Continuity and DifferentiabilityHard

Question

If f(a) = 2, f′(a) = 1, g(a) = -1, g′(a) = 2, then the value of is

Options

A.-5
B.
C.5
D.None of these

Solution

Given, f(a) = 2, f′(a) = 1, g(a) = - 1, g′(a) = 2
∴    
     
            (using L′Hospital′s rule)
      = g′(a) f(a) - g(a) f′(a)
      = 2(2) - (-1)(1) = 5

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