Question

Which of the following real valued functions is / are not even functions?

• A. $f\left(x\right)=x^3\sin x$
• B. $f\left(x\right)=x^2\cos x$
• C. $f\left(x\right)=e^x{x}^3\sin x$
• D. $f\left(x\right)=x-\left[x\right]$, where [x] denotes the greatest integer less than or equal to $x$

Answer 1

Answer: (C), (D)

$f\left(x\right)=x-\left[x\right]$, where [x] denotes the greatest integer less than or equal to $x$
$f\left(x\right)=e^x{x}^3\sin x$

Option A: $f\left(x\right)=x^3\sin x$

$f(-x)=(-x{)}^3\sin (-x)=-x^3(-\sin x)=x^3\sin x=f(x)$

As $f(-x)=f\left(x\right)$, the function is an even function.

Option B: $f\left(x\right)=x^2\cos x$

$f(-x)=(-x{)}^2\cos (-x)=x^2\cos x=f(x)$

As $f(-x)=f\left(x\right)$, the function is an even function.

Option C: $f\left(x\right)=e^x{x}^3\sin x$

Now $x^3\sin x$ is an even function.

Consider, $e^x$.

$e^{-x}=\frac{1}{e^x}$

Hence $f(-x)=e^{-x}{x}^3\sin x$. This is not equal to $f\left(x\right)$ or $-f\left(x\right)$.

Hence $f\left(x\right)$ is neither odd nor even.

Option D : $f\left(x\right)=x-\left[x\right]$

$f(-x)=(-x)-[-x]=-x-(-[x]-1)=-x+\left[x\right]+1$. This is not equal to $f\left(x\right)$ or $-f\left(x\right)$. Hence the function is neither odd nor even.