If a curve passes through the point left( 2, | vedclass

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

Question

(A) The slope of the curve is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int \left( 1 - \frac

Vedclass · Accepted Answer

$-\frac{3}{2}$

Vedclass · Answer

(A) The slope of the curve is given by $\frac{dy}{dx} = 1 - \frac{1}{x^2}$.Integrating both sides with respect to $x$:$\int dy = \int \left( 1 - \frac{1}{x^2} \right) dx$$y = x + \frac{1}{x} + C$Since the curve passes through the point $\left( 2, \frac{7}{2} \right)$,we substitute these values to find $C$:$\frac{7}{2} = 2 + \frac{1}{2} + C$$\frac{7}{2} = \frac{5}{2} + C$$C = 1$Thus,the equation of the curve is $y = x + \frac{1}{x} + 1$.To find the ordinate when the abscissa is $x = -2$:$y = -2 + \frac{1}{-2} + 1$$y = -2 - 0.5 + 1 = -1.5 = -\frac{3}{2}$.

If a curve passes through the point $\left( 2, \frac{7}{2} \right)$ and has slope $\left( 1 - \frac{1}{x^2} \right)$ at any point $(x, y)$ on it,then the ordinate of the point on the curve whose abscissa is $-2$ is

Similar Questions

If $(2+\sin x) \frac{dy}{dx}+(y+1) \cos x=0$ and $y(0)=1$,then $y\left(\frac{\pi}{2}\right)$ is equal to

If a curve passes through $(1, 2)$ and has the slope of its tangent $1 - \frac{1}{x^2}$ at any point $(x, y)$,then the equation of that curve is:

The solution of the differential equation $\frac{dy}{dx} = \frac{1 + y^2}{1 + x^2}$ is

If $\frac{dy}{dx} = e^{-2y}$ and $y = 0$ when $x = 5$,the value of $x$ for $y = 3$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module