- A function f is continuous on the closed interval [1, 4] with f(1) = -2 and f(4) = 7. Which conclusion is guaranteed by the Intermediate Value Theorem?
- There exists a value c in (1, 4) where f(c) = 0
- The function f has a maximum value of 7 on [1, 4]
- The function f is increasing on the interval [1, 4]
- The derivative f'(c) equals 3 for some c in (1, 4)
Correct answer: There exists a value c in (1, 4) where f(c) = 0
The guaranteed conclusion is that f(c) = 0 for some c in (1, 4). Because f is continuous on [1, 4] and 0 lies between f(1) = -2 and f(4) = 7, the Intermediate Value Theorem guarantees the function attains every value between -2 and 7, including 0, somewhere inside the interval. The IVT says nothing about maxima, monotonic behavior, or derivative values; those require additional hypotheses such as differentiability or the Extreme Value Theorem.
- Consider g(x) = (x² - 9) / (x - 3). What type of discontinuity does g have at x = 3?
- A jump discontinuity
- An infinite (vertical asymptote) discontinuity
- A removable discontinuity
- An oscillating discontinuity
Correct answer: A removable discontinuity
The discontinuity at x = 3 is removable. Factoring gives g(x) = (x - 3)(x + 3) / (x - 3), which simplifies to x + 3 for all x not equal to 3, so the limit as x approaches 3 exists and equals 6. Since the limit exists but g(3) is undefined, the gap is a single hole that could be patched by defining g(3) = 6, which is the definition of a removable discontinuity rather than a jump or infinite discontinuity.
- What is the value of the limit of (5x² - 3x + 1) / (2x² + 7) as x approaches infinity?
Correct answer: 5/2
The limit equals 5/2. When evaluating a rational function as x approaches infinity, the end behavior is determined by the leading terms because lower-degree terms become negligible. Since the numerator and denominator have the same degree, the limit equals the ratio of the leading coefficients, 5/2. This horizontal-asymptote reasoning makes the answer 5/2 rather than 0, infinity, or the ratio of constant terms.
- For all x, the inequality 4 - (x - 2)² is less than or equal to h(x), which is less than or equal to 4 + (x - 2)². What is the limit of h(x) as x approaches 2?
Correct answer: 4
The limit of h(x) as x approaches 2 is 4. As x approaches 2, both bounding functions converge to the same value: 4 - (x - 2)² approaches 4 - 0 = 4, and 4 + (x - 2)² also approaches 4 + 0 = 4. By the Squeeze Theorem, since h(x) is trapped between two functions that share the limit 4, h(x) must also approach 4. The bound is forced to that common value, not to 2 or 8.
- The piecewise function f is defined by f(x) = x + 1 for x less than 2 and f(x) = 3x - 3 for x greater than or equal to 2. Is f continuous at x = 2, and why?
- No, because the left-hand and right-hand limits at x = 2 are unequal
- Yes, because the left limit, right limit, and f(2) all equal 3
- No, because f(2) is undefined
- Yes, because the function is a polynomial everywhere
Correct answer: Yes, because the left limit, right limit, and f(2) all equal 3
The function is continuous at x = 2 because the left limit, right limit, and function value all equal 3. The left-hand limit uses x + 1, giving 2 + 1 = 3; the right-hand piece 3x - 3 gives 3(2) - 3 = 3 and also defines f(2) = 3. Since the limit exists (both one-sided limits equal 3) and matches f(2), all three conditions of continuity are satisfied. The pieces meet rather than producing a jump.
- A continuous function f satisfies f(0) = 5, f(2) = 5, and f(1) = -4. A student claims the Intermediate Value Theorem guarantees f has at least two zeros on [0, 2]. Why is this claim justified?
- Because f is continuous, the IVT forces f to be zero exactly once
- Because 0 lies between -4 and 5 on both [0, 1] and [1, 2], guaranteeing a zero in each subinterval
- Because f(0) equals f(2), Rolle's Theorem provides the zeros
- Because a continuous function on a closed interval must have a maximum and minimum
Correct answer: Because 0 lies between -4 and 5 on both [0, 1] and [1, 2], guaranteeing a zero in each subinterval
The claim holds because 0 lies between the endpoint values on both subintervals. On [0, 1], f goes from 5 to -4, and since 0 is between -4 and 5, the IVT guarantees a zero in (0, 1). On [1, 2], f goes from -4 back to 5, so 0 again lies between the values and the IVT guarantees a second zero in (1, 2). Applying the theorem separately to each piece produces at least two distinct zeros; the IVT never limits the count to exactly one.
- What is the value of the limit of (sin(x)) / x as x approaches 0?
Correct answer: 1
The limit equals 1. This is the foundational special trigonometric limit established using the Squeeze Theorem, where sin(x)/x is bounded between cos(x) and 1 near x = 0, both of which approach 1. Although direct substitution yields the indeterminate form 0/0, the squeeze argument confirms the ratio approaches 1 rather than 0 or an undefined value.
- The derivative of a function f at x = a is defined as which of the following limits?
- The limit as h approaches 0 of [f(a + h) - f(a)] / h
- The limit as h approaches 0 of [f(a + h) + f(a)] / h
- The limit as x approaches a of [f(x) - f(a)] times (x - a)
- The limit as h approaches a of [f(h) - f(a)] / h
Correct answer: The limit as h approaches 0 of [f(a + h) - f(a)] / h
The derivative at x = a is the limit as h approaches 0 of [f(a + h) - f(a)] / h. This difference quotient measures the slope of secant lines through (a, f(a)) and a nearby point (a + h, f(a + h)); letting h shrink to 0 yields the instantaneous rate of change, the slope of the tangent line. The numerator must subtract the function values, the denominator must be the change in input h, and the variable approaching 0 must be h.
- If f(x) = 7x⁵, what is f'(x)?
Correct answer: 35x⁴
The derivative is 35x⁴. The power rule states that the derivative of xⁿ is n times xⁿ⁻¹, and a constant coefficient is preserved by the constant multiple rule. Applying this, bring down the exponent 5 to multiply the coefficient 7, giving 35, and reduce the exponent by one to get x⁴. The result is 35x⁴, not 7x⁴ (which forgets to multiply by the exponent) or 35x⁵ (which forgets to reduce the exponent).
- Using the product rule, what is the derivative of h(x) = x² * sin(x)?
- 2x cos(x)
- 2x sin(x) - x² cos(x)
- cos(x)
- 2x sin(x) + x² cos(x)
Correct answer: 2x sin(x) + x² cos(x)
The derivative is 2x sin(x) + x² cos(x). The product rule states that the derivative of a product equals the first factor's derivative times the second, plus the first factor times the second's derivative. Here the derivative of x² is 2x and the derivative of sin(x) is cos(x), so the result is (2x)(sin x) + (x²)(cos x). Simply multiplying the two individual derivatives, 2x cos(x), is incorrect because the product rule requires summing two cross terms.
- What is the derivative of f(x) = tan(x)?
- sec(x) tan(x)
- -csc ²(x)
- sec ²(x)
- cot(x)
Correct answer: sec ²(x)
The derivative of tan(x) is sec ²(x). This is one of the standard trigonometric derivatives memorized for the exam; it can be confirmed by writing tan(x) as sin(x)/cos(x) and applying the quotient rule, which yields [cos(x)cos(x) + sin(x)sin(x)] / cos ²(x) = 1/cos ²(x) = sec ²(x). The expression sec(x)tan(x) is the derivative of sec(x), and -csc ²(x) is the derivative of cot(x), so neither applies here.
- Using the quotient rule, what is the derivative of g(x) = (3x + 2) / (x - 1)?
- -5 / (x - 1)²
- 3 / (x - 1)²
- 5 / (x - 1)²
- (6x - 1) / (x - 1)²
Correct answer: -5 / (x - 1)²
The derivative is -5 / (x - 1)². The quotient rule gives [(low)(derivative of high) - (high)(derivative of low)] / (low)². With high = 3x + 2 (derivative 3) and low = x - 1 (derivative 1), the numerator is (x - 1)(3) - (3x + 2)(1) = 3x - 3 - 3x - 2 = -5, all over (x - 1)². The result is -5 / (x - 1)²; the positive value 5 / (x - 1)² results from subtracting the terms in the wrong order.
- A student uses the definition of the derivative to compute f'(x) for f(x) = x², forming the limit as h approaches 0 of [(x + h)² - x²] / h. After expanding the numerator, why can the limit be evaluated by direct substitution of h = 0?
- Because the limit of a quotient always equals the quotient of the limits regardless of form
- Because the h in the denominator cancels with a factor of h after simplification, removing the 0/0 form
- Because squaring (x + h) eliminates the variable h entirely
- Because the derivative of any polynomial is found without taking a limit
Correct answer: Because the h in the denominator cancels with a factor of h after simplification, removing the 0/0 form
Direct substitution becomes valid because a factor of h cancels, removing the indeterminate 0/0 form. Expanding gives (x² + 2xh + h² - x²) / h = (2xh + h²) / h = 2x + h after canceling the common factor h, valid since h is not yet 0 inside the limit process. Once simplified to 2x + h, substituting h = 0 yields the derivative 2x. The cancellation is the key algebraic step; without it the original quotient is undefined at h = 0.
- If f(x) = 4x³ - 9x¹ᐟ² + 6, what is f'(x)?
- 12x² - (9/2)x⁻¹ᐟ² + 6
- 12x² - (9/2)x⁻¹ᐟ²
- 12x² - 9x⁻¹ᐟ²
- 4x² - (9/2)x⁻¹ᐟ²
Correct answer: 12x² - (9/2)x⁻¹ᐟ²
The derivative is 12x² - (9/2)x⁻¹ᐟ². Differentiating term by term with the power rule: the derivative of 4x³ is 12x²; the derivative of -9x¹ᐟ² is -9 times (1/2) x⁻¹ᐟ² = -(9/2)x⁻¹ᐟ²; and the derivative of the constant 6 is 0. Keeping the +6 or writing the constant's derivative as 6 is the common error, since the derivative of any constant is zero.
- If f(x) = (3x² + 1)⁴, what is f'(x)?
- 24x(3x² + 1)³
- 4(3x² + 1)³
- 24x(3x² + 1)⁴
- 6x(3x² + 1)³
Correct answer: 24x(3x² + 1)³
The derivative is 24x(3x² + 1)³. The chain rule differentiates a composite function by multiplying the derivative of the outer function, evaluated at the inner function, by the derivative of the inner function. The outer power gives 4(3x² + 1)³, and the derivative of the inner expression 3x² + 1 is 6x; multiplying yields 4 times 6x times (3x² + 1)³ = 24x(3x² + 1)³. Omitting the inner derivative 6x gives the incorrect 4(3x² + 1)³.
- For the curve defined implicitly by x² + y² = 25, what is dy/dx in terms of x and y?
Correct answer: -x/y
The derivative dy/dx equals -x/y. Using implicit differentiation, differentiate both sides with respect to x, treating y as a function of x: the derivative of x² is 2x, and the derivative of y² is 2y times dy/dx by the chain rule, while the constant 25 differentiates to 0. This gives 2x + 2y(dy/dx) = 0, and solving for dy/dx yields -2x/(2y) = -x/y. Forgetting to attach dy/dx to the y² term is the common error that produces a wrong sign or expression.
