Q34 / 60 · JEE Advanced 2015 (Paper 1) · Mathematics

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
$y (- 4) = 0$
B
$y (- 2) = 0$
Correct answer
C
$y (x)$ has a critical point in the interval (-1, 0)
D
$y (x)$ has no critical point in the interval (-1, 0)

Solution

(1 + e^{x}) d y + y e^{x} d x = d x

d y + d (e^{x} . y) = d x

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

Now,

x = 0, y = 2

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

\Rightarrow c = 4

Now,

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

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

Now,

\frac{d y}{d x} = \frac{(e^{x} + 1) - (x + 4) e^{x}}{e^{x} + 1}

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

f^{'} (0) = - v e

f^{'} (- 1) = \frac{1 + e - 3 e^{- 1}}{e^{- 1} + 1} = + v e

Critical point in

(- 1, 0)

Hence, the correct answer is options: 'A', 'C'.

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