If $ f(x)=\frac{x}{x-1} $, then $ \frac{f(a)}{f(a+1)} $ is equal to:

$ \mathrm{f}\left(\frac{1}{\mathrm{a}}\right) $

$ \mathrm{f}\left[\frac{-\mathrm{a}}{\mathrm{a}-1}\right] $

JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium

1 of 1

Application +4 | -1

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

, then

\frac{f(a)}{f(a+1)}

is equal to:

View answer and explanation

Correct option: A

\mathrm{f}\left(\mathrm{a}^{2}\right)

Option A (

f(a^2)

) is correct because

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

when

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

Practice All Chapter MCQs →

f:(-\infty, 2] \longrightarrow(-\infty, 4]

f(x)=x(4-x)

f^{-1}(x)

g(f(x))=|\sin x|

f(g(x))=(\sin \sqrt{x})^2

f:\left( {2,\;3} \right) \to \left( {0,\;1} \right)

f\left( x \right) = x - \left[ x \right]

{f^{ - 1}}\left( x \right)

f: R \rightarrow R

f(x)=x^4

f(x)=(x+1)^2-1(x \geq-1)

S=\left\{x: f(x)=f^{-1}(x)\right\}