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Nuclei of radioactive element A are prod...

Nuclei of radioactive element `A` are produced at rate `t^(2)` (where `t` is time) at any time `t`. The element `A` has decay constant `lambda`. Let `N` be the number of nuclei of element `A` at any time `t`. At time `t=t_(0), dN//dt` is minimum. The number of nuclei of element `A` at time `t=t_(0)` is

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The correct Answer is:
`(2t_(0)-lambdat_(0)^(2))/(lambda_(2))`
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