If $f\left( x \right) = \frac{{x - 1}}{{x + 1}},$ then $f\left( {2\;x} \right)$ is

$\frac{{f\left( x \right) + 1}}{{f\left( x \right) + 3}}$

$\frac{{f\left( x \right) + 3}}{{f\left( x \right) + 1}}$

$\frac{{f\left( x \right) + 3}}{{3f\left( x \right) + 1}}$

JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium

1 of 1

Application +4 | -1

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

then

f\left( {2\;x} \right)

View answer and explanation

Correct option: B

\frac{{3f\left( x \right) + 1}}{{f\left( x \right) + 3}}

Option B is correct. By applying componendo-dividendo and substituting the expression for x, we can express f(2x) in terms of f(x).