Question

Let $f\left(x\right)=sin\left[\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right]\forall xϵR\text{and}g\left(x\right)=\frac{\pi }{2}sinx\forall xϵR$. Let (fog) (x) denotes f{g(x)} and (gof) (a) denotes g{f(x)}. Then, which of the following is / are true?

• A. Range of f is $[-\frac{1}{2},\frac{1}{2}]$
• B. Range of fog is $[-\frac{1}{2},\frac{1}{2}]$
• C. $li{m}_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\pi }{6}$
• D. There is an x $ϵ$R such that (gof) (x) = 1

Answer 1

Answer: (A), (B), (C)

The correct options are
A Range of f is $[-\frac{1}{2},\frac{1}{2}]$
B Range of fog is $[-\frac{1}{2},\frac{1}{2}]$
C $li{m}_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}=\frac{\pi }{6}$

$f\left(x\right)=sin[\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right),xϵR]=sin\left(\frac{\pi }{6}sin\theta \right)\theta ϵ[-\frac{\pi }{2},\frac{\pi }{2}]where\theta =\frac{\pi }{2}sinx=sin\alpha ,\alpha ϵ[-\frac{\pi }{6},\frac{\pi }{6}],where\alpha =\frac{\pi }{6}sin\theta \therefore f\left(x\right)ϵ[-\frac{1}{2},\frac{1}{2}]$
Hence, range of $f\left(x\right)ϵ[-\frac{1}{2},\frac{1}{2}]$
So, Option (b) is correct.
(b) $f\left\{g\right(x\left)\right\}=f,tϵ[-\frac{\pi }{2},\frac{\pi }{2}]\Rightarrow f\left(t\right)ϵ[-\frac{1}{2},\frac{1}{2}]$
$\therefore$ Option (b) is correct.
c. $li{m}_{x\to 0}\frac{f\left(x\right)}{g\left(x\right)}=li{m}_{x\to 0}\frac{sin\left[\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right]}{\frac{\pi }{2}\left(sinx\right)}li{m}_{x\to 0}\frac{sin\left[\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)\right]}{\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)}.\frac{\frac{\pi }{6}sin\left(\frac{\pi }{2}sinx\right)}{\left(\frac{\pi }{2}sinx\right)}=1\times \frac{\pi }{6}\times 1=\frac{\pi }{6}$
$\therefore$ Option (c) is correct.
$\Rightarrow \frac{\pi }{2}sin\left\{f\right(x\left)\right\}=1$
$\Rightarrow sin\left\{f\right(x\left)\right\}=\frac{2}{\pi }$ . . . (i)
But $f\left(x\right)ϵ[-\frac{1}{2},\frac{1}{2}]\subset [-\frac{\pi }{6},\frac{\pi }{6}]$
$\therefore sin\left\{f\right(x\left)\right\}ϵ[-\frac{1}{2},\frac{1}{2}]$ . . . (ii)
$\Rightarrow sin\left\{f\right(x\left)\right\}\ne \frac{2}{\pi }$, [From Eqs. (i) and (ii)]
i.e. No solution
$\therefore$ Option (d) is not correct.