CLEP Precalculus Practice Test

Explanation

To find the limit of the function f(x) = (2x - 1) / (x - 1) as x approaches 1, we can factor the numerator and cancel out the common factor (x - 1). This gives us f(x) = (2x - 1) / (x - 1) = 2. As x approaches 1, the function approaches 2. Therefore, the limit is 2. Distractor A is incorrect because the limit does exist. Distractor C is incorrect because the limit is not 1. Distractor D is incorrect because the limit is not 0. The correct answer is B) The limit is 2.

Explanation

To find the sum of the first 10 terms of the series, we need to calculate $a_1 + a_2 + ... + a_{10}$. Using the formula $a_n = 2n^2 + 5n - 3$, we can calculate each term and add them up. The correct sum is $a_1 + a_2 + ... + a_{10} = (2*1^2 + 5*1 - 3) + (2*2^2 + 5*2 - 3) + ... + (2*10^2 + 5*10 - 3) = 2400$. The distractors arise from common algebraic errors. Distractor A targets the misconception of incorrect calculation of the sum of the sequence. Distractor B targets the misconception of forgetting to include the constant term in the calculation. Distractor D targets the misconception of incorrect application of the formula for the sum of a series. This question requires the application of algebraic manipulation to calculate the sum of the series.

Explanation

To find the value of x for which f(g(x)) = 0, we first need to find the composite function f(g(x)). We substitute g(x) = 2x - 1 into f(x) = x^2 - 4, giving f(g(x)) = (2x - 1)^2 - 4. Expanding this expression yields f(g(x)) = 4x^2 - 4x + 1 - 4 = 4x^2 - 4x - 3. To find the value of x for which f(g(x)) = 0, we set 4x^2 - 4x - 3 = 0 and solve for x. Factoring the quadratic equation gives (2x - 3)(2x + 1) = 0, so x = 3/2 or x = -1/2. However, the correct answer x = 3/2 is the only option that matches one of these solutions. Distractor A targets the misconception of incorrectly solving the quadratic equation, distractor C targets the misconception of incorrectly substituting g(x) into f(x), and distractor D targets the misconception of not properly expanding the composite function.

Explanation

To find the doubling time, we need to find when the population size is twice the initial size. So we set up the equation 2 * 1000 = 1000 * (2^(t/10)) and solve for t. Simplifying, we get 2 = 2^(t/10). Taking the logarithm base 2 of both sides, we get 1 = t/10, so t = 10. The correct answer is B) 10 years. Distractor A) 5 years is a common misconception that arises from not properly solving the equation. Distractor C) 20 years is a misconception that arises from misunderstanding the properties of exponential growth. Distractor D) 5 months is a misconception that arises from not converting units properly.

Explanation

The equation of a circle with center (h,k) and radius r is given by (x-h)^2 + (y-k)^2 = r^2. In this case, the center is (2,3) and the radius is 4. Plugging these values into the equation, we get (x-2)^2 + (y-3)^2 = 4^2, which simplifies to (x-2)^2 + (y-3)^2 = 16. Distractor B arises from a common error of using the radius instead of the radius squared. Distractor C arises from a sign error in the center coordinates. Distractor D arises from a misconception about the equation of a circle, using a subtraction instead of an addition.

Explanation

The given function f(x) = x^2 - 4x + 5 can be rewritten in vertex form as f(x) = (x - 2)^2 + 1. This indicates that the vertex of the parabola is at (2, 1). Since the parabola opens upward, it exhibits vertical symmetry about the line x = 2, which passes through the vertex. Distractor A is incorrect because the graph does exhibit symmetry. Distractor B is incorrect because horizontal symmetry would imply that the graph is symmetric about the x-axis, which is not the case for this parabola. Distractor C is incorrect because vertical symmetry about the y-axis would imply that the graph is symmetric about the line x = 0, but the vertex is actually at x = 2.

Explanation

To find the equation of the ellipse, we need to use the standard form of the equation for an ellipse with a horizontal major axis: (x^2/a^2) + (y^2/b^2) = 1, where a is the length of the semi-major axis and b is the length of the semi-minor axis. Given that the major axis has a length of 10, a = 10/2 = 5. Given that the minor axis has a length of 6, b = 6/2 = 3. Substituting these values into the equation gives (x^2/25) + (y^2/9) = 1. Distractor B incorrectly swaps the denominators, which would be the case for a vertical major axis. Distractor C incorrectly calculates the values of a and b. Distractor D also incorrectly calculates the values of a and b, similar to C but with the values swapped.

Explanation

To reflect a graph across the line y = x, we swap the x and y variables in the equation. The original equation is (x-3)^2 + (y-2)^2 = 4. After reflection, the new equation becomes (y-3)^2 + (x-2)^2 = 4. This is because we replace x with y and y with x in the original equation. Distractor B targets the misconception of not swapping the variables correctly, distractor C targets the misconception of changing the sign of the constant term, and distractor D targets the misconception of not swapping the variables in the correct order.

Explanation

To find the magnitude of the resultant velocity vector, we first add the two vectors: (3i + 4j) + (2i - 3j) = (3 + 2)i + (4 - 3)j = 5i + j. Then, we use the formula for the magnitude of a vector: magnitude = sqrt(x^2 + y^2) = sqrt(5^2 + 1^2) = sqrt(25 + 1) = sqrt(26) = 5.1 m/s, which rounds to 7.28 m/s when using the full precision of the calculation, not 5 m/s, 10 m/s, or 12 m/s. Distractor A arises from a sign error in the addition of the vectors. Distractor C arises from adding the magnitudes of the two vectors instead of adding the vectors first. Distractor D arises from a common misconception that the magnitude of the sum of two vectors is always greater than the sum of the magnitudes of the two vectors.

Explanation

To convert the polar equation r = 2sin(θ) to rectangular coordinates, we first use the relation y = rsin(θ) to get y = 2sin^2(θ). Then, we use the trigonometric identity sin^2(θ) = (1 - cos(2θ))/2 to rewrite y = 2(1 - cos(2θ))/2 = 1 - cos(2θ). Next, we use the relation x = rcos(θ) to get x = 2sin(θ)cos(θ) = sin(2θ). Now, we can use the trigonometric identity cos(2θ) = 1 - 2sin^2(θ) to rewrite x^2 + (y - 1)^2 = (sin(2θ))^2 + (1 - cos(2θ) - 1)^2 = sin^2(2θ) + cos^2(2θ) = 1. Therefore, the rectangular equation of the polar curve r = 2sin(θ) is x^2 + (y - 1)^2 = 1. Distractor B targets the misconception of incorrectly applying the conversion formulas. Distractor C targets the misconception of incorrectly using the trigonometric identities. Distractor D targets the misconception of incorrectly applying the conversion formulas and trigonometric identities.