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Question:

Let f:R→(0,∞) and g:R→R be twice differentiable such that f'' and g'' are continuous on R. Suppose f'(2)=g(2)=0, f''(2)≠0 and g'(2)≠0. If limx→2 f(x)g(x)f'(x)g'(x)=1, then

f(x)−f''(x)=0 for at least one x∈R

f has a local minimum at x=2

f''(2)>f(2)

f has a local maximum at x=2

Solution:

limx→2 f(x)g(x)f'(x)g'(x)=1
This is an indeterminate form (0/0) since f'(2)=0, g(2)=0.
Using L.H. rule,
limx→2 [f'(x)g(x) + g'(x)f(x)]/[f''(x)g'(x) + g''(x)f'(x)] = [f'(2)g(2) + g'(2)f(2)]/[f''(2)g'(2) + g''(2)f'(2)] = g'(2)f(2)/[f''(2)g'(2)] = 1
⇒ f''(2) = f(2)
and f'(2) = 0
Since the range of f(x) ∈ (0,∞), so f''(2) = f(2) = +ve
So f(x) has a point of minima at x=2 and f(2) = f''(2)
So f(x) = f''(x) have at least one solution in x∈R.