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Question

# Let f(x)=ex1+x2 and g(x)=f′(x) then

A
g(x) has four points of local extremum
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B
g(x) has two points of local extremum
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C
g(x) has a point of local minimum at x=1
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D
g(x) has a point of local maximum at some x(1,0)
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Solution

## The correct options are B g(x) has two points of local extremum C g(x) has a point of local minimum at x=1 D g(x) has a point of local maximum at some x∈(−1,0)f′(x)=g(x)=(x−1)2ex(1+x2)2g′(x)=(x−1)(x3−3x2+5x+1)ex(1+x2)3 let h(x)=x3−3x2+5x+1 h′(x)=3x2−6x+5, D<0, ∀ x∈R So h′(x)>0, ∀ x ∈ R So, h(x) has only one real root. Now g′(−1)g′(0)<0 so root →(−1,0) g(x) has two points of extrema maxima at x∈(−1,0) minima at x=1

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