- What is the derivative of f(x) = arctan(x)?
- 1 / √(1 - x²)
- -1 / (1 + x²)
- 1 / (1 + x²)
- sec ²(x)
Correct answer: 1 / (1 + x²)
The derivative of arctan(x) is 1 / (1 + x²). This is a standard inverse-trigonometric derivative that can be derived by implicit differentiation: writing y = arctan(x) means tan(y) = x, so sec ²(y) times dy/dx = 1, giving dy/dx = 1/sec ²(y) = 1/(1 + tan ²(y)) = 1/(1 + x²). The expression 1/√(1 - x²) is the derivative of arcsin(x), not arctan(x).
- If g(x) = sin(5x³), what is g'(x)?
- cos(5x³)
- 15x² cos(5x³)
- cos(15x²)
- -15x² cos(5x³)
Correct answer: 15x² cos(5x³)
The derivative is 15x² cos(5x³). The chain rule applies because sin(5x³) is a composition: the outer function sin e differentiates to cosine evaluated at the inner expression, giving cos(5x³), and this is multiplied by the derivative of the inner expression 5x³, which is 15x². Combining gives cos(5x³) times 15x² = 15x² cos(5x³). Writing only cos(5x³) ignores the required inner derivative.
- Let f be a differentiable, invertible function with f(2) = 6 and f'(2) = 4. If g is the inverse function of f, what is g'(6)?
Correct answer: 1/4
The value g'(6) equals 1/4. The derivative-of-an-inverse formula states that g'(x) = 1 / f'(g(x)), where g is the inverse of f. Since f(2) = 6, the inverse satisfies g(6) = 2, so g'(6) = 1 / f'(g(6)) = 1 / f'(2) = 1/4. The inverse derivative is the reciprocal of the original derivative evaluated at the corresponding point, so the answer is 1/4 rather than 4 or a value built from the function outputs.
- A curve is defined implicitly by xy + y³ = 10. Which expression correctly results from differentiating the left side with respect to x before solving for dy/dx?
- Y + 3y²
- X(dy/dx) + 3y²(dy/dx)
- Y + x(dy/dx) + 3y²(dy/dx)
- X + 3y²(dy/dx)
Correct answer: Y + x(dy/dx) + 3y²(dy/dx)
The correct differentiated left side is y + x(dy/dx) + 3y²(dy/dx). The term xy requires the product rule, giving (1)(y) + (x)(dy/dx) = y + x(dy/dx), and the term y³ requires the chain rule, giving 3y²(dy/dx). Combining all pieces produces y + x(dy/dx) + 3y²(dy/dx). Treating xy as a single power or omitting the product rule, as in y + 3y², drops the necessary dy/dx factors that arise because y depends on x.
- To differentiate the function h(x) = e raised to cos(x), which differentiation rule must be applied?
- The product rule, because two functions are multiplied
- The chain rule, because it is a composition of an outer exponential and an inner cosine
- The quotient rule, because the function is a ratio
- The power rule alone, because the exponent is a constant
Correct answer: The chain rule, because it is a composition of an outer exponential and an inner cosine
The chain rule must be applied because e raised to cos(x) is a composite function. The outer function is the exponential e raised to a quantity, and the inner function is cos(x), so the structure is one function nested inside another rather than a product or quotient. Differentiating gives e raised to cos(x) times the derivative of cos(x), which is -sin(x). The power rule does not apply because the exponent is the variable expression cos(x), not a constant power of x.
- A particle moves along a horizontal line so that its position at time t seconds is s(t) = t³ - 6t² + 9t meters. What is the velocity of the particle at t = 4 seconds?
- 9 meters per second
- -3 meters per second
- 12 meters per second
- 24 meters per second
Correct answer: 9 meters per second
The velocity at t = 4 is 9 meters per second. Velocity is the derivative of position, so v(t) = s'(t) = 3t² - 12t + 9. Substituting t = 4 gives 3(16) - 12(4) + 9 = 48 - 48 + 9 = 9. Velocity is found by differentiating position once, not by evaluating the position function or differentiating twice, so the correct value is 9 meters per second.
- Evaluate the limit as x approaches 0 of (e²ˣ - 1) / (3x).
Correct answer: 2/3
The limit equals 2/3. Direct substitution gives the indeterminate form 0/0, so L'Hopital's Rule applies: differentiate the numerator and denominator separately. The derivative of e²ˣ - 1 is 2e²ˣ, and the derivative of 3x is 3, giving the limit of 2e²ˣ/3 as x approaches 0, which is 2(1)/3 = 2/3. Differentiating the top and bottom and then substituting yields 2/3, not 1/3.
- The radius of a circle is increasing at a constant rate of 2 centimeters per second. At the instant when the radius is 5 centimeters, how fast is the area of the circle increasing?
- 10π square centimeters per second
- 20π square centimeters per second
- 25π square centimeters per second
- 4π square centimeters per second
Correct answer: 20π square centimeters per second
The area is increasing at 20π square centimeters per second. The area of a circle is A = π*r², so differentiating both sides with respect to time gives dA/dt = 2π*r*(dr/dt). Substituting r = 5 and dr/dt = 2 yields dA/dt = 2π*(5)*(2) = 20π. Related rates require differentiating the area formula implicitly with respect to time and multiplying by the rate of change of the radius, producing 20π.
- Let f(x) = √x. Using the tangent line to f at x = 9, what is the linear approximation of √(9.2)?
- Approximately 3.0333
- Approximately 3.0167
- Approximately 3.2000
- Approximately 2.9667
Correct answer: Approximately 3.0333
The linear approximation gives √(9.2) is approximately 3.0333. The tangent line at x = 9 uses f(9) = 3 and f'(x) = 1/(2*√x), so f'(9) = 1/6. The linearization L(x) = 3 + (1/6)(x - 9), and evaluating at x = 9.2 gives 3 + (1/6)(0.2) = 3 + 0.0333 = 3.0333. Linear approximation adds the slope times the small change in x to the base value, producing 3.0333.
- A particle moves along a line with velocity v(t) = t² - 4t + 3 for t greater than or equal to 0. At t = 0.5 seconds, is the particle speeding up or slowing down, and why?
- Speeding up, because velocity and acceleration have the same sign
- Slowing down, because velocity and acceleration have opposite signs
- Speeding up, because the velocity is positive
- Slowing down, because the acceleration is zero
Correct answer: Slowing down, because velocity and acceleration have opposite signs
The particle is slowing down because velocity and acceleration have opposite signs at t = 0.5. The velocity is v(0.5) = (0.5)² - 4(0.5) + 3 = 0.25 - 2 + 3 = 1.25, which is positive. The acceleration is a(t) = v'(t) = 2t - 4, so a(0.5) = 2(0.5) - 4 = -3, which is negative. Since velocity is positive while acceleration is negative, the two have opposite signs, and a particle slows down exactly when velocity and acceleration disagree in sign; the determining test is the comparison of the two signs, not the sign of velocity alone.
- Evaluate the limit as x approaches infinity of (ln(x)) / x.
Correct answer: 0
The limit equals 0. As x approaches infinity the expression has the indeterminate form infinity/infinity, so L'Hopital's Rule applies. Differentiating the numerator ln(x) gives 1/x and differentiating the denominator x gives 1, so the limit becomes the limit of (1/x)/1 = 1/x as x approaches infinity, which is 0. Because the denominator grows faster than the logarithm, the ratio approaches 0 rather than 1 or infinity.
- A 13-foot ladder leans against a vertical wall. The bottom of the ladder slides away from the wall at 3 feet per second. When the bottom is 5 feet from the wall, how fast is the top of the ladder sliding down?
- 3 feet per second downward
- 5/4 feet per second downward
- 13/4 feet per second downward
- 5/4 feet per second upward
Correct answer: 5/4 feet per second downward
The top slides down at 5/4 feet per second. With x the distance along the ground and y the height on the wall, x² + y² = 169; when x = 5, y = √(169 - 25) = 12. Differentiating gives 2x(dx/dt) + 2y(dy/dt) = 0, so dy/dt = -x(dx/dt)/y = -(5)(3)/12 = -15/12 = -5/4. The negative sign means the top descends at 5/4 feet per second, found by implicit related-rates differentiation of the Pythagorean relationship.
- The function f is differentiable with f(4) = 10 and f'(4) = -3. Using local linearity, which value best approximates f(4.1)?
Correct answer: 9.7
The local linear approximation gives f(4.1) is approximately 9.7. The tangent line uses the base value f(4) = 10 plus the slope times the change in x: L(4.1) = f(4) + f'(4)(4.1 - 4) = 10 + (-3)(0.1) = 10 - 0.3 = 9.7. Because the slope is negative, the approximation decreases from the base value, so multiplying -3 by the small increment 0.1 and adding to 10 produces 9.7.
- A function f is continuous on the closed interval [2, 6] and differentiable on the open interval (2, 6). Which conclusion is guaranteed by the Mean Value Theorem?
- There exists a value c in (2, 6) where f'(c) equals [f(6) - f(2)] / 4
- The function f attains an absolute maximum on the interval [2, 6]
- There exists a value c in (2, 6) where f'(c) equals 0
- The function f is increasing somewhere on the interval (2, 6)
Correct answer: There exists a value c in (2, 6) where f'(c) equals [f(6) - f(2)] / 4
The guaranteed conclusion is that f'(c) equals [f(6) - f(2)] / 4 for some c in (2, 6). The Mean Value Theorem states that when f is continuous on a closed interval and differentiable on the open interval, there is at least one interior point where the instantaneous rate of change equals the average rate of change over the whole interval. Here the average rate is [f(6) - f(2)] divided by the interval width 6 - 2 = 4. A guaranteed zero derivative requires the extra hypothesis that the endpoint values are equal (Rolle's Theorem), and the existence of a maximum is a separate Extreme Value Theorem result.
- For f(x) = x³ - 3x² - 9x + 5, which of the following are the critical points (x-values) of f?
- X = 0 only
- X = -1 and x = 3
- X = 1 and x = -3
- X = 3 only
Correct answer: X = -1 and x = 3
The critical points occur at x = -1 and x = 3. Critical points are the x-values in the domain where the derivative equals zero or fails to exist. Differentiating gives f'(x) = 3x² - 6x - 9 = 3(x² - 2x - 3) = 3(x - 3)(x + 1), which is zero when x = 3 or x = -1. Because this polynomial derivative exists everywhere, the only critical points come from setting the factored derivative equal to zero, yielding x = -1 and x = 3.
- The derivative of a function f is f'(x) = (x - 2)(x + 4). According to the First Derivative Test, what occurs at x = 2?
