If f(x)=sqrt{2-x^2} and g(x)=ln (1-x) are two | vedclass

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(B) Given $f(x)=\sqrt{2-x^2}$ and $g(x)=\ln (1-x)$.The domain of $(f+g)(x)$ is the intersection of the domains of $f(x)$ and $g(x)$.For $f(x)=\sqrt{2-

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$[-\sqrt{2}, 1)$

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(B) Given $f(x)=\sqrt{2-x^2}$ and $g(x)=\ln (1-x)$.The domain of $(f+g)(x)$ is the intersection of the domains of $f(x)$ and $g(x)$.For $f(x)=\sqrt{2-x^2}$,we require $2-x^2 \geq 0$,which implies $x^2 \leq 2$. Thus,$x \in [-\sqrt{2}, \sqrt{2}]$. So,$D_1 = [-\sqrt{2}, \sqrt{2}]$.For $g(x)=\ln (1-x)$,we require $1-x > 0$,which implies $x The domain of $(f+g)(x)$ is $D_1 \cap D_2 = [-\sqrt{2}, \sqrt{2}] \cap (-\infty, 1) = [-\sqrt{2}, 1)$.

If $f(x)=\sqrt{2-x^2}$ and $g(x)=\ln (1-x)$ are two real-valued functions,then the domain of the function $(f+g)(x)$ is

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