Question

Which of the following functions are even?

• A. $\log \left(\frac{1+x^2}{1-x^2}\right)$
• B. ${\sin }^{2}x-{\cos }^{2}x$
• C. $\log \left(\frac{1+x^3}{1-x^3}\right)$
• D. $\frac{(1+2^x{)}^2}{2^x}$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $\log \left(\frac{1+x^2}{1-x^2}\right)$
B ${\sin }^{2}x-{\cos }^{2}x$
D $\frac{(1+2^x{)}^2}{2^x}$

A. $f\left(x\right)=\log \left(\frac{1+x^2}{1-x^2}\right)$
$\therefore f(-x)=\log \left(\frac{1+(-x{)}^2}{1-(-x{)}^2}\right)=\log \left(\frac{1+x^2}{1-x^2}\right)=f\left(x\right)\Rightarrow f\left(x\right)$ is even.
B. $f\left(x\right)={\sin }^{2}x-{\cos }^{2}x$
$\therefore f(-x)=\left(\sin \right(-x){)}^2-(\cos (-x){)}^2={\sin }^{2}x-{\cos }^{2}x=f\left(x\right)\Rightarrow f\left(x\right)$ is even.
C. $f\left(x\right)=\log \left(\frac{1+x^3}{1-x^3}\right)$
$\therefore f(-x)=\log \left(\frac{1+(-x{)}^3}{1-(-x{)}^3}\right)=\log \left(\frac{1-x^3}{1+x^3}\right)=-\log \left(\frac{1+x^3}{1-x^3}\right)=-f\left(x\right)\Rightarrow f\left(x\right)$ is odd.
D. $f\left(x\right)=\frac{(1+2^x{)}^2}{2^x}$
$\therefore f(-x)=\frac{(1+2^{-x}{)}^2}{2^{-x}}=\frac{(1+2^x{)}^2}{2^{2x}{2}^{-x}}=\frac{(1+2^x{)}^2}{2^x}=f\left(x\right)\Rightarrow f\left(x\right)$ is even.