Continuity and DifferentiabilityHard
Question
If f(a) = 2, f′(a) = 1, g(a) = -1, g′(a) = 2, then the value of 
is
isOptions
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
∴
(using L′Hospital′s rule)
= g′(a) f(a) - g(a) f′(a)
= 2(2) - (-1)(1) = 5
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