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Question

If f(x)=ln(x2+ex2+1) , then range of f(x) is

A
(0,1)
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B
(0,1]
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C
[0,1)
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D
{0,1}
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Solution

The correct option is B (0,1]
f(x)=ln(x2+ex2+1)=ln(x2+11+ex2+1)=ln(1+e1x2+1)
Clearly range is (0, 1]
Hence (B) is correct answer

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