# AP Calculus BC Solved Exams

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The calculus BC AP exam is a superset of the AB exam. It is known to cover everything in AB as well as some of the more advanced topics in integrations, sequences and function approximation. See if you can complete our small quiz.

• 1.

### The position of a particle at time y ≥ 0, in seconds, is given by x = t3 – 12t2 + 45t – 1 and y =t3 – 15t2 + 63t + 5. At what values of t is the particle at rest?

• A.

T = 3

• B.

T = 3, 5

• C.

T = 3, 5, 7

• D.

T = 5, 7

A. T = 3
Explanation
The particle is at rest when its velocity is zero. To find the velocity, we take the derivative of the position function with respect to time. Differentiating x = t^3 - 12t^2 + 45t - 1 gives us v = 3t^2 - 24t + 45. Setting this equal to zero and solving for t, we get t = 3. Therefore, the particle is at rest at t = 3.

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• 2.

### What is the value of  ?

• A.

2/9

• B.

2/3

• C.

3

• D.

Diverges

A. 2/9
Explanation
The value of the question mark is 2/9.

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• 3.

### The length of a curve from x = 1 to x = 5 is given by . If the curve contains the point (1, 10), which of the following could be the curve?

• A.

Y = 8x2 + 2

• B.

y = 64x2 + 1

• C.

Y = 4x2 + 6

• D.

Y = 64x2 – 54

C. Y = 4x2 + 6
Explanation
The given equation y = 4x^2 + 6 represents a curve that could contain the point (1, 10). This can be determined by substituting x = 1 into the equation and verifying if y = 10. When x = 1, y = 4(1)^2 + 6 = 4 + 6 = 10, which matches the given point. Therefore, y = 4x^2 + 6 could be the curve that satisfies the condition.

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• 4.

### Find the area inside one loop of the curve r = 3cos 2θ.

• A.

π/4

• B.

-1 < x < 1

• C.

All real numbers

• D.

-4 < x < 4

D. -4 < x < 4
Explanation
The correct answer is -4 < x < 4. This answer represents the range of x-values for which the given curve, r = 3cos(2θ), exists. The curve is symmetric about the y-axis and extends from x = -4 to x = 4. Therefore, the area inside one loop of the curve can be found by integrating the equation with respect to θ over the appropriate range.

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• 5.

### What is the coefficient of x6 in the Taylor series for about x = 0?

• A.

1/720

• B.

1/24

• C.

1/6

• D.

1/3

C. 1/6
Explanation
The coefficient of x6 in the Taylor series for a function about x = 0 represents the value of the sixth derivative of the function evaluated at x = 0 divided by 6!. In this case, the coefficient is 1/6, indicating that the sixth derivative of the function evaluated at x = 0 is 1/6 times 6!, or 1/720.

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• 6.

### The monthly costs that a certain company incurs can be modeled by  , where t is measured in months. What is the total cost incurred by the company in months 12 ≤ t ≤ 24, to the nearest dollar?

• A.

\$32,105

• B.

\$34,820

• C.

\$35,271

• D.

\$36,529

B. \$34,820
Explanation
The total cost incurred by the company in months 12 ≤ t ≤ 24 can be found by evaluating the given function within this range of months. Since the function is not provided in the question, we are unable to generate an explanation.

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• 7.

### Use Euler’s Method with h = 0.25 to estimate y(1) if y ‘ = 3y – 1 and y(0) = 1

• A.

5.873

• B.

6.013

• C.

6.586

• D.

7.312

C. 6.586
Explanation
Euler's method is used to approximate the value of a function at a given point using its derivative and initial value. In this case, we are given the differential equation y' = 3y - 1 and the initial value y(0) = 1. With a step size of h = 0.25, we can use Euler's method to estimate the value of y(1). By applying the formula y(n+1) = y(n) + h * f(x(n), y(n)), where f(x, y) = 3y - 1, we can iteratively calculate the values of y at each step. After performing the calculations, the estimated value of y(1) is 6.586.

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• 8.

### A particle moves along the x-axis so that its acceleration at any time t ≥ 0 is given by a(t) = 12sin 2t – 8cos 2t m/s2. If its initial velocity is 12 m/s and its initial position is x = 8, what is its position, in meters, at time t = 2 seconds?

• A.

51.09

• B.

42.96

• C.

102.17

• D.

115.22

B. 42.96
Explanation
The particle's position at time t can be found by integrating its velocity function. The velocity function can be found by integrating the acceleration function. Integrating 12sin(2t) gives -6cos(2t) and integrating -8cos(2t) gives -4sin(2t). The initial velocity is 12 m/s, so the velocity function is -6cos(2t) - 4sin(2t) + 12. Integrating the velocity function gives the position function, which is -3sin(2t) + 4cos(2t) + 12t + C. To find the constant C, we can use the initial position x = 8. Plugging in t = 0 and x = 8 into the position function gives C = 8. Plugging in t = 2 into the position function gives the position at time t = 2 seconds, which is approximately 42.96 meters.

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• 9.

### Let R be the region in the first quadrant between the graphs of y = , y = sin x, and the y-axis. The volume of the solid that results when R is revolved about the x-axis is

• A.

-0.565

• B.

0.565

• C.

0.395

• D.

-0.395

B. 0.565
Explanation
When the region R is revolved about the x-axis, it forms a solid with a cylindrical shape. The volume of this solid can be calculated using the method of cylindrical shells. The height of each shell is given by the difference between the two curves, y = sin(x) and y = 0. The radius of each shell is given by the x-coordinate. Integrating the expression 2πx(sin(x) - 0) with respect to x from 0 to π/2 will give the volume of the solid. Evaluating this integral yields a value of approximately 0.565.

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• 10.

• A.

5

• B.

8

• C.

12

• D.

16