Let f   : R → 0 ,   ∞ and g   : R → R be twice differentiable functions…

Let f :R→0, ∞ and g :R→R 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. lim x→2 f( x ) g( x ) f ′  ( x )  g ′ ( x ) =1, then
  1. f has a local minimum at x=2
  2. f has a local maximum at x=2
  3. f″2>f2
  4. fx-f″x= 0 for at least one x∈R
     

Solution

Using L'Hospital Rule ( as  it  is   0 0   form )
lim x→2 f ′ ( x ) g( x )+f( x )  g ′ ( x ) f ″ ( x )  g ′ ( x )+ f ′ ( x )  g ″ ( x ) =1
⇒ f ( 2 )  g ′ ( 2 ) f ″  ( 2 )  g ′ ( 2 ) =1⇒ f ″ ( 2 )=f( 2 )>0
Also  f ′ ( 2 )=0 and  f ″ ( 2 )>0
∴ x=2 is local minima.

Asked in: JEE Advanced 2016 (Paper 2)

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