Question

# The composite mapping ${\text{fog}}$of the map, ${\text{f}}:\mathbb{R} \to \mathbb{R}$, ${\text{f(x) = sinx}}$and ${\text{g}}:\mathbb{R} \to \mathbb{R}$, ${\text{g(x) = }}{{\text{x}}^{\text{2}}}$isA. ${{\text{x}}^{\text{2}}}{\text{sinx}}$B. ${{\text{(sinx)}}^{\text{2}}}$C. ${\text{sin}}{{\text{x}}^{\text{2}}}$D. $\dfrac{{{\text{sinx}}}}{{{{\text{x}}^{\text{2}}}}}$

Hint: As, we are given, ${\text{f}}:\mathbb{R} \to \mathbb{R}$, ${\text{f(x) = sinx}}$and ${\text{g}}:\mathbb{R} \to \mathbb{R}$, ${\text{g(x) = }}{{\text{x}}^{\text{2}}}$, we use ${\text{fog}}$(x) ${\text{ = f(g(x))}}$, to calculate the composite mapping ${\text{fog}}$, it is usually defines as f(g).
It is not also usually equal to ${\text{gof(x)}}$.

Complete step by step solution: We have, ${\text{f}}:\mathbb{R} \to \mathbb{R}$, ${\text{f(x) = sinx}}$and ${\text{g}}:\mathbb{R} \to \mathbb{R}$, ${\text{g(x) = }}{{\text{x}}^{\text{2}}}$ ,
As we know, according to the definition,
${\text{fog}}$(x) ${\text{ = f(g(x))}}$
Now we have, ${\text{g(x) = }}{{\text{x}}^{\text{2}}}$, so,
Going through the process,
${\text{fog}}$(x) ${\text{ = f(g(x))}}$
$\Rightarrow {\text{fog(x) = f(}}{{\text{x}}^{\text{2}}}{\text{)}}$
As ${\text{g}}:\mathbb{R} \to \mathbb{R}$ and ${\text{g(x) = }}{{\text{x}}^{\text{2}}}$,
Now again, we have, ${\text{f}}:\mathbb{R} \to \mathbb{R}$,${\text{f(x) = sinx}}$
So, $\Rightarrow {\text{f(}}{{\text{x}}^{\text{2}}}{\text{) = sin(}}{{\text{x}}^{\text{2}}}{\text{)}}$.
At the end, we find that, ${\text{fog(x) = f(g(x))}}$${\text{ = sin}}{{\text{x}}^{\text{2}}}$.

Hence the correct option is (C).

Note: You need to remember that ${\text{fog(x)}}$is not always equal to ${\text{gof(x)}}$.
In this example only, we can see that,
We have ${\text{fog(x) = f(g(x))}}$${\text{ = sin}}{{\text{x}}^{\text{2}}}$
But if we try to calculate, ${\text{gof(x)}}$ we will get,
${\text{gof(x)}}$${\text{ = g(sinx)}}$as ${\text{f(x) = sinx}}$and
${\text{g(sinx)}}$${\text{ = (sinx}}{{\text{)}}^{\text{2}}}$
Which is not equal to ${\text{fog(x)}}$