# AP Calculus: The Fundamental Theorem Of Calculus! Practice Test!

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Advanced Placement Calculus is not for everybody or do you think it is? Well, Fundamental theorem under AP Calculus basically deals with function, integration and derivation and while many see it as hard but to crack, we think its a fun topic for a start and would really advise you to take this quick test quiz on it just to boost your knowledge of the topic. Enjoy!

• 1.

### What is the inverse function of f(x) = x/x+1?

• A.

F^-1 (x) = –x/x–1

• B.

F²(x) = x/x-2²

• C.

F^-1 (x) = x/x + 1

• D.

F^-1 (x) = x-1/ x + 2

A. F^-1 (x) = –x/x–1
Explanation
The inverse function of f(x) = x/x+1 is f^-1 (x) = -x/x-1. This can be determined by swapping the x and y variables and solving for y.

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• 2.

### Suppose the operation 0 is defined by the relation p 0 q = max {p, q}. Evaluate 304

• A.

41

• B.

4

• C.

40

• D.

400

B. 4
Explanation
The operation 0 is defined as taking the maximum value between two numbers. In this case, we need to evaluate 304 and 41 using this operation. The maximum value between 304 and 41 is 304. Next, we need to evaluate the result (304) and 4 using the same operation. The maximum value between 304 and 4 is 304. Continuing this process, we evaluate 304 and 40, resulting in 304. Finally, we evaluate 304 and 400, resulting in 400. Therefore, the final answer is 400.

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• 3.

### What is the derivative of y = X^7 + cosX + sinX + √x + 3√x²?

• A.

7x6 - sinX + cosX + 1/ 3√x + 2/ 3√x

• B.

7x²- cosX + 1/2x + 3/ 2√x

• C.

7x³ - sinX - 12√2x - cos X

• D.

7x - sinX + cosX - 2√x²

A. 7x6 - sinX + cosX + 1/ 3√x + 2/ 3√x
Explanation
The given function is a sum of different terms, each representing a different type of function. To find the derivative of the function, we apply the power rule for differentiation to the term with x raised to the power of 7, which gives us 7x^6. For the terms involving trigonometric functions (cosX and sinX), we use the differentiation rules for trigonometric functions, which result in -sinX and cosX, respectively. For the term with the square root of x, we use the power rule for differentiation and chain rule, which gives us 1/(2√x). Finally, for the term with the square root of x squared, we again use the power rule for differentiation and chain rule, resulting in 3/(2√x). Therefore, the correct answer is 7x^6 - sinX + cosX + 1/(3√x) + 2/(3√x).

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• 4.

### What is the differential coefficient of y = 2x/4 + x?

• A.

2/ (8 + x)²

• B.

4/ (1 - x)³

• C.

8/ (4 + x)²

• D.

8/ 4 + x²

C. 8/ (4 + x)²
Explanation
The differential coefficient, also known as the derivative, of a function is a measure of its rate of change at a specific point. In this question, the function is y = 2x/4 + x. To find its derivative, we can use the power rule and the sum rule of differentiation. The power rule states that the derivative of x^n is n*x^(n-1), and the sum rule states that the derivative of the sum of two functions is the sum of their derivatives. Applying these rules, we can find that the derivative of y = 2x/4 + x is 2/4 + 1 = 1/2 + 1 = 3/2. However, none of the given answer options match this result, so the correct answer is not provided.

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• 5.

### Find dy/dx of sinX + x / x² + cosX

• A.

X - sinX + cosX + √x + 3√x / x²- cosX

• B.

X²- cosX + 1/2x + 3/ 2√x/ (2x²- sinX)

• C.

X³ - sinX - √2x - cos X / (x² + cosX)²

• D.

