The value of Bob’s car can be calculated from the formula $$V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500$$ where $V$ is the value of the car in pounds (£) and $t$ is the age in years - Edexcel - A-Level Maths Pure - Question 21 - 2012 - Paper 1

To find the exact value of $t$ when $V = 9500$, set up the equation:

$9500 = 17000e^{-0.25t} + 2000e^{-0.5t} + 500$

This simplifies to:

$9500 - 500 = 17000e^{-0.25t} + 2000e^{-0.5t}$

$9000 = 17000e^{-0.25t} + 2000e^{-0.5t}$

Next, isolate the exponentials and rearrange:

$17000e^{-0.25t} + 2000e^{-0.5t} = 9000$

This is a quadratic equation in terms of $e^{-0.25t}$, which can be solved using factoring or the quadratic formula. After some calculations, the exact value of $t$ is determined to be:

t ext{ is approximately } 4 imes ext{ln}igg(rac{1}{2}igg) ext{ or } t ext{ in terms of } ext{ln} .

To find the rate of decrease of the car's value at $t = 8$, we need to differentiate the value formula. Start with:

$V = 17000e^{-0.25t} + 2000e^{-0.5t} + 500$

Using the chain rule, differentiate:

rac{dV}{dt} = -4250e^{-0.25t} - 1000e^{-0.5t}

Now, substitute $t = 8$ to find the rate of change:

rac{dV}{dt}igg|_{t=8} = -4250e^{-0.25 imes 8} - 1000e^{-0.5 imes 8}

Calculating the exponentials gives:

$e^{-2} ext{ and } e^{-4}$

Thus: rac{dV}{dt}igg|_{t=8} ext{ yields approximately } -593 ext{ pounds per year.}