Q. 25.0( 1 Vote )

Let f(x) = x2 + x + 1 and g(x) = sin x. Show that fog ≠ gof.

Answer :

We have, f (x) = x2 + x + 1 and g(x) = sin x


Now,


fog(x) = f(g(x)) = f(sin x)


fog(x) = sin2 x + sin x + 1


Again, gof(x) = g(f(x)) = g (x2 + x + 1)


gof(x) = sin(x2 + x + 1)


Clearly,


fog gof


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