Functions Ques 39
Let $f(x)=\sin [\frac{\pi}{6} \sin (\frac{\pi}{2} \sin x )]$ for all $x \in R$ and $g(x)=\frac{\pi}{2} \sin x$ for all $x \in R$. Let $(f o g)(x)$ denotes $f\{g(x)\}$ and ( $g o f)(x)$ 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) $(x)=1$
Show Answer
Answer:
Correct Answer: 39.(a,b,c)
Solution:
Formula:
- (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) $g\{f(x)\}=1$
$\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}] $
$\therefore \sin \{f(x)\} \in[-\frac{1}{2}, \frac{1}{2}]$
$\Rightarrow \sin \{f(x)\} \neq \frac{2}{\pi}$
[from Eqs. (i) and (ii)]
i.e. No solution.
$\therefore$ Option (d) is not correct.
BETA
