Let y ( x ) be a solution of the differential equation (1 + e ^ x ) y ^ ′ + y e ^ x = 1 . If y(0) = 2 , then which of the following statements is (are) true?

Correct answer: A. y(- 4) = 0 (1 + e ^ x )dy + y e ^ x dx = dx dy + d( e ^ x . y) = dx y e ^ x = x - y + c Now, x = 0, y = 2 2 . e ^ 0 = 0 - 2 + c ⇒ c = 4 Now, y e ^ x = x - y + 4 y = x + 4 e ^ x + 1 Now, (dy)/(dx) = ( e ^ x + 1) - (x + 4) e ^ x e ^ x + 1 = e ^ x + 1 - x e ^ x - 4 e ^ x e ^ x + 1 = 1 - x e ^ x - 3 e ^ x e ^ x + 1 f ^ ′ (0) = - ve f ^ ′ (- 1) = 1 + e - 3 e ^ - 1 e ^ - 1 + 1 = + ve Critical point in (- 1, 0)

JEE Advanced 2015Paper 1

Exam: JEE Advanced
Year: 2015
Paper: Paper 1
Subject: Mathematics
Chapter: Differential Equations

JEE Advanced 2015 · Paper 1 · Mathematics · Differential Equations

Let y ( x ) be a solution of the differential equation (1 + e ^ x )...

Let

y ( x ) be a solution of the differential equation \left(1 + {e}^{x}\right){y}^{′} + y{e}^{x} = 1 .

y\left(0\right) = 2

, then which of the following statements is (are) true?

Options

A $y\left(- 4\right) = 0$
B $y\left(- 2\right) = 0$
C $y ( x )$ has a critical point in the interval (-1, 0)
D $y ( x )$ has no critical point in the interval (-1, 0)

Correct answer

A. $y\left(- 4\right) = 0$

Step-by-step solution

\left(1 + {e}^{x}\right)dy + y{e}^{x}dx = dx

<br >

dy + d\left({e}^{x} . y\right) = dx

<br >

y{e}^{x} = x - y + c

<br >Now,

x = 0, y = 2

<br >

2 . {e}^{0} = 0 - 2 + c

<br >

⇒ c = 4

<br >Now,

y{e}^{x} = x - y + 4

<br >

y = \dfrac{x + 4}{{e}^{x} + 1}

<br >Now,

\dfrac{dy}{dx} = \dfrac{\left({e}^{x} + 1\right) - \left(x + 4\right){e}^{x}}{{e}^{x} + 1}

<br >

= \dfrac{{e}^{x} + 1 - x{e}^{x} - 4{e}^{x}}{{e}^{x} + 1} = \dfrac{1 - x{e}^{x} - 3{e}^{x}}{{e}^{x} + 1}

<br >

{f}^{′}\left(0\right) = - ve

<br >

{f}^{′}\left(- 1\right) = \dfrac{1 + e - 3{e}^{- 1}}{{e}^{- 1} + 1} = + ve

<br >Critical point in \left(- 1, 0\right)

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