$f (x) = l n (l n x) (x > 1)$ is increasing in

(0, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(1, \infty)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(0, 2)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- \infty, 1)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $(1, \infty)$
$f^{'} (x) = \frac{1}{x (l n x)}$

$f^{'} (x) > 0$ If $x > 1$
$f (x)$ is increasing in $(1, \infty)$

Suggest Corrections

Similar questions

List -I	List -II
(A) $f (x) = x^{2} - 2 x + 5$ is increasing in	1) $ϕ$
(B) $f (x) = e^{- x}$ is increasing in	2) $(- \infty, 1) \cup (2, \infty)$
(C) $f (x) = {log}_{e} x$ is increasing in	3) R
(D) $f (x) = \frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 2 x + 5$ is increasing in	4) $(0, \infty)$
	5) $(1, \infty)$

F (x)

\frac{{cos}^{3} x}{{sin}^{2} x + sin x} . x \neq n π

F (x)

1 - 1

(0, π / 2] .

log x

- \infty

0

(0, 1]

sin x

0

1

(0, π / 2]

	List 1		List 2
A.	The set $R e [\frac{2 z i}{1 - z^{2}}]$ : z is a complex number, \|z\|=1, $z \neq \pm 1$ } is	1.	$(- \infty, - 1) \cup (1, \infty)$
B.	The domain of the function $f (x) = {sin}^{- 1} [\frac{8 (3)^{x - 2}}{1 - 3^{2 (x - 1)}}]$ is	2	$(- \infty, 0) \cup (0, \infty)$
C.	If $f (θ) = ∣ ∣ ∣ ∣ \begin{matrix} 1 & tan θ & 1 - tan θ & 1 & tan θ - 1 & - tan θ & 1 \end{matrix} ∣ ∣ ∣ ∣$ then the set ${f (θ) : 0 \leq θ < \frac{π}{2}}$	3.	$[2, \infty)$
D.	lf $f (x) = x^{\frac{3}{2}} (3 x - 10)$ , $x \geq 0$ , then $f (x)$ is increasing in	4.	$(- \infty, - 1] \cup [1, \infty)$
		5.	$(- \infty, 0] \cup [2, \infty)$