JEE Main 2022 · 26 Jun · Shift 2 · Mathematics Q9

← JEE Main 2022 · 26 Jun · Shift 2

MathematicsSingle correct+4 / -1Question 9 of 90

If $y=y(x)$ is the solution of the differential equation $x\dfrac{dy}{dx}+2y=xe^x$ , $y(1)=0$ then the local maximum value of the function $z(x)=x^2y(x)-e^x$ , $x\in\mathbb R$ is :

1-e

\dfrac12

\dfrac4e-e

Answer and worked solution

Official answer: D

Exact form. Multiply by $x$ : $(x^2y)'=x^2e^x$ , so $x^2y=(x^2-2x+2)e^x+C$ ; $y(1)=0$ gives $C=-e$ .

Function. $z=x^2y-e^x=(x-1)^2e^x-e$ , $z'=e^x(x-1)(x+1)$ .

Local maximum at $x=-1$ : $z=\dfrac4e-e$ .

(D)

Topics:Linear first-order equations Maxima and minima

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JEE Main 2022 · 26 Jun · Shift 2, Mathematics question 9