- A relative maximum, because f' changes from positive to negative
- A point of inflection, because f' equals zero
- A relative minimum, because f' changes from negative to positive
- No extremum, because f' does not change sign
Correct answer: A relative minimum, because f' changes from negative to positive
A relative minimum occurs at x = 2 because f' changes from negative to positive there. The First Derivative Test classifies a critical point by the sign change of f' across it. For f'(x) = (x - 2)(x + 4), just to the left of x = 2 (for example at x = 0) the product is (negative)(positive) = negative, and just to the right (for example at x = 3) it is (positive)(positive) = positive. A change from decreasing to increasing identifies a relative minimum; a relative maximum would require the opposite sign change.
- For the function f(x) = x³ - 6x² + 5, on what interval is the graph of f concave up?
- X less than 2
- The entire real line
- 2 less than x less than 4
- X greater than 2
Correct answer: X greater than 2
The graph is concave up for x greater than 2. Concavity is determined by the sign of the second derivative: the curve is concave up where f''(x) is positive. Here f'(x) = 3x² - 12x and f''(x) = 6x - 12 = 6(x - 2), which is positive when x is greater than 2. Since f'' is negative for x less than 2 and positive for x greater than 2, the function is concave up only on the interval where x exceeds 2.
- A function f has a critical point at x = 5 with f'(5) = 0, and its second derivative satisfies f''(5) = -8. What does the Second Derivative Test conclude about x = 5?
- F has a relative maximum at x = 5
- F has a relative minimum at x = 5
- The test is inconclusive at x = 5
- F has a point of inflection at x = 5
Correct answer: F has a relative maximum at x = 5
The Second Derivative Test concludes that f has a relative maximum at x = 5. The test states that at a critical point where f'(c) = 0, a negative second derivative means the graph is concave down there, so the critical point is a relative maximum. Because f''(5) = -8 is negative, the function curves downward at x = 5, producing a peak. A positive second derivative would indicate a relative minimum, and only a zero second derivative would make the test inconclusive.
- A farmer wants to enclose a rectangular field bordering a straight river, fencing only the three sides not along the river, using 400 meters of fencing total. What dimensions maximize the enclosed area?
- 100 meters by 200 meters
- 50 meters by 300 meters
- 133.3 meters by 133.3 meters
- 100 meters parallel to the river by 150 meters perpendicular
Correct answer: 100 meters by 200 meters
The area is maximized with the two perpendicular sides at 100 meters each and the side parallel to the river at 200 meters. Let x be the length of each of the two sides perpendicular to the river and y the side parallel to it; the fencing constraint is 2x + y = 400, so y = 400 - 2x. The area A = xy = x(400 - 2x) = 400x - 2x², and setting A'(x) = 400 - 4x = 0 gives x = 100, making y = 200. Because A''(x) = -4 is negative, this critical point is the maximum, giving dimensions of 100 meters by 200 meters.
- A continuous function f has only one critical point on the closed interval [0, 4], located at x = 1, where f(1) = 9. The endpoint values are f(0) = 2 and f(4) = 11. Using the Candidates Test, what is the absolute maximum value of f on [0, 4]?
Correct answer: 11
The absolute maximum value is 11, occurring at the endpoint x = 4. The Candidates Test finds absolute extrema on a closed interval by evaluating f at every critical point and at both endpoints, then comparing the resulting values. The candidate values here are f(0) = 2, f(1) = 9, and f(4) = 11; the largest of these is 11. The test compares actual function values rather than x-coordinates, so the absolute maximum value is 11, not the location x = 4.
- If a function f is differentiable everywhere and its graph has a point of inflection at x = a, which statement must be true about the second derivative at that point?
- f''(a) must be positive
- f''(a) must be negative
- f'(a) must equal zero
- f''(a) equals zero or the concavity changes sign at x = a
Correct answer: f''(a) equals zero or the concavity changes sign at x = a
At a point of inflection, the second derivative is zero (for a twice-differentiable function) and, essentially, the concavity changes sign there. A point of inflection is defined as a location where the graph switches between concave up and concave down, which requires f'' to change sign. For a differentiable f this sign change forces f''(a) to be zero at that point. The first derivative need not be zero at an inflection point, and the second derivative cannot stay strictly positive or strictly negative through a genuine concavity change.
- A continuous function f satisfies f(1) = 4 and f(7) = 22. The Mean Value Theorem guarantees some interior point where the instantaneous rate of change equals which value?
Correct answer: 3
The Mean Value Theorem guarantees an interior point where the instantaneous rate of change equals 3. That guaranteed value is the average rate of change across the interval, computed as the change in output divided by the change in input: [f(7) - f(1)] / (7 - 1) = (22 - 4) / 6 = 18 / 6 = 3. The theorem promises at least one c in the open interval where f'(c) matches this average slope of 3, not the raw output difference of 18.
- The derivative of a function f is f'(x) = x²(x - 5). Which statement correctly describes the behavior of f at its critical points using the First Derivative Test?
- F has a relative maximum at x = 0 and a relative minimum at x = 5
- F has neither a maximum nor a minimum at x = 0, and a relative minimum at x = 5
- F has a relative minimum at x = 0 and a relative maximum at x = 5
- F has relative minima at both x = 0 and x = 5
Correct answer: F has neither a maximum nor a minimum at x = 0, and a relative minimum at x = 5
The function has no extremum at x = 0 and a relative minimum at x = 5. The critical points come from f'(x) = x²(x - 5) = 0, namely x = 0 and x = 5. Across x = 5, f' changes from negative (for x between 0 and 5) to positive (for x greater than 5), giving a relative minimum. Across x = 0, the factor x² stays non-negative and x - 5 stays negative on both sides, so f' does not change sign, meaning x = 0 is a critical point with no extremum. The First Derivative Test requires an actual sign change to produce an extremum.
- Using a right Riemann sum with 4 subintervals of equal width, approximate the area under f(x) = x² from x = 0 to x = 4.
Correct answer: 30
The right Riemann sum approximation is 30. With 4 equal subintervals on [0, 4], each has width 1, and a right sum uses the right endpoints x = 1, 2, 3, 4. The function values are f(1) = 1, f(2) = 4, f(3) = 9, and f(4) = 16, and multiplying each by the width 1 and adding gives 1 + 4 + 9 + 16 = 30. A right Riemann sum sums the right-endpoint heights times the subinterval width, producing 30.
- What does the Fundamental Theorem of Calculus give for the derivative with respect to x of the integral from 2 to x of (t³ + 1) dt?
- (1/4)x⁴ + x
- 3x²
- x³ + 1
- x³ + 1 minus the value at t = 2
Correct answer: x³ + 1
The derivative is x³ + 1. The first part of the Fundamental Theorem of Calculus states that if a function is defined as the integral of a continuous integrand from a fixed lower limit to a variable upper limit x, then its derivative is simply the integrand evaluated at x. Here the integrand is t³ + 1, so replacing t with x gives x³ + 1. The constant lower limit of 2 does not affect the derivative, so the answer is x³ + 1.
- Evaluate the integral of 2x times cos(x²) dx using substitution.
- 2 sin(x²) + C
- -sin(x²) + C
- x² sin(x²) + C
- sin(x²) + C
Correct answer: sin(x²) + C
The antiderivative is sin(x²) + C. Letting u = x² gives du = 2x dx, so the integral becomes the integral of cos(u) du, which is sin(u) + C. Substituting back u = x² yields sin(x²) + C. Because the 2x factor is exactly the derivative of the inner function x², u-substitution cleanly converts the integral into cos(u) du, giving sin(x²) + C.
- Use integration by parts to evaluate the integral of x times eˣ dx.
- X eˣ - eˣ + C
- X eˣ + eˣ + C
- (1/2)x² eˣ + C
- eˣ(x + 1) + C
Correct answer: X eˣ - eˣ + C
The result is x eˣ - eˣ + C. Integration by parts uses the formula integral of u dv = uv minus integral of v du. Choosing u = x (so du = dx) and dv = eˣ dx (so v = eˣ) gives x eˣ minus the integral of eˣ dx, which is x eˣ - eˣ + C. Selecting the algebraic factor as u and the exponential as dv reduces the remaining integral to a basic one, yielding x eˣ - eˣ + C.
- A table gives values of a continuous function f: f(0) = 3, f(2) = 7, f(4) = 5, and f(6) = 9. Using a trapezoidal sum with the three subintervals [0,2], [2,4], and [4,6], approximate the integral of f from 0 to 6.
Correct answer: 36
The trapezoidal approximation is 36. Each trapezoid has width 2 and area equal to the width times the average of the two endpoint heights. The three trapezoids give 2 times (3 + 7)/2 = 10, 2 times (7 + 5)/2 = 12, and 2 times (5 + 9)/2 = 14, which sum to 10 + 12 + 14 = 36. The trapezoidal rule averages consecutive function values and multiplies by the subinterval width, producing 36.
- Evaluate the improper integral of e⁻ˣ dx from 0 to infinity.
Correct answer: 1
The improper integral converges to 1. Rewriting it as the limit as b approaches infinity of the integral from 0 to b of e⁻ˣ dx, the antiderivative is -e⁻ˣ, evaluated from 0 to b gives -e⁻ᵇ + e⁰ = -e⁻ᵇ + 1. As b approaches infinity, e⁻ᵇ approaches 0, so the value approaches 1. Because the limit exists and is finite, the improper integral converges to 1 rather than diverging.
- Which partial fraction decomposition is correct for 1 / [(x - 1)(x + 3)]?
- (1/4)/(x - 1) + (-1/4)/(x + 3)
- (1/4)/(x - 1) + (1/4)/(x + 3)
- 1/(x - 1) + 1/(x + 3)
- (-1/4)/(x - 1) + (1/4)/(x + 3)
Correct answer: (1/4)/(x - 1) + (-1/4)/(x + 3)
The correct decomposition is (1/4)/(x - 1) + (-1/4)/(x + 3). Writing 1/[(x - 1)(x + 3)] = A/(x - 1) + B/(x + 3) and clearing denominators gives 1 = A(x + 3) + B(x - 1). Substituting x = 1 yields 1 = 4A, so A = 1/4; substituting x = -3 yields 1 = -4B, so B = -1/4. The constants found by the cover-up method give (1/4)/(x - 1) plus (-1/4)/(x + 3).
- Evaluate the definite integral from 1 to 3 of (2x) dx using the Fundamental Theorem of Calculus.
Correct answer: 8
The definite integral equals 8. The second part of the Fundamental Theorem of Calculus says to find an antiderivative and evaluate it at the limits, subtracting. An antiderivative of 2x is x², so evaluating from 1 to 3 gives 3² - 1² = 9 - 1 = 8. Computing the antiderivative at the upper limit minus its value at the lower limit yields 8.
- Evaluate the integral from 0 to 1 of x times (x² + 1)³ dx using substitution.
Correct answer: 15/8
The definite integral equals 15/8. Letting u = x² + 1 gives du = 2x dx, so x dx = (1/2) du; when x = 0, u = 1 and when x = 1, u = 2. The integral becomes (1/2) times the integral from 1 to 2 of u³ du, which is (1/2) times [u⁴ / 4] from 1 to 2 = (1/2)(16/4 - 1/4) = (1/2)(15/4) = 15/8. Changing both the variable and the limits gives 15/8.
- For a function f that is increasing on the interval [a, b], how does a left Riemann sum compare to the exact value of the definite integral of f over [a, b]?
