Functions 2 Question 6
7.
Let $f(x)=x^{2}$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof $)(x)=(\operatorname{gog} f)(x)$, where $(f \circ g)(x)=f(g(x))$, is
(2011)
(a) $\pm \sqrt{n \pi}, n \in{0,1,2, \ldots}$
(b) $\pm \sqrt{n \pi}, n \in{1,2, \ldots}$
(c) $\pi / 2+2 n \pi, n \in{\ldots,-2,-1,0,1,2, \ldots}$
(d) $2 n \pi, n \in{\ldots,-2,-1,0,1,2, \ldots}$
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Answer:
Correct Answer: 7. (b)
Solution:
- $f(x)=x^{2}, g(x)=\sin x$
$ \begin{aligned} (g \circ f)(x) & =\sin x^{2} \\ g o(g \circ f)(x) & =\sin \left(\sin x^{2}\right) \\ (\text { fogogof })(x) & =\left(\sin \left(\sin x^{2}\right)\right)^{2} \end{aligned} $
Again, $(g \circ f)(x)=\sin x^{2}$
$ (\operatorname{gogof})(x)=\sin \left(\sin x^{2}\right) $
Given, $\quad($ fogogof $)(x)=(\operatorname{gogof})(x)$
$\Rightarrow \quad\left(\sin \left(\sin x^{2}\right)\right)^{2}=\sin \left(\sin x^{2}\right)$
$\Rightarrow \sin \left(\sin x^{2}\right){\sin \left(\sin x^{2}\right)-1 }=0$
$\Rightarrow \sin \left(\sin x^{2}\right)=0$ or $\sin \left(\sin x^{2}\right)=1$
$ \Rightarrow \quad \sin x^{2}=0 \quad \text { or } \quad \sin x^{2}=\frac{\pi}{2} $
$ \therefore \quad x^{2}=n \pi $
$ \begin{aligned} & {\left[\sin x^{2}=\frac{\pi}{2} \text { is not possible as }-1 \leq \sin \theta \leq 1\right]} \\ & x= \pm \sqrt{n \pi} \end{aligned} $
