Question

Let $f\left(x\right)$ $=\{\begin{array}{ccc}0& x=0& \\ x^2\sin \left(\frac{\pi }{x}\right)& -1<x<1(x\ne 0)& \\ x|x|& -1>x>1& \end{array}$

• A. $f\left(x\right)$ is an odd function
• B. $f\left(x\right)$ is an even function
• C. $f\left(x\right)$ is neither odd nor even
• D. $f^{\prime }\left(x\right)$ is an even function

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is an odd function
$f^{\prime }\left(x\right)$ is an even function

$f\left(x\right)$ $=\{\begin{array}{ccc}0& x=0& \\ x^2\sin \left(\frac{\pi }{x}\right)& -1<x<1(x\ne 0)& \\ x|x|& -1>x>1& \end{array}$
$f(-x)$ $=\{\begin{array}{ccc}0& -x=0& \\ (-x{)}^2\sin (-\frac{\pi }{x})& -1<-x<1(x\ne 0)& \\ -x|-x|& -1>-x>1& \end{array}$
$=\{\begin{array}{ccc}0& x=0& \\ -x^2\sin \left(\frac{\pi }{x}\right)& -1<-x<1(x\ne 0)& \\ -x|x|& -1>-x>1& \end{array}$
$=-f\left(x\right)$
Hence $f$ is an odd function, and $f^{\prime }$ will be an even function.