Iffx-log1-x-1-x- then fx isA- f x 1 f x 2- f x 1- x 2-B- fx1-fx2-fx1-X2C- Odd functionD- Even function

The correct option is C Odd function Here, f(x)=log (1+x/1 x) and f( x)=log(1 x/1+x)=log(1+x/1 x)^ 1 = log(1 x/1+x)= f(x) ⇒ f(x) is an odd function. ...

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

Byju's Answer

Standard XII

Mathematics

Property 1

Iffx=log1+x/1...

Question

If $f (x) = l o g \frac{1 + x}{1 - x}$ , then f(x) is

Even function

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

f(x₁ f(x₂ ) = f(x₁ + x₂)

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

Odd function

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

Open in App

Solution

The correct option is D Odd function
Here, $f (x) = l o g (\frac{1 + x}{1 - x})$

and $f (- x) = l o g (\frac{1 - x}{1 + x}) = l o g {(\frac{1 + x}{1 - x})}^{- 1}$

$= - l o g (\frac{1 - x}{1 + x}) = - f (x) \Rightarrow f (x)$ is an odd function.

Suggest Corrections

Similar questions

Q. If

f (x_{1}) - f (x_{2}) = f (\frac{x_{1} - x_{2}}{1 - x_{1} x_{2}})

for

x_{1}, x_{2} ϵ

(-1, 1), then f(x) is

Q. If

f (x_{1}) - f (x_{2}) = f (\frac{x_{1} - x_{2}}{1 - x_{1} x_{2}})

for

x_{1}, x_{2} \in [- 1, 1]

then

f (x)

Q. If

f (x_{1}) - f (x_{2}) = f (\frac{x_{1} - x_{2}}{1 - x_{1} x_{2}})

for

x_{1}, x_{2} ϵ

(-1, 1), then f(x) is

Q. I: lf

f (x) = cosh x + sinh x

then

f (x_{1} + x_{2} + \dots . . x_{n}) = f (x_{1}) . f (x_{2}) \dots . f (x_{n})

II.

{tanh}^{- 1} (x) = \frac{1}{2} log (\frac{1 + x}{1 - x})

for

| x | > 1

Q. What is the condition for a function y = f(x) to be a monotonically increasing function

Join BYJU'S Learning Program

Iff(x)=log1+x1−x, then f(x) is

The correct option is D Odd function Here, f(x)=log(1+x1−x) and f(−x)=log(1−x1+x)=log(1+x1−x)−1 =−log(1−x1+x)=−f(x)⇒f(x) is an odd function.

If $f (x) = l o g \frac{1 + x}{1 - x}$ , then f(x) is

The correct option is D Odd function
Here, $f (x) = l o g (\frac{1 + x}{1 - x})$

and $f (- x) = l o g (\frac{1 - x}{1 + x}) = l o g {(\frac{1 + x}{1 - x})}^{- 1}$

$= - l o g (\frac{1 - x}{1 + x}) = - f (x) \Rightarrow f (x)$ is an odd function.