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B
f′(0−)=1
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C
f′(0+)=0
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D
f′(0+)=1
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Solution
The correct options are Bf′(0−)=1 Cf′(0+)=0 We know that, f′(0−)=limx→0−f(x)−f(0)x⇒f′(0−)=limh→0+f(−h)−f(0)−h⇒f′(0−)=limh→0+−h1+e−1/h−h⇒f′(0−)=limh→0+11+e−1/h=1 And f′(0+)=limx→0+f(x)−f(0)x⇒f′(0+)=limh→0+f(h)−f(0)h⇒f′(0+)=limh→0+h1+e1/hh⇒f′(0+)=limh→0+11+e1/h=0