Let u and v be two odd functions- then the uov isAn even functionA periodic functionNeither even nor oddAn odd

Question

Toppr Admin · Accepted Answer

We have- uov-x2212-x-u-v-x2212-x-u-x2212-v-x- -xA0- -xA0- -xA0- -xA0- - -x2235- v is odd-x2212-u-v-x- -xA0- -xA0- -xA0- -xA0-xA0-xA0- - -x2235- u is odd-x2212-uov-x-Thus- uov is an odd function-

Let $u$ and $v$ be two odd functions, then the function uov is
An even function
A periodic function
Neither even nor odd
An odd function

We have, $u o v (- x)$
$= u [v (- x)]$
$= u [- v (x)]$ [ $∵$ v is odd]
$= - u [v (x)]$ [ $∵$ u is odd]
$= - u o v (x)$
Thus, $u o v$ is an odd function.

Let u and v be two odd functions, then the function uov isAn even functionA periodic functionNeither even nor oddAn odd function

We have, uov(−x)=u[v(−x)]=u[−v(x)] [ ∵ v is odd]=−u[v(x)] [ ∵ u is odd]=−uov(x)Thus, uov is an odd function.

Let $u$ and $v$ be two odd functions, then the function uov is
An even function
A periodic function
Neither even nor odd
An odd function

We have, $u o v (- x)$
$= u [v (- x)]$
$= u [- v (x)]$ [ $∵$ v is odd]
$= - u [v (x)]$ [ $∵$ u is odd]
$= - u o v (x)$
Thus, $u o v$ is an odd function.