Let f ( x )= x 2 and g ( x )= sin x for all x ∈ R. Then the set of all x satisfying ( f o g o g o f )( x )=( g o g o f )( x ) where ( f o g )( x )= f ( g ( x )), is

Question

Let f ( x )= x 2 and g ( x )= sin x for all x ∈ R. Then the set of all x satisfying ( f o g o g o f )( x )=( g o g o f )( x ) where ( f o g )( x )= f ( g ( x )), is (A) ± √n π, n ∈ 0,1,2, ldots  (B) ± √n π, n ∈ 1,2, ldots  (C) (π/2)+2 n π, n ∈ ldots,-2,-1,0,1,2, ldots .  (D) 2 n π, n ∈ ldots,-2,-1,0,1,2, ldots

Tardigrade Admin · Accepted Answer

( fogogof )(x)= sin 2( sin x2) ( gogof )(x)= sin ( sin x2) ∴ sin 2( sin x2)= sin ( sin x2) ⇒ sin ( sin x2)[ sin ( sin x2)-1]=0 ⇒ sin ( sin x2)=0 or 1 ⇒ sin x2=n π or 2 m π+π / 2, where m, n ∈ I ⇒ sin x2=0 ⇒ x2=n π ⇒ x=± √n π, n ∈ 0,1,2, ldots

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

$\pm nπ, n \in {0, 1, 2, \dots}$

$\pm nπ, n \in {1, 2, \dots}$

$\frac{π}{2} + 2 nπ, n \in {\dots, - 2, - 1, 0, 1, 2, \dots .}$

$2 nπ, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$

Q. Let f(x)=x2 and g(x)=sinx for all x∈R. Then the set of all x satisfying (fogogof)(x)=(gogof)(x) where (fog)(x)=f(g(x)), is

±nπ​,n∈{0,1,2,…}

±nπ​,n∈{1,2,…}

2π​+2nπ,n∈{…,−2,−1,0,1,2,….}

2nπ,n∈{…,−2,−1,0,1,2,…}

( fogogof )(x)=sin2(sinx2) ( gogof )(x)=sin(sinx2) ∴sin2(sinx2)=sin(sinx2) ⇒sin(sinx2)[sin(sinx2)−1]=0 ⇒sin(sinx2)=0 or 1 ⇒sinx2=nπ or 2mπ+π/2, where m,n∈I ⇒sinx2=0 ⇒x2=nπ⇒x=±nπ​,n∈{0,1,2,…}

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

$\pm nπ, n \in {0, 1, 2, \dots}$

$\pm nπ, n \in {1, 2, \dots}$

$\frac{π}{2} + 2 nπ, n \in {\dots, - 2, - 1, 0, 1, 2, \dots .}$

$2 nπ, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$