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, (f h) (x) = f (x) h (x) = (sin x) (cos x) = \( \dfrac{ 1}{2}\) sin (2x)
Domain of f h is R.
Since range of sin x is [-1, 1], −1 ≤ sin 2x ≤ 1
⇒ \(\dfrac{ -1}{2}\) ≤ sin \( \dfrac{ x}{2}\) ≤ \( \dfrac{ 1}{2}\)
Range of f h = [\( \dfrac{ -1}{2}\), \( \dfrac{ 1}{2}\)]
So, (f h): R → [\( \dfrac{ -1}{2}\),\( \dfrac{ 1}{2}\)]
Clearly, range of f h is a subset of g.
⇒ go (f h): R → R
⇒ Domains of fog and go (f h) are the same.
So, (fog) (x) = f (g (x))
= f (2x)
= sin (2x)
And (go (f h)) (x) = g ((f(x). h(x))
= g (sin x cos x)
= 2sin x cos x
= sin (2x)
⇒ (fog) (x) = (go (f h)) (x), ∀x ∈ R
Hence, fog = go (f h)
Answered by Aaryan | 1 year agoLet A be the set of first five natural and let R be a relation on A defined as follows: (x, y) R x ≤ y
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