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Question

Let f:R(0,) and g:RR be twice differentiable functions such that f′′ and g′′ are continuous functions on R. Suppose f(2)=g(2)=0,f′′(2)0 and g(2)0. If limx2f(x)g(x)f(x)g(x)=1, then

A
f has a local minimum at x=2
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B
f has a local maximum at x=2
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C
f′′(2)>f(2)
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D
f(x)f′′(x)=0 for at least one xR
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Solution

The correct option is D f(x)f′′(x)=0 for at least one xR
f(2)=g(2)=0,f′′(2)0 and g(2)0
limx2f(x)g(x)f(x)g(x)=1

Applying L'Hospital Rule, we get
limx2f(x)g(x)+f(x)g(x)f′′(x)g(x)+f(x)g′′(x)=1f(2)f′′(2)=1f′′(2)=f(2)
f(2)=0 and the co domain of f is (0,)
f′′(2)>0
Hence, f(2) will be local minima.

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