JEE Main 2022 · 25 Jun · Shift 1 · Mathematics Q15

← JEE Main 2022 · 25 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 15 of 90

Let $y=y(x)$ be the solution of the differential equation $(x+1)y'-y=e^{3x}(x+1)^2$ , with $y(0)=\dfrac13$ . Then, the point $x=-\dfrac43$ for the curve $y=y(x)$ is :

not a critical point

a point of local minima

a point of local maxima

a point of inflection

Answer and worked solution

Official answer: B

Exact form. Dividing by $(x+1)^2$ : $\left(\dfrac y{x+1}\right)'=e^{3x}$ , so $\dfrac y{x+1}=\dfrac{e^{3x}}3+C$ ; $y(0)=\frac13$ gives $C=0$ .

Derivative. $y=\dfrac{(x+1)e^{3x}}3$ , $y'=e^{3x}\left(x+\dfrac43\right)$ .

At $x=-\frac43$ . $y'$ changes from negative to positive: a local minimum.

(B)

Topics:Linear first-order equations Maxima and minima

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JEE Main 2022 · 25 Jun · Shift 1, Mathematics question 15