Let f ( x ) = sin ( 6 sin ( 2 sin x ) ) , for all x R and g ( x ) = 2 sin x , 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 (are) true? [2015]

Question

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 (are) true? [2015]

Smart Achievers · Accepted Answer

(1, 2, 3)

$f (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$

$- 1 \leq \sin x \leq 1 \Rightarrow - \frac{π}{2} \leq \frac{π}{2} \sin x \leq \frac{π}{2}$

$\Rightarrow - 1 \leq \sin (\frac{π}{2} \sin x) \leq 1$ $\Rightarrow - \frac{π}{6} \leq \frac{π}{6} \sin (\frac{π}{2} \sin x) \leq \frac{π}{6}$

$\Rightarrow - \frac{1}{2} \leq \sin [\frac{π}{6} \sin (\frac{π}{2} \sin x)] \leq \frac{1}{2}$

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

Now, $(f \circ g) (x) = \sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))$

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

Now, $\lim_{x \to 0} \frac{f (x)}{g (x)} = \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{2} \sin x}$

$= \lim_{x \to 0} \frac{\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x))}{\frac{π}{6} \sin (\frac{π}{2} \sin x)} \times \frac{\frac{π}{6} \sin (\frac{π}{2} \sin x)}{\frac{π}{2} \sin x}$

$= 1 \times \frac{π}{6} = \frac{π}{6}$

Now, $(g \circ f) (x) = \frac{π}{2} \sin (\sin (\frac{π}{6} \sin (\frac{π}{2} \sin x)))$ $- \frac{π}{2} \sin (\frac{1}{2}) \leq g (f (x)) \leq \frac{π}{2} \sin (\frac{1}{2})$

Let $\frac{π}{2} \sin (\frac{1}{2}) = p$

Clearly, $0 < p < 1$

$∴$ $- \frac{π}{2} \sin (\frac{1}{2}) \leq g (f (x)) \leq \frac{π}{2} \sin (\frac{1}{2})$

$- p \leq g (f (x)) \leq p \Rightarrow 0 < p < 1$

$∴$ $(g \circ f) (x) \neq 1 for any x \in R .$

Solution

Q.
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 (are) true [2015]

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

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

3 $\lim_{x \to 0} \frac{f (x)}{g (x)} = \frac{π}{6}$

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

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Solution

Q. Let f(x)=sin(π6sin(π2sinx)), 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 (are) true [2015]

1 Range of f is [-12,12]

2 Range of f∘g is [-12,12]

3 limx→0f(x)g(x)=π6