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Question

Let f(x)=(1+x)n(1+nx), x[1,). Then f

A
has an absolute maximum at x=0
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B
has neither absolute maximum nor absolute minimum at x=0
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C
has an absolute minimum at x=0
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D
does not have absolute minimum at x=0
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Solution

The correct option is B has an absolute minimum at x=0
f(x)=(1+x)n(1+nx)
f(x)=n[(1+x)n11].
Now for x=0,f(0)=0
and for x>0,f(x)>0.
Thus f inereases on [0,) i.e.f(x)f(0)=0
For 1x<0, 01+x<1 so (1+x)n1<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=0.

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