Question -
Answer -
Given that f(x) =sin x, g (x) = 2x and h (x) = cos x
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) = ½ 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 ((f(x). h(x))
= g (sinx cos x)
= 2sin x cos x
= sin (2x)
⇒ (fog) (x) = (go(f h)) (x), ∀x ∈ R
Hence, fog = go (fh)