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JEE Advanced2016MathematicsLimitsActual

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

Options

  1. Af has a local minimum at x = 2
  2. Bf has a local maximum at x = 2
  3. Cf ″ 2 > f 2
  4. Df x - f ″ x = 0 for at least one x ∈ R

Correct answer

A. f has a local minimum at x = 2

Step-by-step solution

Using L'Hospital Rule ( a s i t i s 0 0 f o r m ) 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 A l s o f ′ ( 2 ) = 0 and f ″ ( 2 ) > 0 ∴ x = 2 is local minima.

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