CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image