Let f ( x ) = x 2 and g ( x ) = sin x for all x R . Then the set of all x satisfying ( f ? g ? g ? f ) ( x ) = ( g ? g ? f ) ( x ) , where ( f ? g ) ( x ) = f ( g ( x ) ) , is [2011]

Question

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 [2011]

Smart Achievers · Accepted Answer

(1)

Given : $f (x) = x^{2}$ and $g (x) = \sin x,$ $\forall x \in R$

$∴$ $(g \circ f) (x) = \sin x^{2}$

$\Rightarrow (g \circ g \circ f) (x) = \sin (\sin x^{2})$

$\Rightarrow (f \circ g \circ g \circ f) (x) = \sin^{2} (\sin x^{2})$

Since given that $(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x),$

$∴$ $\sin^{2} (\sin x^{2}) = \sin (\sin x^{2})$

$\Rightarrow \sin (\sin x^{2}) = 0, 1$

$\Rightarrow \sin x^{2} = n π$ or $\frac{(4 n + 1) π}{2},$ where $n \in Z$

$\Rightarrow \sin x^{2} = 0$ $(∵ \sin x^{2} \in [- 1, 1]) \Rightarrow x^{2} = n π$

$∴$ $x = \pm \sqrt{n π},$ where $n \in W$

Solution

Q.
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 [2011]

1 $\pm \sqrt{n π}, n \in {0, 1, 2, \dots}$

2 $\pm \sqrt{n π}, n \in {1, 2, \dots}$

3 $\frac{π}{2} + 2 n π, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$

4 $2 n π, n \in {\dots, - 2, - 1, 0, 1, 2, \dots}$

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Solution

Q. Let f(x)=x2 and g(x)=sinx for all x∈R. Then the set of all x satisfying (f∘g∘g∘f)(x)=(g∘g∘f)(x), where (f∘g)(x)=f(g(x)), is [2011]

1 ±nπ, n∈{0,1,2,…}

2 ±nπ, n∈{1,2,…}

3 π2+2nπ, n∈{…,-2,-1,0,1,2,…}