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JEE AdvancedMedium

Q.A force varies with displacement as F=kx+bx2F=kx+bx^2. The potential energy at displacement x is

a
12kx2+13bx3\frac{1}{2}kx^2+\frac{1}{3}bx^3
b
kx2+bx3kx^2+bx^3
c
12kx2+12bx2\frac{1}{2}kx^2+\frac{1}{2}bx^2
d
kx+bx2kx+bx^2

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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