A curve y = f(x) passing through the point le | vedclass

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(B) Given the differential equation $\frac{dy}{dx} = -x e^{-\frac{x^2}{2}}.$Integrating both sides with respect to $x$,we get $y = \int -x e^{-\frac{x

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$f(x)$ is symmetric with respect to the origin.

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(B) Given the differential equation $\frac{dy}{dx} = -x e^{-\frac{x^2}{2}}.$Integrating both sides with respect to $x$,we get $y = \int -x e^{-\frac{x^2}{2}} dx.$Let $u = -\frac{x^2}{2}$,then $du = -x dx.$Thus,$y = \int e^u du = e^u + C = e^{-\frac{x^2}{2}} + C.$Since the curve passes through $\left(1, \frac{1}{\sqrt{e}}\right)$,we have $\frac{1}{\sqrt{e}} = e^{-\frac{1^2}{2}} + C \implies \frac{1}{\sqrt{e}} = \frac{1}{\sqrt{e}} + C \implies C = 0.$So,$f(x) = e^{-\frac{x^2}{2}}.$This is an even function,meaning it is symmetric with respect to the $y$-axis,not the origin. Thus,statement $B$ is false.$f'(x) = -x e^{-\frac{x^2}{2}}$,which is defined for all $x$,so $f(x)$ is differentiable at $x=0$.$f'(x) > 0$ for $x  0$,so it is increasing for $x  0$.$f''(x) = -e^{-\frac{x^2}{2}} + x^2 e^{-\frac{x^2}{2}} = e^{-\frac{x^2}{2}}(x^2 - 1).$ Setting $f''(x) = 0$ gives $x = \pm 1$,which are the two inflection points.

$A$ curve $y = f(x)$ passing through the point $\left(1, \frac{1}{\sqrt{e}}\right)$ satisfies the differential equation $\frac{dy}{dx} + x e^{-\frac{x^2}{2}} = 0.$ Then which of the following does not hold good?

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