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tejas sanjay akole Grade: 12
        

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

8 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

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



8 years ago
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