Let f(x)=sinπ6sinπ2sinx for all x∈R and g(x)=π2sinx for all x∈R. Let (f∘g)(x) denote f(g(x)) and (g∘f)(x) denote g(f(x))then which of the following is true statement.

We have1) f(x)=sinπ6sinπ2sinx-1≤sinπ2sinx≤1⇒-π6≤π6sinπ2sinx≤π6⇒-12≤sinπ6sinπ2(sinx)≤12 ∴range of fx is -12,122) f(g(x))=fπ2sinx=sinπ6sinπ2sinπ2sinx ∴the range of fogx is -12,123) limx→0fxgx=limx→0sinπ6sinπ2sinxπ2sinx =limx→0cosπ6sinπ2sinxπ6cosπ2sinxπ2cosxπ2cosx =π64) Assume ∃ x∈R such that gofx=1⇒π2sinfx=1 ⇒sinsinπ6sinπ2sinx=2π>12which is not possible ∴options (1), (2) and (3) are correct

Let f(x)=sinπ6sinπ2sinx for all x∈R and g(x)=π2sinx for all x∈R. Let (f∘g)(x) denote f(g(x)) and (g∘f)(x) denote g(f(x))then which of the following is true statement.

We have1) f(x)=sinπ6sinπ2sinx-1≤sinπ2sinx≤1⇒-π6≤π6sinπ2sinx≤π6⇒-12≤sinπ6sinπ2(sinx)≤12 ∴range of fx is -12,122) f(g(x))=fπ2sinx=sinπ6sinπ2sinπ2sinx ∴the range of fogx is -12,123) limx→0fxgx=limx→0sinπ6sinπ2sinxπ2sinx =limx→0cosπ6sinπ2sinxπ6cosπ2sinxπ2cosxπ2cosx =π64) Assume ∃ x∈R such that gofx=1⇒π2sinfx=1 ⇒sinsinπ6sinπ2sinx=2π>12which is not possible ∴options (1), (2) and (3) are correct

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$Let f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)) for all x \in R and g (x) = \frac{π}{2} \sin x for all x \in R$ . $Let (f \circ g) (x) denote f (g (x)) and (g \circ f) (x) denote g (f (x))$
then which of the following is true statement.

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$There is an x \in R such that (g o f) (x) = 1$

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

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

$\lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{π}{6}$

answer is A, B, C.

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Detailed Solution

We have

$\begin{array}{l} 1) f (x) = \sin (\frac{π}{6} \sin ((\frac{π}{2}) \sin x)) \\ - 1 \leq \sin (\frac{π}{2} \sin x) \leq 1 \\ \Rightarrow - \frac{π}{6} \leq \frac{π}{6} \sin (\frac{π}{2} \sin x) \leq \frac{π}{6} \\ \Rightarrow - \frac{1}{2} \leq \sin [\frac{π}{6} (\sin \frac{π}{2} (\sin x))] \leq \frac{1}{2} \end{array} ∴ r a n g e o f f (x) i s [- \frac{1}{2}, \frac{1}{2}]$

$2) f (g (x)) = f (\frac{π}{2} \sin x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin (\frac{π}{2} \sin x))) ∴ t h e r a n g e o f f o g (x) i s [- \frac{1}{2}, \frac{1}{2}]$

$\begin{array}{l} 3) \lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2}) \sin x)}{\frac{π}{2} \sin x} \\ = \lim_{x \to 0} \frac{\cos (\frac{π}{6} \sin (\frac{π}{2} \sin x)) \frac{π}{6} \cos (\frac{π}{2} \sin x) \frac{π}{2} \cos x}{\frac{π}{2} \cos x} \\ = \frac{π}{6} \end{array}$

4) Assume $\exists x \in R s u c h t h a t g o f (x) = 1$

$\Rightarrow \frac{π}{2} \sin (f (x)) = 1 \Rightarrow \sin [\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))] = \frac{2}{π} > \frac{1}{2} w h i c h i s n o t p o s s i b l e$