If gfx - -sin x - and fgx-sin-x2 - then

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

Q. If

f (x) = {\begin{matrix} x - 1, & - 1 \leq x \leq 0 x^{2}, & 0 < x \leq 1 \end{matrix},

g (x) = sin x

and

h (x) = f (| g (x) |) + | f (g (x)) |

then find h(0).

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If g(f(x))=|sinx| and f(g(x))=(sin√x)2 , then

The correct option is A f(x)=sin2x.g(x)=√xg(f(x))=|sinx|=√(sinx)2f(g(x))=(sin√x)2=sin2√x∴f(x)=sin2xandg(x)=√x

If $g (f (x)) = | sin x |$ and $f (g (x)) = (sin \sqrt{x})^{2}$ , then

The correct option is A $f (x) = {sin}^{2} x . g (x) = \sqrt{x}$
$g (f (x)) = | s i n x | = \sqrt{(s i n x)^{2}} f (g (x)) = (s i n \sqrt{x})^{2} = s i n^{2} \sqrt{x} ∴ f (x) = s i n^{2} x a n d g (x) = \sqrt{x}$