JEE Advanced 2015 · Paper 1 · Mathematics Q54

MathematicsMultiple correct+4 / -2Question 54 of 60

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

y(-4)=0

y(-2)=0

y(x)

has a critical point in the interval

(-1,0)

y(x)

has no critical point in the interval

(-1,0)

Answer and worked solution

Official answer: A, C

Solve. The left side is $\dfrac{d}{dx}\left[y(1+e^x)\right]=1$ , so
$y(1+e^x)=x+4\quad(\text{using }y(0)=2),\qquad y=\frac{x+4}{1+e^x}.$

(A) $y(-4)=0$ . True.
(B) $y(-2)\ne0$ . False.
(C), (D) $y'=0$ when $1+e^x=(x+4)e^x$ , i.e. $e^{-x}=x+3$ . Let $g(x)=e^{-x}-x-3$ . Then $g(-1)=e-2>0$ and $g(0)=-2<0$ , so there is a critical point in $(-1,0)$ . (C) is true and (D) is false.

(A), (C)

Topics:Linear first-order equations Monotonicity

JEE Advanced 2015 · Paper 1, Mathematics question 54