(x² + 1) cosX - XsinX - x² + 1 / (x² + cosX)

D. (x² + 1) cosX - XsinX - x² + 1 / (x² + cosX)
Explanation
The given answer is the result of finding the derivative of the given expression. To find the derivative, we can use the quotient rule, which states that the derivative of a quotient of two functions is equal to (f'(x)g(x) - g'(x)f(x))/g(x)^2. By applying the quotient rule to the given expression, we can simplify and rearrange the terms to obtain the answer (x² + 1) cosX - XsinX - x² + 1 / (x² + cosX).

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• 6.

### U = ∅sin∅/ 2∅² + 1 find dy/dx

• A.

(2∅² + 1) ∅cos∅ + (1 - 2∅²) sin∅ / (2∅²+ 1)²

• B.

(3∅² + 1) ∅cos∅ + (1 - 2∅³) sin∅ / (2∅+ 1)

• C.

(2∅² + 1) ∅cos∅ + (1 - 2∅²) sin∅ / (2∅²+ 1)²

• D.

(∅² + 1) cos∅ + (1 - ∅²) sin∅ / (2∅²+ 1)²

C. (2∅² + 1) ∅cos∅ + (1 - 2∅²) sin∅ / (2∅²+ 1)²
Explanation
The given expression represents the derivative of y with respect to x, which is denoted as dy/dx. The expression is in the form of a quotient rule, where the numerator is the derivative of the function inside the sine function, and the denominator is the square of the function inside the sine function plus 1. The correct answer is (2∅² + 1) ∅cos∅ + (1 - 2∅²) sin∅ / (2∅²+ 1)², as it correctly represents the derivative of the given expression.

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• 7.

### Find dy/dx of y = sin²X

• A.

2sinX cosX

• B.

SinX cosX

• C.

Sin²X cos⁴X

• D.

2sinX 2cosX

A. 2sinX cosX
Explanation
The correct answer is 2sinX cosX. To find the derivative of y = sin²X, we can use the chain rule. The derivative of sin²X is equal to 2sinX times the derivative of sinX, which is cosX. Therefore, the derivative of y = sin²X is 2sinX cosX.

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• 8.

### Differentiating y = 1/√x + 2 with respect to x gives:

• A.

1√x³ + 2

• B.

− 1/2√x³

• C.

2 - 1/2√x³

• D.

2/√x³

A. 1√x³ + 2
Explanation
The correct answer is 1/√x³ + 2. When differentiating y = 1/√x + 2 with respect to x, we use the power rule for differentiation. The power rule states that when differentiating a function of the form f(x) = x^n, the derivative is given by f'(x) = n*x^(n-1). Applying this rule, the derivative of 1/√x is -1/2√x³, and the derivative of 2 is 0. Therefore, the derivative of y = 1/√x + 2 is 1/√x³ + 0, which simplifies to 1/√x³ + 2.

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• 9.

### Given f(t) = 3t⁴ − 2, f(t) is equal to:

• A.

12t³ − 2

• B.

3/4 t5 − 2t + c

• C.

12t³

• D.

3t5 − 2

C. 12t³
Explanation
The given function is f(t) = 3t⁴ - 2. To find the value of f(t), we substitute the given value of t into the function. Therefore, f(t) = 3t³ - 2, which simplifies to 12t³.

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• 10.

### An alternating voltage is given by v = 10 sin 300t volts, where t is the time in seconds. The rate of change of voltage when t = 0.01 s is:

• A.

-2996V/s

• B.

157V/s

• C.

-2970V/s

• D.

0.523V/s

C. -2970V/s
Explanation
The given equation represents an alternating voltage that varies with time according to the sine function. To find the rate of change of voltage at a specific time, we need to take the derivative of the voltage equation with respect to time.

Taking the derivative of v = 10 sin 300t with respect to t, we get dv/dt = 300 * 10 cos 300t.

Now, we can substitute the value of t = 0.01s into the derivative equation to find the rate of change of voltage at that time.

dv/dt = 300 * 10 cos (300 * 0.01) = 300 * 10 cos 3 = 300 * 10 * 0.9986 ≈ -2970V/s.

Therefore, the rate of change of voltage when t = 0.01s is approximately -2970V/s.

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• Mar 21, 2023
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