Let f(x) be defined for all xϵR and continuous. Let f(x+y)−f(x−y)=4xy for all x,y∈R and f(0)=0, then
A
The minimum value of f(x) is 0
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B
f(x)=f(1x)=f(x+1x)+2
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C
f(x)+f(1x)=f(x−1x)+2
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D
none of these
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Solution
The correct options are A The minimum value of f(x) is 0 Df(x)+f(1x)=f(x−1x)+2 Substitute y=x in f(x+y)−f(x−y)=4xy f(2x)−f(0)=4x2 However it is given that f(0)=0 Hence,f(2x)=4x2 f(2x)=(2x)2 Therefore f(x)=x2 The minimum value of f(x) is 0. Which also satisfies the equation f(x)+1f(x)=f(x−1x)+2 Hence, options 'A' and 'C' are correct.