- The left Riemann sum overestimates the integral
- The left Riemann sum equals the integral exactly
- The comparison cannot be determined from the information given
- The left Riemann sum underestimates the integral
Correct answer: The left Riemann sum underestimates the integral
A left Riemann sum underestimates the integral when f is increasing. On each subinterval the left endpoint has the smallest function value because the function is rising, so each rectangle's height falls below the curve and the rectangle's area is less than the true area under that piece of the curve. Summing these too-short rectangles produces a total that is below the exact integral, making the left sum an underestimate for an increasing function.
- Evaluate the integral of ln(x) dx using integration by parts.
- X ln(x) - x + C
- X ln(x) + x + C
- 1/x + C
- (ln(x))² / 2 + C
Correct answer: X ln(x) - x + C
The antiderivative is x ln(x) - x + C. Choosing u = ln(x) (so du = (1/x) dx) and dv = dx (so v = x), integration by parts gives uv minus the integral of v du, which is x ln(x) minus the integral of x times (1/x) dx = x ln(x) minus the integral of 1 dx = x ln(x) - x + C. Treating ln(x) as u and dx as dv reduces the problem to integrating a constant, yielding x ln(x) - x + C.
- The improper integral from 1 to infinity of (1/xᵖ) dx converges for which values of p?
- P less than 1
- P greater than 1
- P equal to 1
- All positive values of p
Correct answer: P greater than 1
The integral converges only when p is greater than 1. Evaluating the antiderivative for p not equal to 1 gives a term involving b¹ ⁻ ᵖ as the upper limit b approaches infinity; this limit is finite (equal to 0) exactly when the exponent 1 - p is negative, which means p is greater than 1. When p equals 1 the integral becomes the natural logarithm, which grows without bound, and for p less than 1 the power term diverges. Thus convergence holds precisely for p greater than 1.
- A definite integral from 0 to 8 of a function f represents the total accumulated change of a quantity. If f(t) gives the rate of water flow in liters per minute and the integral from 0 to 8 of f(t) dt equals 240, what does the value 240 represent?
- The flow rate at t = 8 minutes
- The average flow rate over the 8 minutes
- The total liters of water that flowed during the 8 minutes
- The instantaneous change in flow rate
Correct answer: The total liters of water that flowed during the 8 minutes
The value 240 represents the total liters of water that flowed during the 8 minutes. When a rate function is integrated over an interval, the result is the net accumulation of the underlying quantity, by the interpretation of the definite integral as accumulated change. Since f(t) measures liters per minute, integrating that rate over 8 minutes accumulates the total volume, giving 240 liters rather than a rate, an average, or an instantaneous quantity.
- A car's velocity in meters per second is recorded at equal time intervals: at t = 0, 0; at t = 3, 12; at t = 6, 20; at t = 9, 24. Using a trapezoidal sum, approximate the distance traveled from t = 0 to t = 9 seconds.
- 144 meters
- 156 meters
- 168 meters
- 132 meters
Correct answer: 132 meters
The trapezoidal approximation gives 132 meters. Each of the three subintervals has width 3, and each trapezoid area equals the width times the average of its two endpoint velocities. The trapezoids give 3 times (0 + 12)/2 = 18, 3 times (12 + 20)/2 = 48, and 3 times (20 + 24)/2 = 66, which sum to 18 + 48 + 66 = 132. Averaging consecutive velocity readings and multiplying by the time width yields 132 meters.
- Evaluate the integral from 0 to 4 of |x - 2| dx.
Correct answer: 4
The integral equals 4. The absolute value function |x - 2| forms two triangular regions: on [0, 2] the value decreases from 2 to 0, and on [2, 4] it increases from 0 to 2. Each region is a triangle with base 2 and height 2, giving area (1/2)(2)(2) = 2, and the two regions sum to 2 + 2 = 4. Splitting the integral at x = 2 where the expression inside the absolute value changes sign produces a total accumulated area of 4.
- Which choice of u and dv is most strategic for evaluating the integral of x² times ln(x) dx by integration by parts?
- U = x² and dv = ln(x) dx
- U = x² ln(x) and dv = dx
- U = ln(x) and dv = x² dx
- U = x and dv = x ln(x) dx
Correct answer: U = ln(x) and dv = x² dx
The strategic choice is u = ln(x) and dv = x² dx. Integration by parts works best when differentiating u simplifies the expression and dv is easy to integrate. Choosing u = ln(x) gives du = (1/x) dx, which simplifies and cancels neatly against the polynomial factor, while dv = x² dx integrates easily to v = (1/3)x³. The common LIATE guideline places the logarithm before the algebraic term as u, so u = ln(x) and dv = x² dx is the productive setup.
- Evaluate the integral of (cos(x)) / (sin(x)) dx using substitution.
- -ln |cos(x)| + C
- ln |cos(x)| + C
- sin(x) / cos(x) + C
- ln |sin(x)| + C
Correct answer: ln |sin(x)| + C
The antiderivative is ln |sin(x)| + C. Letting u = sin(x) gives du = cos(x) dx, so the integral becomes the integral of (1/u) du, which is ln |u| + C. Substituting back u = sin(x) gives ln |sin(x)| + C. Because the numerator cos(x) is exactly the derivative of the denominator sin(x), the substitution turns the expression into the integral of 1/u, yielding ln |sin(x)| + C.
- Approximate the integral from 0 to 6 of f using a midpoint Riemann sum with 3 subintervals of equal width, given f(1) = 5, f(3) = 8, and f(5) = 6.
Correct answer: 38
The midpoint Riemann sum approximation is 38. With 3 equal subintervals on [0, 6], each has width 2, and the midpoints of [0,2], [2,4], and [4,6] are x = 1, 3, and 5. Multiplying each midpoint function value by the width 2 gives 2(5) + 2(8) + 2(6) = 10 + 16 + 12 = 38. A midpoint Riemann sum evaluates the function at the center of each subinterval and multiplies by the width, producing 38.
- After decomposing (5x - 1) / [(x + 1)(x - 2)] into partial fractions, what is the resulting antiderivative?
- 3 ln |x + 1| + 2 ln |x - 2| + C
- 5 ln |x + 1| - ln |x - 2| + C
- 2 ln |x + 1| + 3 ln |x - 2| + C
- 2 ln |x - 2| - 3 ln |x + 1| + C
Correct answer: 2 ln |x + 1| + 3 ln |x - 2| + C
The antiderivative is 2 ln |x + 1| + 3 ln |x - 2| + C. Setting (5x - 1)/[(x + 1)(x - 2)] = A/(x + 1) + B/(x - 2) and clearing denominators gives 5x - 1 = A(x - 2) + B(x + 1). Substituting x = -1 gives -6 = -3A, so A = 2; substituting x = 2 gives 9 = 3B, so B = 3. Integrating 2/(x + 1) + 3/(x - 2) produces 2 ln |x + 1| + 3 ln |x - 2| + C.
- What is the sum of the geometric series 8 + 4 + 2 + 1 + ... ?
Correct answer: 16
The sum is 16. A geometric series with first term a and common ratio r converges when the absolute value of r is less than 1, and its sum is a/(1 - r). Here a = 8 and r = 1/2, so the sum is 8/(1 - 1/2) = 8/(1/2) = 16. The series converges because |r| = 1/2 is less than 1, so it does not diverge, and 8 or 32 come from misapplying the formula.
- For which values of the common ratio r does an infinite geometric series converge?
- When r is not equal to zero
- When the absolute value of r is less than 1
- When r is greater than 1
- When r is positive
Correct answer: When the absolute value of r is less than 1
An infinite geometric series converges exactly when the absolute value of r is less than 1. In that case the powers rⁿ shrink to zero, allowing the partial sums to approach the finite value a/(1 - r). If |r| is greater than or equal to 1, the terms do not approach zero and the partial sums grow without bound or oscillate, so the series diverges; sign or nonzero conditions alone are insufficient.
- Apply the nth-term test (test for divergence) to the series whose terms are a_n = n/(2n + 1). What does the test conclude?
- The test is inconclusive because the terms approach 0
- The series converges to 1/2
- The series diverges because the terms approach 1/2, not 0
- The series converges because the terms approach 1/2
Correct answer: The series diverges because the terms approach 1/2, not 0
The series diverges because the terms approach 1/2 rather than 0. The nth-term test states that if the limit of a_n is not zero, the series must diverge. Here the limit of n/(2n + 1) as n grows is 1/2, a nonzero value, so the necessary condition for convergence fails and the series diverges. The test never proves convergence and is only inconclusive when the limit equals zero.
- For what values of p does the p-series, the sum of 1/nᵖ, converge?
- When p is greater than or equal to 1
- When p is less than 1
- When p is greater than 0
- When p is greater than 1
Correct answer: When p is greater than 1
A p-series converges precisely when p is greater than 1. This result follows from the integral test applied to 1/xᵖ, whose improper integral converges only for p greater than 1. When p equals 1 the series is the harmonic series, which diverges, and for p less than 1 the terms decrease too slowly, so those cases diverge. Thus strict inequality p greater than 1 is required.
- The harmonic series is the sum of 1/n for n from 1 to infinity. Does it converge or diverge?
- It diverges
- It converges to 1
- It converges to a finite value greater than 2
- It converges because the terms approach 0
Correct answer: It diverges
The harmonic series diverges. Although its terms 1/n approach zero, that condition alone does not guarantee convergence; it is the p-series with p = 1, which is the boundary case that fails to converge. Grouping terms shows the partial sums grow without bound, so the series diverges. The approach of the terms to zero is necessary but not sufficient for convergence, which is why the convergence distractors are false.
- Use the ratio test on the series whose terms are a_n = 2ⁿ / n!. What is the limit of the ratio of consecutive terms, and what does it imply?
- The limit is infinity, so the series diverges
- The limit is 0, so the series converges
- The limit is 2, so the series diverges
- The limit is 1, so the test is inconclusive
Correct answer: The limit is 0, so the series converges
The ratio test gives a limit of 0, so the series converges absolutely. The ratio of consecutive terms is (2ⁿ⁺¹/(n+1)!) divided by (2ⁿ/n!), which simplifies to 2/(n + 1). As n grows without bound this ratio approaches 0, and since 0 is less than 1 the ratio test guarantees convergence. A limit of 1 would be inconclusive and a limit greater than 1 would imply divergence, neither of which occurs here.
- Which condition is required for the Alternating Series Test to guarantee convergence of the sum of (-1)ⁿ * b_n, where b_n is positive?
- The series of b_n converges
- The terms b_n increase without bound
- The terms b_n decrease and approach 0
- The terms b_n approach a nonzero constant
Correct answer: The terms b_n decrease and approach 0
The Alternating Series Test requires that the positive terms b_n be eventually decreasing and approach 0. When an alternating series has terms whose magnitudes shrink monotonically to zero, the partial sums oscillate with diminishing swings and converge. If b_n does not approach zero the series diverges by the nth-term test, and increasing or merely bounded terms fail the test's hypotheses.
- The Maclaurin series for eˣ is the sum from n = 0 to infinity of xⁿ / n!. Using this, what is the first three nonzero terms of the series for eˣ?
