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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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?

Similar Questions

Let $f:(-1,1) \rightarrow R$ be a differentiable function satisfying $(f^{\prime}(x))^4 = 16(f(x))^2$ for all $x \in (-1,1)$ and $f(0)=0$. The number of such functions is:

If the transformation $z = \log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2 y}{d x^2} + \cot x \frac{d y}{d x} + 4 y \operatorname{cosec}^2 x = 0$ into the form $\frac{d^2 y}{d z^2} + k y = 0$,then $k$ is equal to

The curve amongst the family of curves represented by the differential equation,$(x^2 - y^2)dx + 2xy\, dy = 0$ which passes through $(1, 1)$,is

If the curve $y = y(x)$ represented by the solution of the differential equation $(2xy^2 - y)dx + xdy = 0$ passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$,then $|y(1)|$ is equal to ...... .

Vedclass Test Series

Exam Paper Generator

Online Exam Module