### Descriptive Statements:

- Apply trigonometric functions to solve problems involving distance and angles.
- Apply trigonometric functions to solve problems involving the unit circle.
- Manipulate trigonometric expressions and equations using techniques such as trigonometric identities.
- Analyze the relationship between a trigonometric function and its graph.
- Use trigonometric functions to model periodic relationships.

### Sample Item:

Which of the following are the solutions to
2 sin^{2} θ = cos θ + 1 for 0 < θ ≤ 2π2 sine squared theta equals cosine theta plus 1 for 0 is less than theta is less than or equal to 2pi?

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**B. ** This question requires the examinee to manipulate trigonometric expressions and equations using techniques such as trigonometric identities. Since sin^{2} θ = 1 – cos^{2} θ, 2 sin^{2} θ = cos θ + 1 ⇒ 2(1 – cos^{2} θ) = cos θ + 1 ⇒ 2 cos^{2} θ + cos θ –1 = 0 ⇒ (2 cos θ – 1)(cos θ + 1) = 0 ⇒ cos θ = ^{1}/_{2} or cos θ = –1. Thus for 0 < θ ≤ 2π, θ = ^{π}/_{3}, ^{5π}/_{3}, or πsine squared theta equals 1 minus cosine squared theta, 2 sine squared theta equals cosine theta plus 1 which becomes 2 times the quantity 1 minus cosine squared theta equals cosine theta plus 1 which becomes 2 cosine squared theta plus cosine theta negative 1 equals 0 which becomes the quantity 2 cosine theta minus 1 times the quantity cosine theta plus 1 equals 0 which becomes cos theta equals 1 half or cosine theta equals negative 1. Thus for 0 is less than theta is less than or equal to 2pi, theta equals pi over 3, 5pi over 3, or pi.

### Descriptive Statements:

- Evaluate limits.
- Demonstrate knowledge of continuity.
- Analyze the derivative as the slope of a tangent line and as the limit of the difference quotient.
- Calculate the derivatives of functions (e.g., polynomial, exponential, logarithmic).
- Apply differentiation to analyze the graphs of functions.
- Apply differentiation to solve real-world problems involving rates of change and optimization.

### Sample Item:

If *f*(*x*) = 3*x*^{4} – 8*x*^{2} + 6, what is the value of *f* of *x* equals 3*x* to the fourth minus 8*x* squared plus 6, what is the value of the limit as *h* approaches 0 of *f* of the quantity 1 plus *h* minus *f* of 1 all over *h*?

- –4negative 4
- –1negative 1
- 1
- 4

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**A. ** This question requires the examinee to analyze the derivative as the slope of a tangent line and as the limit of the difference quotient. The limit expression is equivalent to the derivative *f*'(1)*f* of 1. Since it is much easier to evaluate the derivative of a polynomial, this is preferred over evaluating the limit expression. *f*'(*x*)= 12*x*^{3} – 16*x*, so *f*'(1) = 12 – 16 = –4.*f* of *x* equals 12*x* cubed minus 16*x*, so *f* of 1 equals 12 minus 16 equals negative 4

### Descriptive Statements:

- Analyze the integral as the area under a curve and as the limit of the Riemann sum.
- Calculate the integrals of functions (e.g., polynomial, exponential, logarithmic).
- Apply integration to analyze the graphs of functions.
- Apply integration to solve real-world problems.

### Sample Item:

A sum of $20002000 dollars is invested in a savings account. The amount of money in the account in dollars after *t* years is given by the equation *A* = 2000e^{0.05t}*A* equals 2000e to the *0.05t*. What is the approximate average value of the account over the first two years?

- $21032103 dollars
- $21052105 dollars
- $22062206 dollars
- $22102210 dollars

Correct Response and Explanation (Show Correct ResponseHide Correct Response)

**A. ** This question requires the examinee to apply integration to solve real-world problems. The average value of a continuous function *f*(*x*) over an interval [*a*, *b*] is *f*(*x*)*dx**f* of *x* over an interval from a to b is 1 over b minus a times the integral from a to b of *f* of *x* *dx*. Since the independent variable *t* represents the number of years, the average daily balance over 2 years will be ^{1}/_{2}1 half of the integral of the function evaluated from 0 to 2: 2000*e*^{0.05t}*dt* = (*e*^{0.1} – 1) ≈ 21031 half the integral from 0 to 2 of 2000 e to the 0.05*t* *dt* equals 1000 over 0.05 times e to the 0.1 minus 1, which is apporximately equal to 2103.