What is the differentiation of e^-alpha*t

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Q. The nuclei of a radioactive element is produced at constant rate

α

and they decay with decay constant

λ

. At

t = 0

, number of nuclei is zero, then the number of nuclei at time

t

If nuclei of a radioactive element is produced at constant rate $α$ and they decay with decay constant $λ$ . At t = 0, number of nuclei is zero than the number of nuclei at time t is

Q. The option(s) with the values of

α

and

L

that satisfies the following equation is (are)

\frac{\int_{0}^{4 π} e^{t} ({sin}^{6} α t + {cos}^{4} α t) d t}{\int_{0}^{π} e^{t} ({sin}^{6} α t + {cos}^{4} α t) d t} = L,

is/are

Q. Differentiate :

e^{x} + e^{- x} = t

Q. Nuclei of a radioactive element is produced at constant rate

α

and they decay with decay constant

λ

. At

t = 0

, the number of nuclei is 0. The number of active nuclei at time

t

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