Let f(x)=x2 and g(x)=sin⁡x for all x∈RThen the set of all x satisfying( fogogof )(x) = (gogof )(x) ,where (fog)(x) = f(g(x)), is

f(x)=x2 and g(x)=sin⁡x for all x∈R( fogogof )(x) = (gogof )(x)fgf(x)=ggf(x)∴sin2⁡sin⁡x2=sin⁡sin⁡x2⇒sin⁡sin⁡x2sin⁡sin⁡x2−1=0⇒sin⁡x2=0⇒x2=nπ⇒x=±nπ,n∈{0,1,2,…}

Let f(x)=x2 and g(x)=sin⁡x for all x∈RThen the set of all x satisfying( fogogof )(x) = (gogof )(x) ,where (fog)(x) = f(g(x)), is

f(x)=x2 and g(x)=sin⁡x for all x∈R( fogogof )(x) = (gogof )(x)fgf(x)=ggf(x)∴sin2⁡sin⁡x2=sin⁡sin⁡x2⇒sin⁡sin⁡x2sin⁡sin⁡x2−1=0⇒sin⁡x2=0⇒x2=nπ⇒x=±nπ,n∈{0,1,2,…}

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

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

$\pm \sqrt{nπ}, n \in {1, 2, \dots}$

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

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

answer is A.

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Detailed Solution

$f (x) = x^{2} and g (x) = \sin x for all x \in R$
( fogogof )(x) = (gogof )(x)
$f (g (f (x))) = g (g (f (x)))$
$\begin{array}{l} ∴ \sin^{2} (\sin x^{2}) = \sin (\sin x^{2}) \\ \Rightarrow \sin (\sin x^{2}) [\sin (\sin x^{2}) - 1] = 0 \Rightarrow \sin x^{2} = 0 \\ \Rightarrow x^{2} = nπ \Rightarrow x = \pm \sqrt{nπ}, n \in {0, 1, 2, \dots} \end{array}$