- 1 + x + x²
- X + x²/2 + x³/6
- 1 + x/1! + x²/3!
- 1 + x + x²/2
Correct answer: 1 + x + x²/2
The first three nonzero terms are 1 + x + x²/2. Substituting n = 0, 1, and 2 into xⁿ/n! gives x⁰/0! = 1, x¹/1! = x, and x²/2! = x²/2 since 2! equals 2. The term x² without the factorial denominator is incorrect, starting at x omits the constant term 1, and using 3! in the third term applies the wrong factorial.
- Find the radius of convergence of the power series the sum of xⁿ / 3ⁿ.
Correct answer: 3
The radius of convergence is 3. Applying the ratio test to the absolute value of xⁿ⁺¹/3ⁿ⁺¹ divided by xⁿ/3ⁿ gives |x|/3, and the series converges when this is less than 1, that is when |x| is less than 3. Therefore the radius of convergence is 3. The value 1/3 inverts the ratio, and 1 ignores the base 3 in the denominator.
- What is the second-degree Taylor polynomial for f(x) = cos(x) centered at x = 0?
- 1 - x²
- 1 - x²/2
- 1 + x²/2
- X - x³/6
Correct answer: 1 - x²/2
The second-degree Taylor polynomial for cos(x) at 0 is 1 - x²/2. The Maclaurin series for cosine begins 1 - x²/2! + x⁴/4! - ..., and truncating after the degree-two term gives 1 - x²/2 since 2! = 2. A positive sign on the quadratic term contradicts the alternating pattern of cosine, the form with x and x³ belongs to sin e, and 1 - x² omits the factorial denominator.
- A telescoping series has terms a_n = 1/n - 1/(n + 1) summed from n = 1 to infinity. What is its sum?
Correct answer: 1
The sum of the telescoping series is 1. Writing out the partial sums, most terms cancel: (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + ... leaves only the first term 1 minus the vanishing tail 1/(n + 1). As n grows, 1/(n + 1) approaches 0, so the sum converges to 1 - 0 = 1. The series converges, so it does not diverge, and 0 or 1/2 misread the cancellation.
- A series of positive terms converges, but the series formed by the absolute values of an alternating version also converges. This situation is best described as which type of convergence?
- Conditional convergence
- Divergence
- Oscillating convergence
- Absolute convergence
Correct answer: Absolute convergence
When the series of absolute values converges, the original series is absolutely convergent. A series is absolutely convergent if the sum of the absolute values of its terms converges, which automatically guarantees the original series converges as well. Conditional convergence would apply only if the original series converged while the series of absolute values diverged, so that label does not fit here.
- Which series is conditionally convergent?
- The alternating harmonic series, the sum of (-1)ⁿ⁺¹/n
- The sum of 1/n²
- The sum of (-1)ⁿ / n²
- The sum of 1/n
Correct answer: The alternating harmonic series, the sum of (-1)ⁿ⁺¹/n
The alternating harmonic series, the sum of (-1)ⁿ⁺¹/n, is conditionally convergent. It converges by the Alternating Series Test because its terms decrease to zero, yet the series of absolute values is the harmonic series, which diverges; convergence that fails when absolute values are taken is exactly conditional convergence. The sum of 1/n² and the sum of (-1)ⁿ/n² both converge absolutely, and the sum of 1/n diverges outright.
- The Lagrange error bound estimates the remainder of a Taylor polynomial approximation. The bound is expressed in terms of which quantity?
- The radius of convergence of the series
- A bound on the next derivative not used in the polynomial
- The value of the function at the center
- The sum of all included polynomial terms
Correct answer: A bound on the next derivative not used in the polynomial
The Lagrange error bound depends on a bound on the next higher-order derivative that the polynomial omits. The remainder after an nth-degree Taylor polynomial is bounded by the maximum value of the (n+1)st derivative on the interval times |x - c|ⁿ⁺¹ divided by (n+1) factorial. The center value, the included terms, and the radius of convergence do not directly bound the truncation error; the next derivative does.
- Use the integral test to determine the behavior of the series the sum of 1/(n ln n) for n starting at 2. What is the conclusion?
- The integral test cannot be applied to this series
- The series converges to ln 2
- The series diverges because the corresponding integral diverges
- The series converges because the corresponding integral converges
Correct answer: The series diverges because the corresponding integral diverges
The series diverges because the corresponding improper integral diverges. The function 1/(x ln x) is positive, continuous, and decreasing for x at least 2, so the integral test applies; the integral of 1/(x ln x) dx is ln(lnx), which grows without bound as x approaches infinity. Since the integral diverges, the integral test forces the series to diverge as well, and it does not converge to any finite value.
- Using the geometric series 1/(1 - x) equal to the sum of xⁿ for |x| less than 1, what is the power series representation of 1/(1 + x²)?
- The sum of x²ⁿ
- The sum of (-1)ⁿ * xⁿ
- The sum of xⁿ / n!
- The sum of (-1)ⁿ * x²ⁿ
Correct answer: The sum of (-1)ⁿ * x²ⁿ
The power series for 1/(1 + x²) is the sum of (-1)ⁿ * x²ⁿ. Rewriting the denominator as 1 - (-x²) lets us substitute -x² in place of x in the geometric series 1/(1 - u) = sum of uⁿ, giving the sum of (-x²)ⁿ = the sum of (-1)ⁿ * x²ⁿ. Omitting the alternating sign ignores the negative substitution, using xⁿ misses that the substitution squares x, and the factorial form belongs to eˣ.
- For the differential equation dy/dx = 6x², find the general solution y in terms of x.
- Y = 2x³ + C
- Y = 12x + C
- Y = 6x³ + C
- Y = 3x³ + C
Correct answer: Y = 2x³ + C
The general solution is y = 2x³ + C. Antidifferentiating both sides of dy/dx = 6x² with respect to x means applying the power rule for integration: the antiderivative of 6x² is 6 times x³/3, which equals 2x³, and an arbitrary constant C must be added because infinitely many functions share this derivative. The choice 6x³ forgets to divide by the new exponent, 12x differentiates instead of integrates, and 3x³ uses the wrong coefficient.
- The amount of a radioactive substance decays according to dA/dt = -0.03A, where A is in grams and t is in years. If the half-life is the time for the amount to fall to half its initial value, which equation determines the half-life T?
- 0.5A = -0.03T
- 0.5 = e⁻⁰·⁰³ᵀ
- 0.5 = e⁰·⁰³ᵀ
- 0.5 = -0.03T
Correct answer: 0.5 = e⁻⁰·⁰³ᵀ
The half-life satisfies 0.5 = e⁻⁰·⁰³ᵀ. The solution of dA/dt = -0.03A is A(t) = A(0)e⁻⁰·⁰³ᵗ, so the amount reaches half its start when A(0)e⁻⁰·⁰³ᵀ = 0.5A(0); dividing both sides by A(0) gives 0.5 = e⁻⁰·⁰³ᵀ. A positive exponent describes growth rather than decay, and the linear forms ignore that exponential decay is governed by e raised to the rate times time.
- Verify which function is a solution to the differential equation dy/dx = 3y.
- Y = x³
- Y = 3x
- Y = e³ˣ
- Y = 3eˣ
Correct answer: Y = e³ˣ
The solution is y = e³ˣ. Differentiating y = e³ˣ using the chain rule gives dy/dx = 3e³ˣ, which equals 3 times the original y, satisfying dy/dx = 3y exactly. For y = 3eˣ the derivative is 3eˣ = y, not 3y; for y = x³ the derivative is 3x², not 3 times x³; and for y = 3x the derivative is 3, not 3 times 3x. Only the exponential whose rate constant matches the coefficient works.
- Given dy/dx = (cos x)/(2y) with y(0) = 1, what is the particular solution y in terms of x?
- Y = √(sin x)
- Y = sin x + 1
- Y = (1/2) sin x + 1
- Y = √(sin x + 1)
Correct answer: Y = √(sin x + 1)
The particular solution is y = √(sin x + 1). Separating variables gives 2y dy = cos x dx; integrating both sides yields y² = sin x + C. Applying y(0) = 1 gives 1 = sin 0 + C = C, so C = 1 and y² = sin x + 1, and taking the positive root consistent with y(0) = 1 gives y = √(sin x + 1). Omitting the constant or failing to square-root the result produces the other choices.
- A slope field for dy/dx = x/y is undefined along which set of points?
- The x-axis, where y = 0
- The y-axis, where x = 0
- The line y = x
- The origin only
Correct answer: The x-axis, where y = 0
The slope field is undefined along the x-axis, where y = 0. The expression dy/dx = x/y divides by y, so wherever y equals zero the slope cannot be computed and no segment can be drawn; this occurs all along the x-axis. The y-axis (x = 0) instead produces horizontal segments with slope 0, the line y = x gives a defined slope of 1, and restricting the trouble to only the origin ignores every other point where y = 0.
- Newton's law of cooling gives dT/dt = -k(T - 70) for an object cooling in a 70-degree room. What does the equilibrium (constant) solution of this differential equation represent?
- T = -70, the negative of the surroundings
- T = 70, the room temperature the object approaches
- T = 0, where cooling stops
- T = k, the cooling constant
Correct answer: T = 70, the room temperature the object approaches
The equilibrium solution is T = 70, the room temperature. A constant solution occurs where dT/dt = 0, which requires T - 70 = 0, so T = 70; physically the object's temperature levels off at the surrounding room temperature over time. Setting T to 0, to the constant k, or to -70 does not make the factor T - 70 vanish and therefore does not produce a steady state.
- Find the area of the region enclosed between the curves y = x² and y = 2x.
Correct answer: 4/3
The enclosed area is 4/3. The curves intersect where x² = 2x, giving x = 0 and x = 2, and on that interval the line y = 2x lies above the parabola y = x². The area equals the integral from 0 to 2 of (2x - x²) dx = [x² - x³/3] from 0 to 2 = (4 - 8/3) = 4/3. Reversing the order of subtraction or mis-evaluating the bounds produces the larger or sign-flipped distractors.
- The region bounded by y = √x, the x-axis, and x = 4 is revolved about the x-axis. What is the volume of the resulting solid using the disk method?
Correct answer: 8π
The volume is 8π. The disk method integrates π times the radius squared, where the radius is y = √x, so the volume is the integral from 0 to 4 of π*(√x)² dx = π times the integral of x dx = π*[x²/2] from 0 to 4 = π*(16/2) = 8π. Forgetting to square the radius or mishandling the bounds gives 4π or 16π.
- A solid has a base bounded by the region between y = x and y = 0 from x = 0 to x = 3. Cross-sections perpendicular to the x-axis are squares. What is the volume of the solid?
Correct answer: 9
The volume is 9. Each square cross-section has side length equal to the height of the region, which is x, so its area is x². Integrating the cross-sectional area from 0 to 3 gives the integral of x² dx = [x³/3] from 0 to 3 = 27/3 = 9. Introducing π treats the cross-sections as circles, and 27 forgets to divide by 3 during integration.
- What is the average value of the function f(x) = 3x² on the interval [0, 2]?
