Questions
a
has an absolute maximum at x = 0
b
has neither absolute maximum nor absolute minimum at x = 0
c
has an absolute minimum at x = 0
d
does not have absolute minimum at x = 0
detailed solution
Correct option is C
f(x)=(1+x)n−(1+nx)⇒f′(x)=n(1+x)n−1−1. Now for x=0,f′(0)=0 and for x>0,f′(x)>0. Thus _f increases on [0,∞) i.e. f(x)≥f(0) for x∈[0,∞). For −1≤x<0,0≤1+x<1 so (1+x)n−1<0. So f decreases on [-1. 0) i.e. f(x)≥f(0)=0 for x∈[−1,0). Hence f has an absolute minimum at x = 0Talk to our academic expert!
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