1. A car enters a toll road at point A at 2:40pm and exits the toll road at poin
ID: 2891385 • Letter: 1
Question
1. A car enters a toll road at point A at 2:40pm and exits the toll road at point B at 5:00pm. If point A and point B are 168 miles apart and the speed limit on the toll road is 65 miles per hour, prove that the car was speeding at some point during its journey. Do we know at what time the car was speeding? 2. Use the following ideas to show that the equation 2x + sin x = 1 has exactly one solution. (a) Use the Intermediate Value Theorem to show that there is at least one solution (b) Assuming there are two solutions, what would have to exist from the Mean Value in the interval [0, Theorem? Why can that not exist?Explanation / Answer
1) Distance = 168 miles
Time taken = Time at point B - Time at point A = 5:00 pm - 2:40 pm = 2 hour 20 minutes = 140 minutes
Average Speed of the car in this range = Total Distance Travelled/ Total time taken = 168/140 = 1.2 miles/minute
For converting the speed into miles/hour = miles/minute * 60 = 1.2 mi/min * 60 min/1 hr = 72 mi/hr
Since the average speed is greater than the speed limit on the highway, hence the car is speeding at some point in the journey
With this information provided, we cannot find the exact point where the car is speeding up
Q2)
a) Using the intermediate value theorem, if the function takes value f(a) and f(b) at the end points then it will take any value between f(a) and f(b) in the range of (a,b)
f(x) = 2x + sinx - 1
f(0) = 2(0) + sin(0) - 1 = -1
f(pi) = 2(pi) + sin(pi) - 1 = 2*pi -1
Since f(0) and f(pi) has different signs hence there will be a cross over point where the function will have the zero value
Hence using the IVT there exists atleast one solution
b)
f(x) = 2x + sinx - 1
f'(x) = 2 + cosx
f'(c) = f(b)-f(a)/(b-a) = 2*pi-2/(pi-0) =2 - 2/pi
2 -2/pi = 2 + cosx
cos(x) = -2/pi
Hence the x value will 129.564 and one more existing in 2pi
Therefore there are two c values that satisfy the equation
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