Question

If the Laplace transform of a function \(f(t)\) is given by \[ \frac{s+3}{(s+1)(s+2)}, \] then the value of \(f(\ln2)\) is

• A. \(0\)
• B. \(3\)
• C. \(2\)
• D. \(\dfrac34\)

Hint:

Remember these standard inverse Laplace transforms: \[ \boxed{ L^{-1}\left\{\frac1{s+a}\right\}=e^{-at} } \] After finding \(f(t)\), substitute the required value of \(t\) carefully using \[ e^{-\ln a}=\frac1a. \]

Answer 1

The Correct Option is D

Concept: To find the original function from its Laplace transform, we first express the given transform using partial fractions and then apply the inverse Laplace transform. Useful formulas are \[ \boxed{ L^{-1}\left\{\frac1{s+a}\right\}=e^{-at} } \]

Step 1: Resolve into partial fractions.
Let \[ \frac{s+3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}. \] Multiplying throughout, \[ s+3=A(s+2)+B(s+1). \] Putting \(s=-1\), \[ 2=A. \] Putting \(s=-2\), \[ 1=-B \] or \[ B=-1. \] Hence, \[ \frac{s+3}{(s+1)(s+2)} = \frac2{s+1} - \frac1{s+2}. \]

Step 2: Take inverse Laplace transform.
Therefore, \[ f(t) = 2e^{-t}-e^{-2t}. \]

Step 3: Substitute \(t=\ln2\).
Since \[ e^{-\ln2} =\frac12, \] and \[ e^{-2\ln2} =\frac14, \] we get \[ f(\ln2) = 2\left(\frac12\right)-\frac14 = 1-\frac14 = \frac34. \] Hence, \[ \boxed{ f(\ln2)=\frac34 } \]

Step 4: Choose the correct option.
\[ \boxed{Option (D) is correct \]