Let f(x)=sin⁡ π6 sin⁡π2sin⁡x ∀x∈R and g(x)=π2sinx, ∀x∈R. Let (fog)(x) denote f(g(x)) and (gof)(x) denote g(f(x)). Then which of the following is (are) true?

R a n g e o f f i s - 1 2 , 1 2 , R a n g e o f f o g i s - 1 2 , 1 2 , lim x → 0 ⁡ f ( x ) g ( x ) = π 6

Let f(x)=sin⁡ π6 sin⁡π2sin⁡x ∀x∈R and g(x)=π2sinx, ∀x∈R. Let (fog)(x) denote f(g(x)) and (gof)(x) denote g(f(x)). Then which of the following is (are) true?

Given f ( x ) = sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x − 1 ≤ sin ⁡ x ≤ 1 ⇒ − π 2 ≤ π 2 sin ⁡ x ≤ π 2 ⇒ − 1 ≤ sin ⁡ π 2 sin ⁡ x ≤ 1 ⇒ − π 6 ≤ π 6 sin ⁡ π 2 sin ⁡ x ≤ π 6 ⇒ − 1 2 ≤ sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x ≤ 1 2 ∴ Range of f = − 1 2 , 1 2 Range of fog = − 1 2 , 1 2 Now, lim x → 0 ⁡ f ( x ) g ( x ) = lim x → 0 sin ⁡ π 6 sin ⁡ π 2 sin ⁡ x π 2 sin ⁡ x x x = 2 π ⋅ lim x → 0 sin ⁡ π 6 ⋅ sin ⁡ π 2 sin ⁡ x x ⋅ x x = 2 π ⋅ lim x → 0 sin ⁡ π 6 sin ⁡ ( πx ) 2 x x = 2 π ⋅ lim x → 0 sin ⁡ π 6 ⋅ π 2 ⋅ x x = 2 π × π 2 12 = π 6

R a n g e o f f i s - 1 2 , 1 2

R a n g e o f f o g i s - 1 2 , 1 2

lim x → 0 ⁡ f ( x ) g ( x ) = π 6

T h e r e i s a n x ∈ R s u c h t h a t ( g o f ) ( x ) = 1

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Let $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)) \forall x \in R$ and $g (x) = \frac{π}{2} \sin x, \forall x \in R .$ Let $(f o g) (x)$ denote $f (g (x))$ and $(g o f) (x)$ denote $g (f (x)) .$ Then which of the following is (are) true?

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$R a n g e o f f i s [- \frac{1}{2}, \frac{1}{2}]$

$R a n g e o f f o g i s [- \frac{1}{2}, \frac{1}{2}]$

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

$T h e r e i s a n x \in R s u c h t h a t (g o f) (x) = 1$

answer is A, B, C.

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Detailed Solution

Given $f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$
$\begin{array}{l} - 1 \leq \sin x \leq 1 \Rightarrow \frac{- π}{2} \leq \frac{π}{2} \sin x \leq \frac{π}{2} \Rightarrow - 1 \leq \sin (\frac{π}{2} \sin x) \leq 1 \\ \Rightarrow \frac{- π}{6} \leq \frac{π}{6} \sin (\frac{π}{2} \sin x) \leq \frac{π}{6} \Rightarrow \frac{- 1}{2} \leq \sin [\frac{π}{6} \sin (\frac{π}{2} \sin x)] \leq \frac{1}{2} \\ ∴ Range of f = [\frac{- 1}{2}, \frac{1}{2}] \end{array}$
$Range of fog = [\frac{- 1}{2}, \frac{1}{2}]$
Now, $\lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{2} \frac{\sin x}{x} x}$
$\begin{array}{l} = \frac{2}{π} \cdot \lim_{x \to 0} \frac{\sin (\frac{π}{6} \cdot \sin (\frac{π}{2} \frac{\sin x}{x} \cdot x))}{x} \\ = \frac{2}{π} \cdot \lim_{x \to 0} \frac{\sin (\frac{π}{6} \frac{\sin (πx)}{2} x)}{x} \\ = \frac{2}{π} \cdot \lim_{x \to 0} \frac{\sin (\frac{π}{6} \cdot \frac{π}{2} \cdot x)}{x} \\ = \frac{2}{π} \times \frac{π^{2}}{12} = \frac{π}{6} \end{array}$

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