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

Answer Verified Verified
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 \[{\text{fog(x)}}\]