Let a curve y = y (x) be given by the solution of the differential equationcos ((1)/(2) ^ - 1 ( e ^ - x ))dx = ( e ^ 2x - 1 )dy . If it intersects y -axis at y = - 1 , and the intersection point of the curve with x

Correct answer: 0 We have, ((1)/(2) ^ - 1 ( e ^ - x ))dx = ( e ^ 2x - 1 )dy . . . (i) Put ^ - 1 ( e ^ - x ) = θ, θ ∈ [0, π] ⇒ cosθ = e ^ - x ⇒ 2 ^ 2 (θ)/(2) - 1 = e ^ - x ⇒ ((θ)/(2)) = e ^ - x + 1 2 = e ^ x + 1 2 e ^ x Hence, by (i) , we have ((θ)/(2))dx = ( e ^ 2x - 1 )dy ⇒ e ^ x + 1 2 e ^ x dx = e ^ 2x - 1 dy ⇒ ( e ^ x + 1 2 e ^ x )dx = ( e ^ x - 1 )( e ^ x + 1 )dy ⇒ 1 2 ∫ dx e ^ x e ^ x - 1 = ∫ dy Put e ^ x = t ⇒ dt = e ^ x dx 1 2 ∫ dt e ^ x e ^ x e ^ x - 1 = ∫ dy ∫ dt t t ^ 2 - t = 2 ∫ dy Put t = (1)/(z) ⇒ (dt)/(dz) = - 1 z ^ 2 ∫ - dz z ^ 2 (1)/(z) 1 z ^ 2 - (1)/(z) = 2 ∫ dy ⇒ - ∫ dz 1 - z = 2 ∫ dy ⇒ - 2 ( 1 - z ) ^ 1 / 2 - 1 = 2 y + c ⇒ 2 (1 - (1)/(t)) ^ 1 / 2 = 2 y + c ⇒ 2 (1 - e ^ - x ) ^ 1 / 2 = 2 y + c Now, it meets y - axis at (0, - 1) , hence 0 = - 2 + c ⇒ c = 2 Hence, 2 (1 - e ^ - x ) ^ 1 / 2 = 2 (y + 1) It passes through (α, 0) 2 (1 - e ^ - α ) ^ 1 / 2 = 2 ⇒ 1 - e ^ - α = 1 2 ⇒ 1 - e ^ - α = (1)/(2) ⇒ e ^ - α = (1)/(2) ⇒ e ^ α = 2

JEE Main 202120 JulyEvening Shift

Exam: JEE Main
Year: 2021
Date: 20 July
Shift: Evening Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2021 · 20 July · Evening Shift · Mathematics · Differential Equations

Let a curve y = y (x) be given by the solution of the differential...

Let a curve

y = y \left(x\right) be given by

the solution of the differential equationcos \left(\dfrac{1}{2}\cos^{- 1}\left({e}^{- x}\right)\right)dx = \left(\sqrt{{e}^{2x} - 1}\right)dy

. If it intersects

-axis at

y = - 1

, and the intersection point of the curve with

x -

axis is \left(\alpha, 0\right)

, then

{e}^{\alpha}

is equal to

Correct answer

Step-by-step solution

We have,

$\cos \left(\dfrac{1}{2}\cos^{- 1}\left({e}^{- x}\right)\right)dx = \left(\sqrt{{e}^{2x} - 1}\right)dy . . . \left(i\right)$

Put $\cos^{- 1}\left({e}^{- x}\right) = \theta, \theta \in \left[0, \pi\right]$

$⇒ cos\theta = {e}^{- x}$

$⇒ 2\cos^{2}\dfrac{\theta}{2} - 1 = {e}^{- x}$

$⇒ \cos \left(\dfrac{\theta}{2}\right) = \sqrt{\dfrac{{e}^{- x} + 1}{2}} = \sqrt{\dfrac{{e}^{x} + 1}{2{e}^{x}}}$

Hence, by \left(i\right) $, we have$ \cos \left(\dfrac{\theta}{2}\right)dx = \left(\sqrt{{e}^{2x} - 1}\right)dy $$ ⇒ \sqrt{\dfrac{{e}^{x} + 1}{2{e}^{x}}}dx = \sqrt{{e}^{2x} - 1}dy $$ ⇒ \left(\sqrt{\dfrac{{e}^{x} + 1}{2{e}^{x}}}\right)dx = \left(\sqrt{{e}^{x} - 1}\right)\left(\sqrt{{e}^{x} + 1}\right)dy $$ ⇒ \dfrac{1}{\sqrt{2}} ∫ \dfrac{dx}{\sqrt{{e}^{x}}\sqrt{{e}^{x} - 1}} = ∫ dy $Put$ {e}^{x} = t ⇒ dt = {e}^{x}dx $$ \dfrac{1}{\sqrt{2}} ∫ \dfrac{dt}{{e}^{x}\sqrt{{e}^{x}}\sqrt{{e}^{x} - 1}} = ∫ dy $$ ∫ \dfrac{dt}{t\sqrt{{t}^{2} - t}} = \sqrt{2} ∫ dy $Put$ t = \dfrac{1}{z} ⇒ \dfrac{dt}{dz} = - \dfrac{1}{{z}^{2}} $$ ∫ \dfrac{- \dfrac{dz}{{z}^{2}}}{\dfrac{1}{z}\sqrt{\dfrac{1}{{z}^{2}} - \dfrac{1}{z}}} = \sqrt{2} ∫ dy $$ ⇒ - ∫ \dfrac{dz}{\sqrt{1 - z}} = \sqrt{2} ∫ dy $$ ⇒ \dfrac{- 2 ( 1 - z{)}^{1 / 2}}{- 1} = \sqrt{2}y + c $$ ⇒ 2{\left(1 - \dfrac{1}{t}\right)}^{1 / 2} = \sqrt{2}y + c $$ ⇒ 2{\left(1 - {e}^{- x}\right)}^{1 / 2} = \sqrt{2}y + c $Now, it meets$ y - \text{axis} $at \left(0, - 1\right)$ , hence

$0 = - \sqrt{2} + c ⇒ c = \sqrt{2}$

Hence,

$2{\left(1 - {e}^{- x}\right)}^{1 / 2} = \sqrt{2}\left(y + 1\right)$

It passes through \left(\alpha, 0\right) $$ 2{\left(1 - {e}^{- \alpha}\right)}^{1 / 2} = \sqrt{2} $$ ⇒ \sqrt{1 - {e}^{- \alpha}} = \dfrac{1}{\sqrt{2}} $$ ⇒ 1 - {e}^{- \alpha} = \dfrac{1}{2} $$ ⇒ {e}^{- \alpha} = \dfrac{1}{2} ⇒ {e}^{\alpha} = 2

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