Let f(x)=x^2 and g(x)=sin x for all x in R. T | vedclass

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(A) Given,$f(x) = x^2$ and $g(x) = \sin x$ for all $x \in R$. First,calculate $(f \circ g \circ g \circ f)(x)$: $(f \circ g \circ g \circ f)(x) = f(g(

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$\pm \sqrt{n \pi}, n \in \{0, 1, 2, \ldots\}$

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(A) Given,$f(x) = x^2$ and $g(x) = \sin x$ for all $x \in R$. First,calculate $(f \circ g \circ g \circ f)(x)$: $(f \circ g \circ g \circ f)(x) = f(g(g(f(x)))) = f(g(g(x^2))) = f(g(\sin x^2)) = f(\sin(\sin x^2)) = (\sin(\sin x^2))^2$. Next,calculate $(g \circ g \circ f)(x)$: $(g \circ g \circ f)(x) = g(g(f(x))) = g(g(x^2)) = g(\sin x^2) = \sin(\sin x^2)$. Equating the two expressions: $(\sin(\sin x^2))^2 = \sin(\sin x^2)$. Let $u = \sin(\sin x^2)$. Then $u^2 = u$,which implies $u^2 - u = 0$,so $u(u - 1) = 0$. This gives $u = 0$ or $u = 1$. Case $1$: $\sin(\sin x^2) = 0$. This implies $\sin x^2 = n \pi$ for some integer $n$. Since the range of $\sin x^2$ is $[-1, 1]$,the only possible value for $n \pi$ is $0$ (when $n = 0$). So,$\sin x^2 = 0$,which implies $x^2 = n \pi$ for $n \in \{0, 1, 2, \ldots\}$. Thus,$x = \pm \sqrt{n \pi}$ for $n \in \{0, 1, 2, \ldots\}$. Case $2$: $\sin(\sin x^2) = 1$. This implies $\sin x^2 = \frac{\pi}{2}$. Since $\frac{\pi}{2} \approx 1.57 > 1$,this has no real solution for $x$. Therefore,the set of all $x$ is $\pm \sqrt{n \pi}$ for $n \in \{0, 1, 2, \ldots\}$.

Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then the set of all $x$ satisfying $(f \circ g \circ g \circ f)(x)=(g \circ g \circ f)(x)$,where $(f \circ g)(x)=f(g(x))$,is

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