f(x) = 2x/log x is increasing inA. (0, 1)B. (1, e)C. (e, ∞)D. (-∞,

1.	f(x) = 2x/log x is increasing inA. (0, 1)B. (1, e)C. (e, ∞)D. (-∞, e)
Answer» 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, ∞)

f(x) = 2x/log x is increasing inA. (0, 1)B. (1, e)C. (e, ∞)D. (-∞, e)

Answer»

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, ∞)