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  • The slope of the tangent to a curve at ay point <img alt="\mathrm{(x, y)}" src="/latex-image/?%5Cmathrm%7B%28

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

Option: 1

\frac{3+\sqrt{2}}{3-\sqrt{2}} \\


Option: 2

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


Option: 3

\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right) \\


Option: 4

\frac{\sqrt{2}+1}{\sqrt{2}-1}


Answers (1)

best_answer

\mathrm{\frac{d y}{d x}=\frac{2 e^{2 x}-6 x^{-x}+9}{2+9 e^{-2 x}}}

\mathrm{\frac{d y}{d x}=e^{2 x}-\frac{6 e^{x}}{2 e^{2 x}+9}} \\

\mathrm{\int d y=\int \cdot\left(e^{2 x}-\frac{6 e^{x}}{2 e^{2 x}+9}\right) d x} \\

\mathrm{y=\frac{e^{2 x}}{2}-\tan ^{-1}\left(\frac{\sqrt{2} e^{x}}{3}\right)+c}

If c passes through the point \mathrm{\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)}

\mathrm{c=\frac{-\pi}{4}-\tan ^{-1} \frac{\sqrt{2}}{3}}

Again c passes through the point \mathrm{\left(\alpha, \frac{1}{2} e^{2 \alpha}\right)}

then \mathrm{e^{\alpha}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)}

Hence correct option is 2

Posted by

Suraj Bhandari

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