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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

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*3n-1


4]3n+1

Grade:12

1 Answers

Badiuddin askIITians.ismu Expert
148 Points
12 years ago

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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