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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]2 n+1 2]6*3 n 3]2*3 n-1 4]3 n+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
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
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