The slope of the tangent to a curve C:y=y(x) at any point [x,y) on it is 2e^2x-6e^-x+9 2+9e^-2x. If C passes through the points (0, 1 2+ π 2√2) and…

Given,

$$\frac{dy}{dx}=\frac{2{e}^{2x}-6{e}^{-x}+9}{2+9{e}^{-2x}}$$dxdy​=2+9e−2x2e2x−6e−x+9​

On rearranging we get,

$$\frac{dy}{dx}={e}^{2x}-\frac{6{e}^{x}}{2{e}^{2x}+9}$$dxdy​=e2x−2e2x+96ex​

Integrating both side we get,

$$y=\frac{{e}^{2x}}{2}-\sqrt{2}{\mathrm{tan}}^{-1}(\frac{\sqrt{2}{e}^{x}}{3})+c$$y=2e2x​−2​tan−1(32​ex​)+c

If the curve passes through the point $(0,\frac{1}{2}+\frac{\pi }{2\sqrt{2}})$(0,21​+22​π​)

Then $c=\sqrt{2}(\frac{\pi }{4}+{\mathrm{tan}}^{-1}\frac{\sqrt{2}}{3})$c=2​(4π​+tan−132​​)

So, curve will be $y=\frac{{e}^{2x}}{2}-\sqrt{2}({\mathrm{tan}}^{-1}(\frac{\sqrt{2}{e}^{x}}{3})-\frac{\pi }{4}-{\mathrm{tan}}^{-1}\frac{\sqrt{2}}{3})$y=2e2x​−2​(tan−1(32​ex​)−4π​−tan−132​​)

Again curve passes through the point

$$(\alpha ,\frac{1}{2}{e}^{2\alpha })$$(α,21​e2α)

Putting the value in curve equation we get,

$$\frac{{e}^{2\alpha }}{2}=\frac{{e}^{2\alpha }}{2}-\sqrt{2}({\mathrm{tan}}^{-1}(\frac{\sqrt{2}{e}^{\alpha }}{3})-\frac{\pi }{4}-{\mathrm{tan}}^{-1}\frac{\sqrt{2}}{3})$$2e2α​=2e2α​−2​(tan−1(32​eα​)−4π​−tan−132​​)

$$\Rightarrow {\mathrm{tan}}^{-1}(\frac{\sqrt{2}{e}^{\alpha }}{3})=\frac{\pi }{4}+{\mathrm{tan}}^{-1}\frac{\sqrt{2}}{3}$$⇒tan−1(32​eα​)=4π​+tan−132​​

$$\Rightarrow \frac{\sqrt{2}{e}^{\alpha }}{3}=\mathrm{tan}(\frac{\pi }{4}+{\mathrm{tan}}^{-1}\frac{\sqrt{2}}{3})$$⇒32​eα​=tan(4π​+tan−132​​)

$$\Rightarrow \frac{\sqrt{2}{e}^{\alpha }}{3}=(\frac{\mathrm{tan}\frac{\pi }{4}+{\mathrm{tantan}}^{-1}\frac{\sqrt{2}}{3}}{1-\mathrm{tan}\frac{\pi }{4}\times {\mathrm{tantan}}^{-1}\frac{\sqrt{2}}{3}})$$⇒32​eα​=(1−tan4π​×tantan−132​​tan4π​+tantan−132​​​)

$$\Rightarrow \frac{\sqrt{2}{e}^{\alpha }}{3}=(\frac{1+\frac{\sqrt{2}}{3}}{1-\frac{\sqrt{2}}{3}})$$⇒32​eα​=(1−32​​1+32​​​)

$$\Rightarrow {e}^{\alpha }=\frac{3}{\sqrt{2}}(\frac{3+\sqrt{2}}{3-\sqrt{2}})$$⇒eα=2​3​(3−2​3+2​​)