If f(x) = log left[ frac{1 + x}{1 - x} right] | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$.We need to find $f\left( \frac{2x}{1 + x^2} \right)$.Substituting $\frac{2x}{1 + x^2}$ for

Vedclass · Accepted Answer

$2f(x)$

Vedclass · Answer

(C) Given $f(x) = \log \left( \frac{1 + x}{1 - x} \right)$.We need to find $f\left( \frac{2x}{1 + x^2} \right)$.Substituting $\frac{2x}{1 + x^2}$ for $x$ in the function $f(x)$:$f\left( \frac{2x}{1 + x^2} \right) = \log \left[ \frac{1 + \frac{2x}{1 + x^2}}{1 - \frac{2x}{1 + x^2}} \right]$Simplify the expression inside the logarithm:$= \log \left[ \frac{\frac{1 + x^2 + 2x}{1 + x^2}}{\frac{1 + x^2 - 2x}{1 + x^2}} \right]$$= \log \left[ \frac{1 + x^2 + 2x}{1 + x^2 - 2x} \right]$Recognizing the squares:$= \log \left[ \frac{(1 + x)^2}{(1 - x)^2} \right]$$= \log \left[ \frac{1 + x}{1 - x} \right]^2$Using the property $\log(a^n) = n \log a$:$= 2 \log \left[ \frac{1 + x}{1 - x} \right]$$= 2f(x)$.

If $f(x) = \log \left[ \frac{1 + x}{1 - x} \right]$,then $f\left[ \frac{2x}{1 + x^2} \right]$ is equal to

Similar Questions

The function $f(x) = |px - q| + r|x|$,$x \in (-\infty, \infty)$,where $p > 0, q > 0, r > 0$ assumes its minimum value only at one point,if

Let $f(x) = [x]^2 - [x+3] - 3, x \in \mathbb{R}$, where $[\bullet]$ is the greatest integer function. Then:

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

If $f(x) = \frac{x - |x|}{|x|}$,then $f(-1) = $

If $f(x) = \cos(\log x)$,then the value of $f(x^2) \cdot f(y^2) - \frac{1}{2} \left[ f\left(\frac{x^2}{y^2}\right) + f(x^2 y^2) \right]$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module