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).
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
has exactly three points of inflection.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
has exactly two points of local maxima.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
has a negative point of local minimum.
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution
The correct option is 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
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.