The slope of the tangent to a curve C : y = y | vedclass

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

Question

(B) Given the slope of the tangent $\frac{dy}{dx} = \frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$.Multiply numerator and denominator by $e^{2x}$:$\frac{

Vedclass · Accepted Answer

$\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)$

Vedclass · Answer

(B) Given the slope of the tangent $\frac{dy}{dx} = \frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$.Multiply numerator and denominator by $e^{2x}$:$\frac{dy}{dx} = \frac{2e^{4x} - 6e^x + 9e^{2x}}{2e^{2x} + 9}$.This simplifies to $\frac{dy}{dx} = e^{2x} - \frac{6e^x}{2e^{2x} + 9}$.Integrating both sides with respect to $x$:$y = \int e^{2x} dx - \int \frac{6e^x}{2e^{2x} + 9} dx$.Let $u = \sqrt{2}e^x$,then $du = \sqrt{2}e^x dx$,so $e^x dx = \frac{du}{\sqrt{2}}$.$y = \frac{1}{2}e^{2x} - \int \frac{6}{\sqrt{2}(u^2 + 3^2)} du = \frac{1}{2}e^{2x} - \frac{6}{\sqrt{2} \cdot 3} 	an^{-1}(\frac{u}{3}) + C$.$y = \frac{1}{2}e^{2x} - \sqrt{2} 	an^{-1}(\frac{\sqrt{2}e^x}{3}) + C$.Using the point $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$:$\frac{1}{2} + \frac{\pi}{2\sqrt{2}} = \frac{1}{2} - \sqrt{2} 	an^{-1}(\frac{\sqrt{2}}{3}) + C \implies C = \frac{\pi}{2\sqrt{2}} + \sqrt{2} 	an^{-1}(\frac{\sqrt{2}}{3})$.Using the point $(\alpha, \frac{1}{2}e^{2\alpha})$:$\frac{1}{2}e^{2\alpha} = \frac{1}{2}e^{2\alpha} - \sqrt{2} 	an^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) + C$.$\sqrt{2} 	an^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) = \frac{\pi}{2\sqrt{2}} + \sqrt{2} 	an^{-1}(\frac{\sqrt{2}}{3})$.$	an^{-1}(\frac{\sqrt{2}e^{\alpha}}{3}) = \frac{\pi}{4} + 	an^{-1}(\frac{\sqrt{2}}{3})$.Taking $	an$ on both sides:$\frac{\sqrt{2}e^{\alpha}}{3} = 	an(\frac{\pi}{4} + 	an^{-1}(\frac{\sqrt{2}}{3})) = \frac{1 + \frac{\sqrt{2}}{3}}{1 - \frac{\sqrt{2}}{3}} = \frac{3+\sqrt{2}}{3-\sqrt{2}}$.$e^{\alpha} = \frac{3}{\sqrt{2}} \left(\frac{3+\sqrt{2}}{3-\sqrt{2}}ight)$.

The slope of the tangent to a curve $C : y = y(x)$ at any point $(x, y)$ on it is $\frac{2e^{2x} - 6e^{-x} + 9}{2 + 9e^{-2x}}$. If $C$ passes through the points $(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}})$ and $(\alpha, \frac{1}{2}e^{2\alpha})$,then $e^{\alpha}$ is equal to:

Similar Questions

Verify that the given function $y = \sqrt{a^{2} - x^{2}}$,where $x \in (-a, a)$,is a solution of the differential equation $x + y \frac{dy}{dx} = 0$ (where $y \neq 0$).

The microorganisms double themselves in $3$ hours. Assuming that the quantity increases at a rate proportional to itself,then the number of times it multiplies itself in $18$ hours is:

The bacteria increase at a rate proportional to the number of bacteria present. If the original number $N$ doubles in $8$ hours,then the number of bacteria in $24$ hours will be (in $N$)

The solution of the differential equation $y \frac{dy}{dx} + x = k$ represents . . . . . . .

Let $f$ be a twice differentiable non-negative function such that $(f(x))^2 = 25 + \int_{0}^{x} ((f(t))^2 + (f'(t))^2) dt$. Then the mean of $f(\log_e(1)), f(\log_e(2)), \ldots, f(\log_e(625))$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module