SATHEE: Functions 2 Question 12

Functions 2 Question 12

13. 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 $(f \circ g) (x)$ denotes $f g (x)$ and ( $g \circ f) (x)$ denotes $g f (x)$ . Then, which of the following is/are true?

(2015 Adv.)

(a) Range of $f$ is $- \frac{1}{2}, \frac{1}{2}$

(b) Range of fog is $- \frac{1}{2}, \frac{1}{2}$

(d) There is an $x \in R$ such that (gof) $(x) = 1$

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Answer:

Correct Answer: 13. (d)

Solution:

(a) $f (x) = \sin \frac{π}{6} \sin \frac{π}{2} \sin x, x \in R$ $= \sin \frac{π}{6} \sin θ, θ \in - \frac{π}{2}, \frac{π}{2}$ , where $θ = \frac{π}{2} \sin x$ $= \sin α, α \in - \frac{π}{6}, \frac{π}{6}$ , where $α = \frac{π}{6} \sin θ$

$∴ f (x) \in - \frac{1}{2}, \frac{1}{2}$

Hence, range of $f (x) \in - \frac{1}{2}, \frac{1}{2}$

So, option (a) is correct.

(b) $f g (x) = f (t), t \in - \frac{π}{2}, \frac{π}{2} \Rightarrow f (t) \in - \frac{1}{2}, \frac{1}{2}$

$∴$ Option (b) is correct. (c) $lim_{x \to 0} \frac{f (x)}{g (x)} = lim_{x \to 0} \frac{\sin \frac{π}{6} \sin \frac{π}{2} \sin x}{\frac{π}{2} (\sin x)}$

$\begin{aligned} = lim_{x \to 0} \frac{\sin \frac{π}{6} \sin \frac{π}{2} \sin x}{\frac{π}{6} \sin \frac{π}{2} \sin x} \cdot \frac{\frac{π}{6} \sin \frac{π}{2} \sin x}{\frac{π}{2} \sin x} \\ = 1 \times \frac{π}{6} \times 1 = \frac{π}{6} \end{aligned}$

$∴$ Option (c) is correct.

(d) $g f (x) = 1$

$\begin{array}{lrl} \Rightarrow & \frac{π}{2} \sin f (x) = 1 \\ \Rightarrow & \sin f (x) = \frac{2}{π} \\ But & f (x) \in - \frac{1}{2}, \frac{1}{2} \subset - \frac{π}{6}, \frac{π}{6} \\ ∴ & \sin f (x) \in - \frac{1}{2}, \frac{1}{2} \end{array}$