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Given f(t)=L⁻¹ [(3s+1)/(s³+4s²+(K-3)s) ]. If lim_(t→∞) f(t)=1, then the value of K is

1. 1
2. 2
3. 3
4. 4

Correct answer: 4

Solution

By final value theorem, lim f(t)=lim_{s->0} sF(s)=(3s+1)/(s^2+4s+K-3) at s=0 = 1/(K-3)=1, so K-3=1 and K=4. Stored value 2 is wrong.

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