If x-1-x-1 and gx-x-1-x-1- then the domain of f - gx is

The correct option is C {x∈ R|x≠ 1} nbsp;g(x) is not defined at x= 1 So, (f ∘ g)(x) is also not defined at x= 1 Also, f(x) is not defined at x=1 So, we find ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Definition of Functions

If x=1/x-1 an...

Question

If $f (x) = \frac{1}{x - 1}$ and $g (x) = \frac{x - 1}{x + 1}$ , then the domain of $(f \circ g) (x)$ is

{x \in R | x \neq - 1}

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

{x \in R | x \neq 1}

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

{x \in R | x \neq 1, - 1}

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

{x \in R | x \neq 0, - 1, 1}

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

Open in App

Solution

The correct option is A ${x \in R | x \neq - 1}$
$g (x)$ is not defined at $x = - 1$
So, $(f \circ g) (x)$ is also not defined at $x = - 1$

Also, $f (x)$ is not defined at $x = 1$
So, we find $x$ for which $g (x) = 1$
$\Rightarrow \frac{x - 1}{x + 1} = 1$ which has no solution.
Therefore, domain of $(f \circ g) (x)$ is $R - {- 1}$

Suggest Corrections

Similar questions

Q. If f(x) = x + x + tan x, and f is inverse of g, then g'(x) equal to

Q. If

f (x) = \frac{x^{2} - 1}{x}

and

g (x) = \frac{x + 2}{x - 3},

then the domain of

\frac{f (x)}{g (x)}

Q. If

f (x) = \frac{1}{2} - tan (\frac{π x}{2}) - 1 < x < 1

and

g (x) = \sqrt{(3 + 4 x - 4 x^{2})}

. Find domain of

(f + g) (x)

Q. If

f (x) = \frac{x}{\sqrt{1 - x^{2}}}, g (x) = \frac{x}{\sqrt{1 + x^{2}}}

then

(f \circ g) (x) =

Q. If

f (x) = \frac{x + 1}{x - 1}

and

g (x) = 2 x - 1, f [g (x)] =

Join BYJU'S Learning Program

If f(x)=1x−1 and g(x)=x−1x+1, then the domain of (f∘g)(x) is

The correct option is A {x∈R|x≠−1} g(x) is not defined at x=−1 So, (f∘g)(x) is also not defined at x=−1 Also, f(x) is not defined at x=1 So, we find x for which g(x)=1 ⇒x−1x+1=1 which has no solution. Therefore, domain of (f∘g)(x) is R−{−1}

If $f (x) = \frac{1}{x - 1}$ and $g (x) = \frac{x - 1}{x + 1}$ , then the domain of $(f \circ g) (x)$ is