Correct answer: 4
The average value is 4. The average value of a function on [a, b] equals 1/(b - a) times the integral from a to b of f(x) dx. Here that is 1/(2 - 0) times the integral from 0 to 2 of 3x² dx = (1/2)*[x³] from 0 to 2 = (1/2)*(8) = 4. The value 8 is the integral itself before dividing by the interval length, and 6 or 12 result from arithmetic errors in the antiderivative.
- A particle moves along a line with velocity v(t) = t - 3 for t in [0, 5], where t is in seconds and v is in meters per second. What is the total distance traveled by the particle over the interval?
Correct answer: 6.5
The total distance traveled is 6.5 meters. Total distance integrates the absolute value of velocity, and v(t) = t - 3 changes sign at t = 3. From 0 to 3 the velocity is negative, so distance is the integral of (3 - t) dt = [3t - t²/2] from 0 to 3 = 9 - 4.5 = 4.5; from 3 to 5 the velocity is positive, so distance is the integral of (t - 3) dt = [t²/2 - 3t] from 3 to 5 = (12.5 - 15) - (4.5 - 9) = -2.5 + 4.5 = 2.0. Summing the two positive pieces gives 4.5 + 2.0 = 6.5. Using signed displacement instead of absolute value yields the smaller distractor 2.0.
- For a smooth curve given by y = f(x) from x = a to x = b, which integral gives the arc length of the curve?
- The integral from a to b of sqrt((f'(x))² - 1) dx
- The integral from a to b of sqrt(1 + (f'(x))²) dx
- The integral from a to b of √(1 + f(x)) dx
- The integral from a to b of (1 + (f'(x))²) dx
Correct answer: The integral from a to b of sqrt(1 + (f'(x))²) dx
The arc length is the integral from a to b of sqrt(1 + (f'(x))²) dx. The formula comes from summing infinitesimal segments of length √((dx)² + (dy)²); factoring out dx gives √(1 + (dy/dx)²) dx, where dy/dx is f'(x). Omitting the square root leaves an area-type integrand, using f(x) instead of its derivative is incorrect, and subtracting 1 instead of adding it reverses the Pythagorean relationship.
- A curve is given parametrically by x = t² and y = t³. What is dy/dx in terms of t?
Correct answer: (3/2)t
The derivative dy/dx equals (3/2)t. For parametric equations, dy/dx is found as (dy/dt) divided by (dx/dt). Here dy/dt = 3t² and dx/dt = 2t, so dy/dx = 3t² / (2t) = (3/2)t after simplifying. The reciprocal (2/3)t inverts the ratio, and t/2 drops the factors of t incorrectly; only (3/2)t correctly divides the two parametric derivatives.
- For the parametric curve x = 2t and y = t² + 1, what is the slope of the tangent line at t = 3?
Correct answer: 3
The slope of the tangent line at t = 3 is 3. The parametric slope dy/dx is (dy/dt)/(dx/dt); here dy/dt = 2t and dx/dt = 2, so dy/dx = 2t/2 = t. Evaluating at t = 3 gives a slope of 3. The value 6 mistakenly uses dy/dt alone, 2 uses dx/dt, and 9 squares t instead of leaving it linear.
- A particle has position given by the vector-valued function r(t) = (cos t, sin t). What is the speed of the particle at any time t?
Correct answer: 1
The speed is 1 at all times. Speed equals the magnitude of the velocity vector, where velocity is the derivative r'(t) = (-sin t, cos t). The magnitude is √((-sin t)² + (cos t)²) = √(sin ² t + cos ² t) = √1 = 1, using the Pythagorean identity. Adding the components instead of using the magnitude formula gives sin t + cos t, and √2 would only arise if both components were 1 simultaneously.
- Which integral gives the arc length of the parametric curve defined by x(t) and y(t) for t from a to b?
- The integral from a to b of ((dx/dt)² + (dy/dt)²) dt
- The integral from a to b of √((dx/dt)² - (dy/dt)²) dt
- The integral from a to b of (dx/dt + dy/dt) dt
- The integral from a to b of √((dx/dt)² + (dy/dt)²) dt
Correct answer: The integral from a to b of √((dx/dt)² + (dy/dt)²) dt
The arc length is the integral from a to b of √((dx/dt)² + (dy/dt)²) dt. This comes from integrating the speed, the magnitude of the velocity vector, over the parameter interval. Omitting the square root, subtracting the squared terms, or simply adding the derivatives all fail to give the Euclidean magnitude that arc length requires.
- Convert the polar equation r = 4 cos(θ) to identify the curve it represents.
- A circle of radius 2 centered at (2, 0)
- A circle of radius 4 centered at the origin
- A line through the origin with slope 4
- A cardioid
Correct answer: A circle of radius 2 centered at (2, 0)
The equation r = 4 cos(θ) represents a circle of radius 2 centered at (2, 0). Multiplying both sides by r gives r² = 4r cos(θ), and substituting r² = x² + y² and r cos(θ) = x yields x² + y² = 4x, which rearranges to (x - 2)² + y² = 4, a circle of radius 2 centered at (2, 0). A circle centered at the origin would require r equal to a constant, and r = 4 cos(θ) is not a line or cardioid.
- What is the area enclosed by one petal of the polar rose r = 2 sin(2 θ)?
Correct answer: π/2
The area of one petal is π/2. Polar area is (1/2) times the integral of r² dtheta, and one petal of r = 2 sin(2 θ) is traced as theta goes from 0 to π/2. The integral is (1/2) times the integral from 0 to π/2 of 4 sin ²(2 θ) dtheta = 2 times the integral of sin ²(2 θ), which using the identity averages to 2*(π/4) = π/2. Forgetting the one-half factor or the squared amplitude leads to the larger distractors.
- A particle moves with position vector r(t) = (t², 4t). What is the acceleration vector at t = 1?
- (2t, 4)
- (0, 0)
- (2, 0)
- (2, 4)
Correct answer: (2, 0)
The acceleration vector at t = 1 is (2, 0). Acceleration is the second derivative of position. The velocity is r'(t) = (2t, 4), and differentiating again gives r''(t) = (2, 0), which is constant and equal to (2, 0) at every time including t = 1. The vector (2, 4) is the velocity at t = 1, and (2t, 4) is the unevaluated velocity, not the acceleration.
- For the polar curve r = 1 + cos(θ), what is the value of r when theta = 0?
Correct answer: 2
At theta = 0, r equals 2. Substituting theta = 0 into r = 1 + cos(θ) gives r = 1 + cos(0) = 1 + 1 = 2, since the cosine of 0 is 1. This point lies on the positive x-axis at distance 2 from the pole, the maximum of this cardioid. The value 1 ignores the cosine term, and 0 would require cos(θ) = -1, which occurs at theta = π, not theta = 0.
- A particle's velocity vector is v(t) = (3, 2t). If its position at t = 0 is (1, 0), what is the position vector at t = 2?
Correct answer: (7, 4)
The position at t = 2 is (7, 4). Position is found by antidifferentiating each velocity component and applying the initial condition. The x-component integrates to 3t + C, and with x(0) = 1 gives x(t) = 3t + 1, so x(2) = 7; the y-component integrates to t² + C, and with y(0) = 0 gives y(t) = t², so y(2) = 4. The result is (7, 4). Forgetting the initial x-value gives 6, and mishandling the y-integral gives 2.
- Which integral expression gives the area of the region inside the polar curve r = f(theta) swept from theta = alpha to theta = beta?
- The integral from alpha to beta of (f(θ))² dtheta
- (1/2) times the integral from alpha to beta of (f(θ))² dtheta
- The integral from alpha to beta of f(theta) dtheta
- (1/2) times the integral from alpha to beta of f(theta) dtheta
Correct answer: (1/2) times the integral from alpha to beta of (f(θ))² dtheta
The polar area is (1/2) times the integral from alpha to beta of (f(θ))² dtheta. This formula sums infinitesimal circular sectors, each with area (1/2) r² dtheta, where r = f(theta). Omitting the one-half factor, leaving r unsquared, or both errors together fail to match the sector-area derivation, so only the expression with both the one-half and the squared radius is correct.
- A sequence is defined by a_n = (3n + 5)/(n - 2). What is the limit of the sequence as n approaches infinity?
Correct answer: 3
The limit of the sequence is 3. For a rational expression in n, the limit as n grows without bound is the ratio of the leading coefficients of the numerator and denominator when their degrees match; here both numerator and denominator are degree one, giving 3/1 = 3. The value 5 misreads the constant term, 0 would require the denominator degree to exceed the numerator degree, and the sequence does converge rather than diverge.
- Consider the sequence a_n = (-1)ⁿ. Does this sequence converge, and if so to what value?
- It converges to 1
- It diverges because it oscillates between 1 and -1
- It converges to -1
- It converges to 0
Correct answer: It diverges because it oscillates between 1 and -1
The sequence diverges because it oscillates between 1 and -1. A sequence converges only if its terms approach a single value; the terms of (-1)ⁿ alternate forever between 1 (for even n) and -1 (for odd n) and never settle, so no single limit exists. It therefore does not converge to 1, -1, or 0, since each of those would require the terms to eventually stay near that one number.
- What is the sum of the infinite geometric series the sum from n = 0 of 5 times (1/3)ⁿ?
Correct answer: 15/2
The sum is 15/2. An infinite geometric series with first term a and ratio r converges to a/(1 - r) when the absolute value of r is less than 1; here a = 5 and r = 1/3, so the sum is 5/(1 - 1/3) = 5/(2/3) = 15/2. The value 5 ignores the ratio entirely, 15 inverts the fraction, and the series converges because |1/3| is less than 1, so it does not diverge.
- The repeating decimal 0.444... can be written as an infinite geometric series. What rational number does it equal?
Correct answer: 4/9
The repeating decimal 0.444... equals 4/9. It is the geometric series 4/10 + 4/100 + 4/1000 + ..., with first term a = 4/10 and ratio r = 1/10, summing to (4/10)/(1 - 1/10) = (4/10)/(9/10) = 4/9. The value 4/10 is only the first term, 4/99 corresponds to a two-digit repeat like 0.0404..., and 2/5 equals 0.4 exactly rather than the repeating decimal.
- Apply the Direct Comparison Test to the series the sum of 1/(n² + 1). Comparing to which convergent series shows it converges?
- The sum of 1/n²
- The sum of 1/n
- The sum of 1/(n + 1)
- The sum of (1/2)ⁿ only
Correct answer: The sum of 1/n²
Comparing to the sum of 1/n² shows the series converges. Since 1/(n² + 1) is less than 1/n² for every n, and the p-series the sum of 1/n² (with p = 2) converges, the Direct Comparison Test forces the smaller positive series the sum of 1/(n² + 1) to converge as well. Comparing to the harmonic series the sum of 1/n or to the sum of 1/(n + 1) is useless because those diverge, and a single geometric term is not a valid comparison series.
- Use the Limit Comparison Test on the series the sum of (2n + 1)/(n³ + 4). Which series is the natural choice to compare with, and what is the conclusion?
