Let I be the purchase value of an equipment and V(t) be the value after it has been used for t years. The value V(t) depreciates at a rate given by differential equation `(d V(t)/(dt)=-k(T-t)` , where `k"">""0` is a constant and T is the total life in years of the equipment. Then the scrap value V(T) of the equipment is :
(1) `T^2-1/k`
(2) `I-(k T^2)/2`
(3) `I-(k(T-t)^2)/2`
(4) `e^(-k T)`

Options

A. `e^(-kT)`
B. `T^(2)-I/k`
C. `I-(kT^(2))/2`
D. `I-(k(T-t)^(2))/(2)`

Answer

C

Explanation

Since total life is T, scrap value is V(T).
We have
`(dV)(t)=-k(T-t)dt`
`rArr int_(I)^(V(t))dV(t) = int_(t=0)^(T)-k(T-t)dt`
`rArr V(T)-I = k[((T-t)^(2))/(2)]_(0)^(T)`
`rArr V(T)-I = -k[t^(2)/2]`
`rArr V(T)=I-(kT^(2))/2`