The domain of f(x) is (0, 1). Then the domain of $f (e^{x}) + f (l n | x |)$ is

$(- 1, e)$

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

$(1, e)$

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

$(- e, - 1)$

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

$(- e, 1)$

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

Open in App

Solution

The correct option is C
$(- e, - 1)$

To find the domain of $f (e^{x}) + f (l n | x |)$ .
Given that the domain of f(x) is (0, 1)
$\Rightarrow 0 < e^{x} < 1 a n d 0 < l n | x | < 1 0 < e^{x} < 1 \Rightarrow x ϵ (- \infty, 0) . . . . . (i) 0 < l n | x | < 1 \Rightarrow 1 < | x | < e \Rightarrow x ϵ (- e, - 1) \cup (1, e) . . . . (i i)$

Combining (i) and (ii), the domain of $f (e^{x}) + f (l n | x |)$ is (-e,-1)

Suggest Corrections

Similar questions

The domain of f(x) is (0, 1). Then the domain of $f (e^{x}) + f (l n | x |)$ is

f (x)

(0, 1)

f (e^{x}) + f (ln | x |)

f (e^{x}) + f (l n | x |)

f (x)

[0, 1]

y = f (e^{x}) + f (| [x] |)

f (x)

(- 1, 1)

g (x) = f (e^{x}) + f ({log}_{e} | x |))

Join BYJU'S Learning Program