JEE Main 8 Apr 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 6 of 90

Marks: +1, -0

Let f(x) be a positive function such that the area bounded by y=f(x),y=0 from x=0 to x=a>0 is e−a+4a2+a−1. Then the differential equation, whose general solution is y=c1f(x)+c2, where c1 and c2 are arbitrary constants, is :

[8 Apr 2024 Shift 1]

(8ex−1)‌
d2y
dx2
+‌
dy
dx
=0
(8ex+1)‌
d2y
dx2
−‌
dy
dx
=0
(8ex+1)‌
d2y
dx2
+‌
dy
dx
=0
(8ex−1)‌
d2y
dx2
−‌
dy
dx
=0

Validate

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