Question

let f and g be two odd functions then the function of fog is
a. an even function
b. an odd function
c. neither even nor odd
d. a periodic function​

Answer 1

Answer:-

f is even (given)

∴f(−x)=f(x)

g is odd (given)

∴g(−x)=−g(x)

So,

According to question,

fog=f[g(x)]=f(y)             Let [g(x)]=y

and also,

fog=f[g(−x)]        ∵g(x) is odd

        =f[−g(x)]

        =f(−y)

        =f(y)

⇒fog(x)=fog(−x)

∴fog is an even function

Step-by-step explanation:

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Answer 2

Given:

Functions $f$ and $g$ are two odd functions

To Find:

Nature of the function $fog$

Solution:

We know that a function $f$ is odd if $f(-x)=-f(x)$.

The function $fog$ is the composition of the functions $f$ and $g$

i.e.  $fog(x)=f(g(x))$

We will check what is the value of $fog(-x)$

$fog(-x)=f(g(-x))$

We know that $g$ is odd

∴  $g(-x)=-g(x)$

⇒ $fog(-x)=f(-g(x))$

Similarly $f$ is also an odd function

∴ $f(-x)=-f(x)$

⇒ $fog(-x)=-f(g(x))$

Hence, when $f$ and $g$ are odd functions, $fog$ is also odd.

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