Discussion
If the failure rate function \(r\) of a component is \(r(t) = \dfrac{t}{1 + t}\), \(t \ge 0\); then its survival function is :
Answer: D
The survival function is \((1 + t)e^{-t}\).
The survival function is \(S(t) = \exp\left(-\int_{0}^{t} r(u)\,du\right)\). Since \(\dfrac{u}{1 + u} = 1 - \dfrac{1}{1 + u}\),
\[ \int_{0}^{t}\frac{u}{1 + u}du = t - \ln(1 + t) \]
so \(S(t) = e^{-t + \ln(1 + t)} = (1 + t)e^{-t}\). Check: \(S(0) = 1\), and \(-S'(t)/S(t) = t/(1 + t)\).
\(e^{-t}\) has constant hazard 1, \(te^{-t}\) is 0 at \(t = 0\), and \(1 - e^{-t}\) is a distribution function, not a survival function. (This is the gamma(2, 1) lifetime.)
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