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Question
Let f, g, h be real functions given by f(x) = sin x, g (x) = 2x and h (x) = cos x. Prove that fog = go (fh).
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Solution
We know that f : R→[−1, 1] and g : R→R
Clearly, the range of g is a subset of the domain of f.
fog : R → R
Now, (fh) (x)=f (x)h (x) = (sin x) (cos x) =`1/2`sin (2x)
Domain of fh is R.
Since range of sin x is [-1,1],
−1 ≤ sin 2x ≤ 1
⇒ ` (-1)/2 ≤ sin x/2 ≤ 1/2`
Range of fh = `[(-1)/2 ","1/2]`
So, (fh) : R →`[(-1)/2 ","1/2]`
Clearly, range of fh is a subset of g.
⇒ go (fh) : R → R
⇒ domains of fog and go (fh) are the same .
So, (fog) (x)=f (g (x)) = f (2x) = sin (2x)
and ( go (fh)) (x) = g ((fh) (x)) = g (sinx cos x) = 2sin x cos x = sin (2x)
⇒ (fog) (x) = ( go(fh)) (x), ∀x ∈ R
Hence, fog = go (fh)
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