- Compare with the sum of 1/n²; the series converges
- Compare with the sum of 1/n; the series diverges
- Compare with the sum of 1/n³; the series diverges
- Compare with the sum of 2ⁿ; the series converges
Correct answer: Compare with the sum of 1/n²; the series converges
Comparing with the sum of 1/n² shows the series converges. For large n the term (2n + 1)/(n³ + 4) behaves like 2n/n³ = 2/n², so the natural comparison series is the convergent p-series the sum of 1/n². The limit of the ratio of the terms is the finite positive number 2, so by the Limit Comparison Test both series share the same behavior and converge. Comparing with 1/n or 1/n³ picks the wrong dominant power, and 2ⁿ is unrelated.
- Use the Ratio Test on the series the sum of n! / 10ⁿ. What does the test conclude?
- The limit is infinity, so the series diverges
- The limit is 1/10, so the series converges
- The limit is 1, so the test is inconclusive
- The limit is 0, so the series converges
Correct answer: The limit is infinity, so the series diverges
The Ratio Test gives a limit of infinity, so the series diverges. The ratio of consecutive terms is ((n+1)!/10ⁿ⁺¹) divided by (n!/10ⁿ), which simplifies to (n + 1)/10. As n grows without bound this ratio increases without bound, and a limit greater than 1 (here infinity) means the series diverges. A finite limit such as 1/10 would wrongly suggest convergence, and the factorial in the numerator makes the terms grow too fast.
- Apply the Root Test to the series the sum of (n/(2n + 1))ⁿ. What is the value of the relevant limit and the conclusion?
- The limit is 1/2, so the series converges
- The limit is 2, so the series diverges
- The limit is 1, so the test is inconclusive
- The limit is 0, so the series converges
Correct answer: The limit is 1/2, so the series converges
The Root Test gives a limit of 1/2, so the series converges. Taking the nth root of the absolute value of (n/(2n + 1))ⁿ leaves n/(2n + 1), whose limit as n grows is 1/2. Since 1/2 is less than 1, the Root Test guarantees convergence. A limit of 1 would be inconclusive and a limit greater than 1 would imply divergence, but neither occurs because the base approaches one-half.
- For the convergent alternating series the sum of (-1)ⁿ⁺¹/n², the Alternating Series Estimation Theorem bounds the error when using a partial sum. If the sum is approximated by the first three terms, the error is at most which value?
- The fourth term in absolute value, 1/16
- The third term, 1/9
- The first term, 1
- The sum of all remaining terms exactly
Correct answer: The fourth term in absolute value, 1/16
The error is at most the absolute value of the fourth term, 1/16. The Alternating Series Estimation Theorem states that for a convergent alternating series with decreasing terms, the error after summing the first N terms is no larger than the absolute value of the first omitted term. After three terms the first omitted term is the fourth, (-1)⁵/4², whose absolute value is 1/16. Using the third term or the first term picks the wrong omitted term, and an exact remainder is not the bound the theorem provides.
- Determine whether the series the sum of (-1)ⁿ times n/(n + 1) converges or diverges.
- It diverges because the terms do not approach 0
- It converges conditionally by the Alternating Series Test
- It converges absolutely
- It converges to 1/2
Correct answer: It diverges because the terms do not approach 0
The series diverges because its terms do not approach 0. The nth-term test for divergence requires that the underlying magnitude n/(n + 1) approach zero for any chance of convergence, but n/(n + 1) approaches 1, so the terms (-1)ⁿ times n/(n+1) oscillate without shrinking. The Alternating Series Test cannot save it because that test also requires the terms to go to zero, so it neither converges conditionally nor absolutely.
- Find the interval of convergence for the power series the sum from n = 1 of xⁿ / n.
- From -1 to 1, including -1 but not 1
- From -1 to 1, including both endpoints
- All real numbers
- Only x = 0
Correct answer: From -1 to 1, including -1 but not 1
The interval of convergence is from -1 to 1, including -1 but excluding 1. The Ratio Test gives convergence for the absolute value of x less than 1, so the radius is 1; testing the endpoints, at x = 1 the series becomes the harmonic series the sum of 1/n, which diverges, while at x = -1 it becomes the alternating harmonic series, which converges. Thus -1 is included and 1 is excluded. The full closed interval, all reals, and a single point all misjudge the endpoints or radius.
- What is the center and radius of convergence of the power series the sum of (x - 4)ⁿ / 2ⁿ?
- Center 4, radius 2
- Center 2, radius 4
- Center 4, radius 1/2
- Center -4, radius 2
Correct answer: Center 4, radius 2
The center is 4 and the radius of convergence is 2. A power series in (x - c)ⁿ is centered at c, so here c = 4. Applying the Ratio Test to the absolute value of (x - 4)ⁿ⁺¹/2ⁿ⁺¹ divided by (x - 4)ⁿ/2ⁿ gives the absolute value of (x - 4)/2, which is less than 1 when the absolute value of x - 4 is less than 2; hence the radius is 2. Swapping center and radius or inverting the base gives the other options.
- The Maclaurin series for sin(x) is the sum from n = 0 of (-1)ⁿ x²ⁿ⁺¹/(2n+1)!. What are the first three nonzero terms?
- X - x³/6 + x⁵/120
- 1 - x²/2 + x⁴/24
- X + x³/6 + x⁵/120
- X - x²/2 + x³/6
Correct answer: X - x³/6 + x⁵/120
The first three nonzero terms of the Maclaurin series for sin(x) are x - x³/6 + x⁵/120. Substituting n = 0, 1, 2 into (-1)ⁿ x²ⁿ⁺¹/(2n+1)! gives x¹/1! = x, then -x³/3! = -x³/6, then x⁵/5! = x⁵/120. The series with even powers belongs to cosine, the all-positive version drops the alternating sign, and the form mixing x² and x³ uses wrong powers and factorials.
- The Maclaurin series for ln(1 + x) is the sum from n = 1 of (-1)ⁿ⁺¹ xⁿ / n. What are the first three nonzero terms?
- X - x²/2 + x³/3
- X + x²/2 + x³/3
- 1 - x + x²
- X - x³/6 + x⁵/120
Correct answer: X - x²/2 + x³/3
The first three nonzero terms of the series for ln(1 + x) are x - x²/2 + x³/3. Substituting n = 1, 2, 3 into (-1)ⁿ⁺¹ xⁿ/n gives x¹/1 = x, then -x²/2, then x³/3. The all-positive version drops the alternating sign, the form starting with the constant 1 is incorrect because ln(1) = 0, and the series with x³/6 and x⁵/120 is the sin e series.
- Find a power series for f(x) = 1/(1 - 2x) using the geometric series 1/(1 - u) equal to the sum of uⁿ, valid for the absolute value of 2x less than 1.
- The sum of 2ⁿ xⁿ
- The sum of 2 xⁿ
- The sum of xⁿ / 2ⁿ
- The sum of (-1)ⁿ 2ⁿ xⁿ
Correct answer: The sum of 2ⁿ xⁿ
The power series is the sum of 2ⁿ xⁿ. Substituting u = 2x into the geometric series 1/(1 - u) = the sum of uⁿ gives the sum of (2x)ⁿ = the sum of 2ⁿ xⁿ, valid when the absolute value of 2x is less than 1, that is the absolute value of x less than 1/2. Pulling the 2 out only once ignores the exponent, dividing by 2ⁿ inverts the substitution, and the alternating sign would arise from a plus sign in the denominator, not a minus.
- By integrating the geometric series for 1/(1 + x) term by term, which function's power series is obtained?
- ln(1 + x)
- arctan(x)
- eˣ
- 1/(1 + x)²
Correct answer: ln(1 + x)
Integrating the series for 1/(1 + x) term by term produces the power series for ln(1 + x). Because the antiderivative of 1/(1 + x) is ln(1 + x), integrating the geometric series the sum of (-1)ⁿ xⁿ term by term yields the sum of (-1)ⁿ xⁿ⁺¹/(n + 1), which is the Maclaurin series for ln(1 + x). The series for arctan(x) comes from integrating 1/(1 + x²), eˣ is unrelated, and differentiating rather than integrating would give 1/(1 + x)².
- By differentiating the geometric series the sum of xⁿ term by term, which function's power series is obtained?
- 1/(1 - x)²
- ln(1 - x)
- 1/(1 + x)
- X/(1 - x)
Correct answer: 1/(1 - x)²
Differentiating the series for 1/(1 - x) term by term gives the power series for 1/(1 - x)². Since the derivative of 1/(1 - x) is 1/(1 - x)², differentiating the geometric series the sum of xⁿ term by term yields the sum of n xⁿ⁻¹, which represents 1/(1 - x)². Integrating instead would give ln(1 - x) type results, and the other options correspond to different operations or denominators.
- Using the known Maclaurin series for eˣ, what is the Maclaurin series for e raised to x²?
- The sum of x²ⁿ/n!
- The sum of xⁿ/n!
- The sum of x²ⁿ/(2n)!
- The sum of (x²)ⁿ / (2n)!
Correct answer: The sum of x²ⁿ/n!
The Maclaurin series for e raised to x² is the sum of x²ⁿ/n!. Substituting x² in place of x in the series for eˣ, which is the sum of xⁿ/n!, gives the sum of (x²)ⁿ/n! = the sum of x²ⁿ/n!. Leaving xⁿ is the original eˣ series without substitution, and using (2n)! in the denominator wrongly applies the cosine or hyperbolic factorial pattern; only the index inside the power is doubled, not the factorial.
- What is the coefficient of the x⁴ term in the Maclaurin series of f(x) = cos(x)?
Correct answer: 1/24
The coefficient of x⁴ in the Maclaurin series for cos(x) is 1/24. The series for cosine is 1 - x²/2! + x⁴/4! - x⁶/6! + ..., so the x⁴ term is +x⁴/4! = x⁴/24, giving a positive coefficient of 1/24 because the signs alternate starting positive and the fourth-power term lands on a plus. The value -1/24 uses the wrong sign, and 1/2 or -1/2 belong to the x² term.
- The Taylor series of f centered at x = 2 has coefficient of (x - 2)³ equal to f'''(2)/3!. If f'''(2) = 12, what is that coefficient?
Correct answer: 2
The coefficient of (x - 2)³ is 2. A Taylor series coefficient of the (x - c)ⁿ term is the nth derivative at c divided by n factorial; here n = 3, so the coefficient is f'''(2)/3! = 12/6 = 2. The value 12 forgets to divide by the factorial, 4 divides by the wrong factorial, and 6 reports the value of 3! rather than the quotient.
- Use the Lagrange error bound to estimate the maximum error when approximating e⁰·¹ with the second-degree Maclaurin polynomial of eˣ. The bound uses the next derivative; on the interval from 0 to 0.1 a usable bound for eˣ is about which value?
- A small value near e⁰·¹ times (0.1)³ / 3!
- Exactly (0.1)³
- The value of the third term in the polynomial
- Zero, because the polynomial is exact
Correct answer: A small value near e⁰·¹ times (0.1)³ / 3!
