Let f(x)=2x3+3x2+6x+12 and g'(x)=(x-1)(x-2)(x-3)2x2-1x2-9 . If n1,n2,n3denote the number of local minimum, local maximum, points of inflection respectively of the function fgx then
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a
n1+n2=4
b
n1+n3=3
c
n2+n3=3
d
n1+n3=4
answer is A.
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Detailed Solution
f′(x)=6x2+x+1>0,∀x∈Rg1(x)=(x+3)(x+1)(x−1)2(x−2)(x−3)3ddx(f(g)(x))=f'(g(x))⋅g′(x) therefore fgx has local minimum at x=-1,3 ⇒n1=2 and has local maximum at x=-3,2 ⇒n2=2 and point of inflection at x=1 ⇒n3=1