Let f(x)=sqrt{2-x-x^2} and g(x)=cos x. Which | vedclass

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(A) Given $f(x) = \sqrt{2-x-x^2}$ and $g(x) = \cos x$.For $f(x)$ to be defined,$2-x-x^2 \geq 0 \Rightarrow x^2+x-2 \leq 0 \Rightarrow (x+2)(x-1) \leq

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Only $I$

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(A) Given $f(x) = \sqrt{2-x-x^2}$ and $g(x) = \cos x$.For $f(x)$ to be defined,$2-x-x^2 \geq 0 \Rightarrow x^2+x-2 \leq 0 \Rightarrow (x+2)(x-1) \leq 0 \Rightarrow x \in [-2, 1]$.For $f(g(x))$,we need $g(x) \in [-2, 1]$. Since $-1 \leq \cos x \leq 1$,this is always true for all $x \in \mathbb{R}$. Thus,$	ext{Domain of } f(g(x)) = \mathbb{R}$.For $f((g(x))^2)$,we need $(g(x))^2 \in [-2, 1]$. Since $0 \leq \cos^2 x \leq 1$,this is always true for all $x \in \mathbb{R}$. Thus,$	ext{Domain of } f((g(x))^2) = \mathbb{R}$.Therefore,$	ext{Domain of } f((g(x))^2) = 	ext{Domain of } f(g(x))$,so statement $I$ is true.For $g(f(x))$,we need $f(x)$ to be defined,so $x \in [-2, 1]$. Thus,$	ext{Domain of } g(f(x)) = [-2, 1]$.Since $	ext{Domain of } f(g(x)) = \mathbb{R}$ and $	ext{Domain of } g(f(x)) = [-2, 1]$,statements $II$ and $III$ are false.For $g((f(x))^3)$,we need $f(x)$ to be defined,so $x \in [-2, 1]$. Thus,$	ext{Domain of } g((f(x))^3) = [-2, 1]$. This is not equal to $\mathbb{R}$,so statement $IV$ is false.Hence,only statement $I$ is true.

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