Let f(x) = sin \textrm{ } \left(\right. \frac{\pi}{6} sin \left(\right. \frac{\pi}{2} sinx \left.\right) \left.\right) for all x ∈ R and g(x) = \frac{\pi}{2} sin x for all x ∈ R. Let (fog)(x) denote f (g(x)) and (g o f )(x) denote g (f(x)). Then which of the following is (are) true?

Engineering

Mathematics

Composition of Functions

Question

Let f(x) = $\sin (\frac{π}{6} \sin (\frac{π}{2} sinx))$ for all x ∈ R and g(x) = $\frac{π}{2}$ sin x for all x ∈ R. Let (fog)(x) denote f (g(x)) and (g o f )(x) denote g (f(x)). Then which of the following is (are) true?

$f (x)$ is many one function

Range of f o g is $[\frac{- 1}{2}, \frac{1}{2}]$

There is an x ∈ R such that (g o f )(x) = 1

Range of f is $[\frac{- 1}{2}, \frac{1}{2}]$

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(A) Range of f (x) is $[\frac{- 1}{2}, \frac{1}{2}]$

(B) Range of g (x) is $[\frac{- π}{2}, \frac{π}{2}]$

f o g (x) = f [ g (x)] = $[\frac{- 1}{2}, \frac{1}{2}]$

(D) $g o f (x) = g (f (x)) = \frac{π}{2} sin (f (x))$

$∵ f (x) \in [\frac{- 1}{2}, \frac{1}{2}] = sinf (x) < \frac{1}{2}$

$∴$ g(f(x)) can not be equal to for any x ∈ R.