1.

If `f:R -> R, g:R -> R` defined as `f(x) = sin x and g(x) = x^2`, then find the value of `(gof)(x) and (fog)(x) `and also prove that `gof != fog`.

Answer» (fog) (x) = f[(g(x)]
= `f(x^(2)) = sin x^(2)`Ans.
and (gof)(x)= g[f(x)]
=g(sinx) = `(sinx)^(2)= sin^(2) x` Ans.
Therefore, fog `ne` gof . Hence Proved.


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