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Question

# Let f(x) be defined for all x ϵ R 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(x1x)+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 D f(x)+f(1x)=f(x−1x)+2Substitute y=x in f(x+y)−f(x−y)=4xyf(2x)−f(0)=4x2However it is given that f(0)=0Hence,f(2x)=4x2f(2x)=(2x)2Thereforef(x)=x2The minimum value of f(x) is 0.Which also satisfies the equation f(x)+1f(x)=f(x−1x)+2Hence, options 'A' and 'C' are correct.

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