Question

Which of the function is not even:

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

Answer 1

The correct option is C $\log (\frac{1+x^3}{1-x^3})$

A function $f\left(x\right)$ is said to be even if $f(-x)=f\left(x\right),\forall x$.

Now for $f\left(x\right)=\log \left(\frac{1+x^2}{1-x^2}\right)$, $f(-x)=\log \left(\frac{1+x^2}{1-x^2}\right)=f\left(x\right)\forall x$.

So, $\log \left(\frac{1+x^2}{1-x^2}\right)$ is a even function.

Again for $f\left(x\right)={\sin }^{2}x+{\cos }^{2}x$, $f(-x)={\sin }^{2}x+{\cos }^{2}x=f\left(x\right)\forall x$.

So ${\sin }^{2}x+{\cos }^{2}x$ is also an even function.

Again for $f\left(x\right)=\log \left(\frac{1+x^3}{1-x^3}\right)$, $f(-x)=\log \left(\frac{1-x^3}{1+x^3}\right)=-\log \left(\frac{1+x^3}{1-x^3}\right)=-f\left(x\right)\forall x$.

So, $\log \left(\frac{1+x^3}{1-x^3}\right)$ is not an even function.

Again for $f\left(x\right)=\frac{(1+2^x{)}^2}{2^x}$, $f(-x)=\frac{(2^x+1{)}^2}{2^x}=f\left(x\right)\forall x$.

So $\frac{(1+2^x{)}^2}{2^x}$ is even function.

So option (C) is correct.