If fx=log_e(1-x)/(1+x), x< 1, then f2x/(1+x^2) is equal to

Correct answer: (D) 2fx. fx=ln(1-x)/(1+x) ⇒f2x/(1+x^2)=ln(1-2x)/(1+x^2)1+2x/(1+x^2) =ln(1+x^2-2x)/(1+x^2+2x) =ln1-x1+x^2 =2ln(1-x)/(1+x) =2ln(1-x)/(1+x) ∵x<1 =2fx

Q36 / 90 · JEE Main 2019 (08 Apr Shift 1) · Mathematics

If $f (x) = \log_{e} (\frac{1 - x}{1 + x}), |x| < 1,$ then $f (\frac{2 x}{1 + x^{2}})$ is equal to

A
${(f (x))}^{2}$
B
$2 f {(x^{2})}^{}$
C
$- 2 f (x)$
Correct answer
D
$2 f (x)$

Solution

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

$\Rightarrow f (\frac{2 x}{1 + x^{2}}) = l n (\frac{1 - \frac{2 x}{1 + x^{2}}}{1 + \frac{2 x}{1 + x^{2}}})$

$= l n (\frac{1 + x^{2} - 2 x}{1 + x^{2} + 2 x})$

$= l n ({(\frac{1 - x}{1 + x})}^{2})$

$= 2 l n |\frac{1 - x}{1 + x}|$

$= 2 l n (\frac{1 - x}{1 + x}) (∵ |x| < 1)$

$= 2 f (x)$

Hence, the correct answer is option: 'D'.

If fx=loge1-x1+x, x< 1, then f2x1+x2 is equal to

Solution

Relevant Questions

If $f (x) = \log_{e} (\frac{1 - x}{1 + x}), |x| < 1,$ then $f (\frac{2 x}{1 + x^{2}})$ is equal to