Question

If \( (x-1)^{2018}(x-2)^{2019} \) is the derivative of \( f(x) \) and \( f(x)=\log _{e} g(x) \) then at \( x=1, g(x) \) has local maximum at \( x=1, g(x) \) has a local minimum at \( x=2, g(x) \) has neither maximum nor minimum at \( x=2, g(x) \) has minimum

Answer 1

Given that f¹ (x) = (x-1)²⁰¹⁸ (x-2)²⁰¹⁹ 

And f (x) = log_e g(x)

Then f¹ (x) = \(\frac {1}{g(x)}f^1(x)\)

= g¹(x) = g(x) f¹ (x)

= e^f(x) f¹(x) (∵ f(x) = log_e g(x) = g(x) = e^7(x))

= e^f(x) (x-1)²⁰¹⁸ (x-2)²⁰¹⁹

g¹(x) = 0 gives x = 1 or x =2

∵ g¹(x) = e^f(x) (x-1)²⁰¹⁸ (x-2)²⁰¹⁹

e^f(x) >0

g¹(x) does not change sign passing through x = 1

∴ x = 1 is a point of reflection

Hence, at x = 1, g(x) has neither maximum nor minimum 

∴ g¹(x) change sign -ve to +ve while passing through x = 2

∴ At x = 2, g(x) has minimum.

Hence, at x = 2, g(x) has local minimum and x = 1, g(x) has neither minimum nor maximum.