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Question

# Let f(x) be a twice differentiable function and has no critical point and 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 function h(x)=f2(x)+(f′(x))2

A
is monotonic increasing in (2,4)
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B
has exactly 3 point of inflection
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C
has exactly two points local maxima
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D
has a negative point of local minima.
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Solution

## The correct options are A is monotonic increasing in (–2,4) B has exactly 3 point of inflection C has exactly two points local maxima D has a negative point of local minima.Given, h(x)=f2(x)+(f′(x))2 differentiation with respect to x ⇒h′(x)=2f(x)f′(x)+2f′(x)f′′(x) ⇒h′(x)=2f′(x)(f(x)+f′′(x)) ⇒h′(x)=2f′(x)(−g(x)f′(x)) ⇒h′(x)=−2g(x)(f′(x))2 ⇒h′(x)=−2(f′(x))2(x+6)2009(x+1)2010(x+2)2011(x–3)2012(x–4)2013(x–5)2014 Sign of h′(x) →h′(x)>0 in (−2,4)⇒h(x) is increasing. →x=−1,3 and 5 are points of inflection. →x=−6 and x=4 are points of local minima. →x=−2 is the only point of local minima.

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