Question

# Let $$f(x)=x^{2}$$ and $$g(x)=\sin x$$ for all $$x\in \mathbb{R}$$. 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

A
±nπ, n{0,1,2,}
B
±nπ, n{1,2,}
C
π2+2nπ , n{, 2, 1,0,1,2, }
D
2nπ, n{, 2, 1,0,1,2, }

Solution

## The correct option is A $$\pm\sqrt{n\pi},\ n\in\{0,1,2,\ldots\}$$Given, $$f\left( x \right) ={ x }^{ 2 }$$ for all $$x\in R$$           $$g\left( x \right)=\sin { x }$$ for all $$x\in R$$Now, $$(fogogof)(x) = f(g(g(f(x)))$$                                $$= f(g(g(x^2)))$$                                $$= f(g(\sin { { x }^{ 2 } } )))$$                                $$= f( \sin { \sin { { x }^{ 2 } } })$$       $$(fogogof)(x) = { (\sin { \sin { { x }^{ 2 } } } ) }^{ 2 }$$     Now, $$(gogof)(x) = g(g(f(x))$$                               $$= g(g({ x }^{ 2 }))$$                          $$= g( \sin { { x }^{ 2 } })$$             $$(gogof)(x) =\sin { \sin { { x }^{ 2 } } }$$Given ,$$(fogogof)(x) = (gogof)(x)$$So, $${ (\sin { \sin { { x }^{ 2 } } } ) }^{ 2 } = \sin { \sin { { x }^{ 2 } } }$$       $${ (\sin { \sin { { x }^{ 2 } } } ) }^{ 2 } - \sin { \sin { { x }^{ 2 } } } = 0$$       $$\sin { \sin { { x }^{ 2 } } } (\sin { \sin { { x }^{ 2 } } } - 1) = 0$$ So, $$\sin { \sin { { x }^{ 2 } } } = 0$$  or  $$\sin { \sin { { x }^{ 2 } } } = 1 \Rightarrow \sin x^2=\dfrac {\pi}{2}$$ it is not possible   $$\sin {x} ^{2} = 0$$ implies $${x}^2 = n\pi$$ or $$x = \pm\sqrt { n\pi} , n\in \{ 0,1,2,3.....\}$$$$x =\pm \sqrt { n\pi } \quad \quad n\in \{ 0,1,2,3.....\}$$Maths

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