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Question

Let f:(,+1]R, g:[1,)R be such that f(x)=1x and g(x)=1+x, then f(x)+1g(x) exist if x

A
[1,1]
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B
(,1)
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C
(1,)
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D
(1,1]
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Solution

The correct option is D (1,1]
f(x)+1g(x)=1x+11+x

Now, 1x0,1+x>0
x1, x>1x(1,1]

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