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Question

Let f(x) be a twice differentiable function and has no critical points. Let g(x)=(x+6)2009(x+1)2010(x+2)2011(x−3)2012(x−4)2013(x−5)2014 be such that f(x)+g(x)f′(x)+f′′(x)=0. Then the function h(x)=f2(x)+(f′(x))2

A
is monotonically increasing in (2,4).
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B
has exactly three points of inflection.
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C
has exactly two points of local maxima.
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D
has a negative point of local minimum.
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Solution

The correct options are
A is monotonically increasing in (2,4).
B has exactly three points of inflection.
C has exactly two points of local maxima.
D has a negative point of local minimum.
h(x)=f2(x)+(f(x))2
h(x)=2f(x)f(x)+2f(x)f′′(x)
=2f(x)[f(x)+f′′(x)]
=2f(x)[g(x)f(x)]
=2g(x)(f(x))2

g(x)=(x+6)2009(x+1)2010(x+2)2011(x3)2012(x4)2013(x5)2014

Sign of h(x)



h(x)>0 in (2,4)
h(x) is monotonically increasing in (2,4)
x=1,3 and 5 are points of inflection.
x=6 and x=4 are points of local maxima.
x=2 is only the point of local minimum.

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