If L\left\{f(t)\right\}=F(s) , then L{f(4t)} (a)frac(1)(4) F(s) (b)F(frac(s)(4)) (c)4F(s) (d)frac(1)(4) F(frac(s)(4))

If
a) $\frac{1}{4}F\left(s\right)$
b) $F\left(\frac{s}{4}\right)$
c) $4F\left(s\right)$
d) $\frac{1}{4}F\left(\frac{s}{4}\right)$

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Step 1
Here we solve this from the definition of Laplace transformation.
step 2
$L\left(f\left(t\right)\right)=F\left(s\right)={\int }_{0}^{\mathrm{\infty }}f\left(t\right){e}^{-st}dt$
$\to L\left(f\left(4t\right)\right)={\int }_{0}^{\mathrm{\infty }}f\left(4t\right){e}^{-st}dt$
Step 3
Let us take transformation $z=4t\to dt=\frac{dz}{4}$
Then the integration becomes
$L\left(f\left(4t\right)\right)={\int }_{0}^{\mathrm{\infty }}f\left(4t\right){e}^{-st}dt={\int }_{0}^{\mathrm{\infty }}f\left(z\right){e}^{-\frac{sz}{4}}\frac{dz}{4}$
$=\frac{1}{4}{\int }_{0}^{\mathrm{\infty }}f\left(z\right){e}^{\frac{s}{4}z}dz=\frac{1}{4}F\left(\frac{s}{4}\right)$
Step 4
The final answer is $L\left(f\left(4t\right)\right)=\frac{1}{4}F\left(\frac{s}{4}\right)$ so 4th option is correct.