Question

f(x) = 2x/log x is increasing in

A. (0, 1)

B. (1, e)

C. (e, ∞)

D. (-∞, e)

Answer 1

Answer is : C. (e, ∞)

⇒ f(x) = \(\frac{2x}{log\,x}\)

⇒ f'(x) = \(\frac{2.log\,x-2}{log^2x}\)

Put f’(x) = 0

We get

⇒ \(\frac{2.log\,x-2}{log^2x}\) = 0

⇒ 2.log x = 2

log x = 1

⇒ x = e

We only have one critical point

So, we can directly say x > e f(x) would be increasing

∴ f(x) will be increasing in (e, ∞)