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JEE MAIN 2022
25-6-2022 S1
Question
Let $y=y(x)$ be the solution of the differential equation $(x+1) y^{\prime}-y=e^{3 x}(x+1)^2$, with $y(0)=\frac{1}{3}$. Then, the point $x =-\frac{4}{3}$ for the curve $y=y(x)$ is :
Select the correct option:
A
not a critical point
B
a point of local minima
C
a point of local maxima
D
a point of inflection
✓ Correct! Well done.
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Solution
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Question Tags
JEE Main
Mathematics
Medium
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