of functions f(x) are

1]2ⁿ⁺¹

2]6*3ⁿ

3]2*3^n-1

4]3ⁿ⁺¹

You can use short cut to solve this problem

put n=1

   now our domain is [0,1]

then f(x) should be discontinuous at 0 and 1

given (f(x))² =1

         f(x) =1 or -1

 case 1

 let f(x)=1 at x=0

 then  f(x) =-1 for x ε (0,1)

and f(x) = 1  at x=1

second case

let f(x)= -1 at x=0

then  f(x) =1 for x ε (0,1)

and f(x) = -1  at x=1

so there are only 2 possibilities

noe check the option for n=1

3rd option is correct