Question

Given function $f(x)=\left(\frac{e^{2 x}-1}{e^{2 x}+1}\right)$ is.

• A. increasing
• B. decreasing
• C. even
• D. none of these

Answer 1

Answer: (A)

increasing

Answer 2

$f(x)=\frac{e^{2 x}-1}{e^{2 x}+1}$
$f(x)=\frac{\left[e^{2 x}+1\right]\left[2 e^{2 x}\right]-\left[e^{2 x}-1\right]\left[2 e^{2 x}\right]}{\left[e^{2 x}+1\right]^{2}}$
$f ^{1}( x )=\frac{2 e ^{2 x }\left[ e ^{2 x }+1- e ^{2 x }+1\right]}{\left[ e ^{2 x }+1\right]^{2}}$
$f ^{1}( x )=\frac{4 e ^{2 x }}{\left[ e ^{2 x }+1\right]^{2}}$
Now, $\left[ e ^{2 x }+1\right]^{2}>0$ and $e ^{2 x }>0$
$\therefore f ^{1}( x )>0$
$\therefore f(x)$ is increasing function.