Functions 2 Question 12
13. Let $f(x)=\sin \frac{\pi}{6} \sin \frac{\pi}{2} \sin x \quad$ for all $x \in R$ and $g(x)=\frac{\pi}{2} \sin x$ 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 $-\frac{1}{2}, \frac{1}{2}$
(c) $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\frac{\pi}{6}$
(d) There is an $x \in R$ such that (gof) $(x)=1$
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Answer:
Correct Answer: 13. (d)
Solution:
- (a) $f(x)=\sin \frac{\pi}{6} \sin \frac{\pi}{2} \sin x \quad, x \in R$ $=\sin \frac{\pi}{6} \sin \theta, \theta \in-\frac{\pi}{2}, \frac{\pi}{2}$, where $\theta=\frac{\pi}{2} \sin x$ $=\sin \alpha, \alpha \in-\frac{\pi}{6}, \frac{\pi}{6}$, where $\alpha=\frac{\pi}{6} \sin \theta$
$$ \therefore \quad f(x) \in-\frac{1}{2}, \frac{1}{2} $$
Hence, range of $f(x) \in-\frac{1}{2}, \frac{1}{2}$
So, option (a) is correct.
(b) $f{g(x)}=f(t), t \in-\frac{\pi}{2}, \frac{\pi}{2} \Rightarrow f(t) \in-\frac{1}{2}, \frac{1}{2}$
$\therefore \quad$ Option (b) is correct. (c) $\lim _{x \rightarrow 0} \frac{f(x)}{g(x)}=\lim _{x \rightarrow 0} \frac{\sin \frac{\pi}{6} \sin \frac{\pi}{2} \sin x}{\frac{\pi}{2}(\sin x)}$
$$ \begin{aligned} & =\lim _{x \rightarrow 0} \frac{\sin \frac{\pi}{6} \sin \frac{\pi}{2} \sin x}{\frac{\pi}{6} \sin \frac{\pi}{2} \sin x} \cdot \frac{\frac{\pi}{6} \sin \frac{\pi}{2} \sin x}{\frac{\pi}{2} \sin x} \\ & =1 \times \frac{\pi}{6} \times 1=\frac{\pi}{6} \end{aligned} $$
$\therefore$ Option (c) is correct.
(d) $g{f(x)}=1$
$$ \begin{array}{lrl} \Rightarrow & \frac{\pi}{2} \sin {f(x)}=1 \\ \Rightarrow & \sin {f(x)}=\frac{2}{\pi} \\ \text { But } & f(x) \in-\frac{1}{2}, \frac{1}{2} \subset-\frac{\pi}{6}, \frac{\pi}{6} \\ \therefore & \sin {f(x)} \in-\frac{1}{2}, \frac{1}{2} \end{array} $$
$\Rightarrow \sin {f(x)} \neq \frac{2}{\pi}$
[from Eqs. (i) and (ii)]
i.e. No solution.
$\therefore$ Option (d) is not correct.
