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If $y = x - \log (1 + x)$ , then minimum value of $y$ is

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Correct option is B

y'(x)=1-11+x=x1+x The domain y is x>-1. so y'(x)=0⇒x=0y''(x)=1(1+x)2>0. Hence y has min at x = 0 and ymin= 0-log1=0

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If y=x-log(1+x), then minimum value of y is

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If $y = x - \log (1 + x)$ , then minimum value of $y$ is