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QJEEM22Maad88: JEE Main 2022 Mathematics - Class 12 JEEM22Maad88 Solution: The slope of the tangent to a curve at any point on it is . If…

The slope of the tangent to a curve C:y=y(x)C: y=y(x) at any point (x,y)(x, y) on it is 2e2x−6e−x+92+9e−2x\frac{2 \mathrm{e}^{2 x}-6 \mathrm{e}^{-x}+9}{2+9 \mathrm{e}^{-2 x}} . If CC passes through the points (0,12+π22)\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right) and (α,12e2α)\left(\alpha, \frac{1}{2} \mathrm{e}^{2 \alpha}\right) , then eα\mathrm{e}^{\alpha} is equal to :

Multiple Choice Options

  • A 3+23−2\frac{3+\sqrt{2}}{3-\sqrt{2}}
  • B 32(3+23−2)\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)
  • C 12(2+12−1)\frac{1}{\sqrt{2}}\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)
  • D 2+12−1\frac{\sqrt{2}+1}{\sqrt{2}-1}

Step-by-Step Verified Solution

dydx=2e2x−6e−x+92+9e−2x=e2x−6e−x2+9e−2x\frac{d y}{d x}=\frac{2 e^{2 x}-6 e^{-x}+9}{2+9 e^{-2 x}}=e^{2 x}-\frac{6 e^{-x}}{2+9 e^{-2 x}}

amp;∫dy=∫e2xdx−3∫1+(3e−x2)2⏟put e−x=tdxamp;=e2x2+3∫dt1+(3t2)2amp;=e2x2+2tan⁡−13t2+C\begin{aligned}&\int d y=\int e^{2 x} d x-3 \int \underbrace{1+\left(\frac{3 e^{-x}}{\sqrt{2}}\right)^{2}}_{\text {put } e^{-x}=t} d x \\\\&=\frac{e^{2 x}}{2}+3 \int \frac{d t}{1+\left(\frac{3 t}{\sqrt{2}}\right)^{2}} \\\\&=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1} \frac{3 t}{\sqrt{2}}+C\end{aligned}

y=e2x2+2tan⁡−1(3e−x2)+Cy=\frac{e^{2 x}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-x}}{\sqrt{2}}\right)+C

It is given that the curve passes through

(0,12+π22)\left(0, \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}\right)

amp;12+π22=12+2tan⁡−1(32)+C\begin{aligned}& \frac{1}{2}+\frac{\pi}{2 \sqrt{2}}=\frac{1}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)+C\end{aligned}

⇒C=π22−2tan⁡−1(32)\Rightarrow \quad C=\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)

Now if (α,12e2α)\left(\alpha, \frac{1}{2} e^{2 \alpha}\right) satisfies the curve, then

12e2α=e2α2+2tan⁡−1(3e−α2)+π22−2tan⁡−1(32)\frac{1}{2} e^{2 \alpha}=\frac{e^{2 \alpha}}{2}+\sqrt{2} \tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)+\frac{\pi}{2 \sqrt{2}}-\sqrt{2} \tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)

tan⁡−1(32)−tan⁡−1(3e−α2)=π22×12=π4\tan ^{-1}\left(\frac{3}{\sqrt{2}}\right)-\tan ^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right)=\frac{\pi}{2 \sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{\pi}{4}

32−3e−α21+92e−α=1\frac{\frac{3}{\sqrt{2}}-\frac{3 e^{-\alpha}}{\sqrt{2}}}{1+\frac{9}{2} e^{-\alpha}}=1

32eα−32=eα+92\frac{3}{\sqrt{2}} e^{\alpha}-\frac{3}{\sqrt{2}}=e^{\alpha}+\frac{9}{2}

eα=92+3232−1=32(3+23−2)e^{\alpha}=\frac{\frac{9}{2}+\frac{3}{\sqrt{2}}}{\frac{3}{\sqrt{2}}-1}=\frac{3}{\sqrt{2}}\left(\frac{3+\sqrt{2}}{3-\sqrt{2}}\right)

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