The maximum error is bounded by about e⁰·¹ times (0.1)³ divided by 3!. The Lagrange error bound after a degree-2 polynomial is the maximum of the third derivative on the interval times the absolute value of (x - 0)³ divided by 3!; for eˣ every derivative is eˣ, whose maximum on the interval is e⁰·¹, giving the stated bound. Using (0.1)³ alone omits the derivative bound and factorial, the polynomial's own term is not an error bound, and the approximation is not exact.
- The Maclaurin series for arctan(x) is the sum from n = 0 of (-1)ⁿ x²ⁿ⁺¹/(2n + 1). What is the radius of convergence of this series?
Correct answer: 1
The radius of convergence of the arctan(x) series is 1. This series is obtained by integrating the geometric series for 1/(1 + x²), which converges only for the absolute value of x less than 1; term-by-term integration preserves the radius of convergence, so the radius remains 1. An infinite radius would apply to series like eˣ or sin e, and the values 1/2 or 2 do not match the geometric-series origin of this expansion.
- Evaluate the infinite series the sum from n = 0 of (-1)ⁿ / (2n + 1), which is a famous series. To what value does it converge?
Correct answer: π/4
The series the sum of (-1)ⁿ/(2n + 1) converges to π/4. Substituting x = 1 into the Maclaurin series for arctan(x), which is the sum of (-1)ⁿ x²ⁿ⁺¹/(2n + 1), gives arctan(1) = π/4, and the resulting numerical series is exactly 1 - 1/3 + 1/5 - 1/7 + .... The value ln 2 comes from the alternating harmonic series, π/2 doubles the result, and 1 is only the first term.
- Consider the sequence defined recursively by a_1 = 2 and a_(n+1) = (a_n)/2 + 1. Assuming it converges, what is its limit L?
Correct answer: 2
The limit of the sequence is 2. If a convergent sequence has limit L, then both a_n and a_(n+1) approach L, so the recursive rule becomes L = L/2 + 1; solving gives L - L/2 = 1, so L/2 = 1 and L = 2. The value 1 solves the equation incorrectly, 4 doubles the result, and 0 ignores the constant term in the recursion.
- A series the sum of a_n is conditionally convergent. According to the Riemann rearrangement theorem, what can happen if its terms are rearranged?
- The rearranged series can be made to converge to any chosen value or to diverge
- The sum stays exactly the same no matter the order
- The series always becomes absolutely convergent
- The series always diverges after rearrangement
Correct answer: The rearranged series can be made to converge to any chosen value or to diverge
For a conditionally convergent series, rearranging the terms can make it converge to any chosen value or even diverge. This is the Riemann rearrangement theorem: because the positive and negative parts each diverge separately in a conditionally convergent series, reordering can steer the partial sums anywhere. The sum is not order-independent (that property holds only for absolutely convergent series), and rearrangement does not force absolute convergence or guaranteed divergence.
- Which statement about absolute and conditional convergence is correct?
- Absolute convergence implies convergence, but convergence does not imply absolute convergence
- Convergence always implies absolute convergence
- Absolute convergence and conditional convergence are the same thing
- A series can be both absolutely and conditionally convergent at once
Correct answer: Absolute convergence implies convergence, but convergence does not imply absolute convergence
The correct statement is that absolute convergence implies convergence, but convergence does not imply absolute convergence. If the series of absolute values converges, the original series must converge; however, a series like the alternating harmonic series converges while its absolute-value series diverges, showing convergence alone is weaker. Conditional and absolute convergence are distinct, and a single series cannot be both because conditional convergence specifically requires the absolute-value series to diverge.
- Estimate the sum of the series the sum of 1/n² using the first two terms, then identify the error bound from the integral test remainder. The remainder after two terms is bounded above by which integral expression?
- The integral from 2 to infinity of 1/x² dx, equal to 1/2
- The integral from 0 to 2 of 1/x² dx
- The second term 1/4 itself
- Zero, since two terms give the exact sum
Correct answer: The integral from 2 to infinity of 1/x² dx, equal to 1/2
The remainder after two terms is bounded above by the integral from 2 to infinity of 1/x² dx, which equals 1/2. The integral test remainder estimate states that the tail after the Nth term is at most the integral of the function from N to infinity; with N = 2 and f(x) = 1/x², this integral evaluates to 1/2. Integrating from 0 to 2 reverses the region, the single term 1/4 is not the tail bound, and two terms do not give the exact sum since the true sum is π²/6.
- Determine whether the series the sum of cos(n · π)/n converges, and classify the type of convergence.
- It converges conditionally
- It converges absolutely
- It diverges
- It converges to ln 2 absolutely
Correct answer: It converges conditionally
The series converges conditionally. Since cos(n · π) equals (-1)ⁿ, the series is the alternating series the sum of (-1)ⁿ/n, which converges by the Alternating Series Test because 1/n decreases to zero; however, the series of absolute values is the harmonic series the sum of 1/n, which diverges. Convergence that fails when absolute values are taken is conditional convergence, so it is not absolutely convergent and it does not diverge.
- Evaluate the indefinite integral of (3x² - 4x + 5) dx.
- 6x - 4 + C
- x³ - 4x² + 5x + C
- 3x³ - 2x² + 5x + C
- x³ - 2x² + 5x + C
Correct answer: x³ - 2x² + 5x + C
The antiderivative is x³ - 2x² + 5x + C. The power rule for antiderivatives raises each exponent by one and divides by the new exponent, so 3x² becomes x³, -4x becomes -2x², and the constant 5 becomes 5x. Adding the constant of integration C gives x³ - 2x² + 5x + C.
- Evaluate the indefinite integral of (1/x) dx for x greater than 0.
- -1/x² + C
- 1 + C
- X ln(x) + C
- ln(x) + C
Correct answer: ln(x) + C
The antiderivative is ln(x) + C. The reciprocal function 1/x does not fit the standard power rule because that would require dividing by a zero exponent, so its antiderivative is the natural logarithm. For x greater than 0 the absolute value is unnecessary, giving ln(x) + C.
- Evaluate the indefinite integral of sec ²(x) dx.
- sec(x) + C
- -cot(x) + C
- sec(x) tan(x) + C
- tan(x) + C
Correct answer: tan(x) + C
The antiderivative is tan(x) + C. Since the derivative of tan(x) is sec ²(x), reversing that derivative shows the antiderivative of sec ²(x) is tan(x). Adding the constant of integration gives tan(x) + C.
- Evaluate the definite integral from 0 to π/2 of cos(x) dx.
Correct answer: 1
The definite integral equals 1. An antiderivative of cos(x) is sin(x), and evaluating from 0 to π/2 gives sin(π/2) - sin(0) = 1 - 0 = 1. Computing the antiderivative at the upper limit minus its value at the lower limit yields 1.
- If the integral from 1 to 5 of f(x) dx equals 12 and the integral from 1 to 5 of g(x) dx equals 3, what is the integral from 1 to 5 of [2 f(x) - g(x)] dx?
Correct answer: 21
The value is 21. By the linearity properties of definite integrals, constants factor out and sums split apart, so the integral of 2 f(x) - g(x) equals 2 times 12 minus 3. That gives 24 - 3 = 21.
- Given that the integral from 0 to 3 of f(x) dx equals 10 and the integral from 3 to 7 of f(x) dx equals 4, what is the integral from 0 to 7 of f(x) dx?
Correct answer: 14
The value is 14. The additivity property of definite integrals lets adjacent intervals be combined, so the integral over [0, 7] equals the integral over [0, 3] plus the integral over [3, 7]. That gives 10 + 4 = 14.
- What is the value of the integral from 4 to 4 of (x³ + 7x) dx?
Correct answer: 0
The value is 0. A definite integral whose upper and lower limits are identical encloses no width, so it represents zero accumulation regardless of the integrand. Therefore the integral from 4 to 4 of any function equals 0.
- If the integral from 2 to 6 of f(x) dx equals 9, what is the integral from 6 to 2 of f(x) dx?
Correct answer: -9
The value is -9. Reversing the limits of a definite integral changes its sign, so the integral from 6 to 2 equals the negative of the integral from 2 to 6. That gives -9.
- Find the derivative with respect to x of the integral from 0 to x² of sin(t) dt.
- sin(x²)
- 2x sin(x²)
- cos(x²)
- 2x cos(x²)
Correct answer: 2x sin(x²)
The derivative is 2x sin(x²). The Fundamental Theorem combined with the chain rule requires evaluating the integrand at the upper limit and multiplying by the derivative of that upper limit. Here the integrand sin(t) becomes sin(x²), and the derivative of x² is 2x, giving 2x sin(x²).
- Evaluate the integral of (2x + 1)⁵ dx using substitution.
- (2x + 1)⁶ / 6 + C
- (2x + 1)⁶ / 12 + C
- 5(2x + 1)⁴ + C
- (2x + 1)⁶ / 2 + C
Correct answer: (2x + 1)⁶ / 12 + C
The antiderivative is (2x + 1)⁶ / 12 + C. Letting u = 2x + 1 gives du = 2 dx, so dx = (1/2) du, and the integral becomes (1/2) times the integral of u⁵ du = (1/2)(u⁶ / 6) = u⁶ / 12. Substituting back gives (2x + 1)⁶ / 12 + C.
- A particle moves along a line with velocity v(t) = 3t² meters per second. What is the total displacement from t = 0 to t = 2 seconds?
- 12 meters
- 8 meters
- 6 meters
- 24 meters
Correct answer: 8 meters
The displacement is 8 meters. Displacement equals the definite integral of velocity over the time interval, so it is the integral from 0 to 2 of 3t² dt. An antiderivative is t³, and evaluating from 0 to 2 gives 8 - 0 = 8 meters.
- Water flows into a tan k at a rate of r(t) = 4t gallons per hour. How many gallons enter the tan k during the first 3 hours?
- 12 gallons
- 18 gallons
- 36 gallons
- 24 gallons
Correct answer: 18 gallons
The amount is 18 gallons. The total accumulated amount equals the definite integral of the rate over the interval, so it is the integral from 0 to 3 of 4t dt. An antiderivative is 2t², and evaluating from 0 to 3 gives 2(9) - 0 = 18 gallons.
- A population grows at a rate of P'(t) = 50e⁰·¹ᵗ people per year. If the population is 800 at t = 0, which expression gives the population at t = 10?
- 800 + the integral from 0 to 10 of 50e⁰·¹ᵗ dt
- The integral from 0 to 10 of 50e⁰·¹ᵗ dt
- 800 times 50e¹
- 50e⁰·¹(10)
Correct answer: 800 + the integral from 0 to 10 of 50e⁰·¹ᵗ dt
The correct expression is 800 plus the integral from 0 to 10 of 50e⁰·¹ᵗ dt. The accumulation function adds the net change over the interval to the initial value, and the net change is the integral of the rate. Adding the starting population 800 to that accumulated change gives the population at t = 10.
- Evaluate the integral of x² sin(x) dx using integration by parts.
- -x² cos(x) + 2x sin(x) + 2 cos(x) + C
- x² cos(x) - 2x sin(x) + C
- -x² cos(x) - 2x sin(x) + C
- x² sin(x) - 2x cos(x) + C
Correct answer: -x² cos(x) + 2x sin(x) + 2 cos(x) + C
The antiderivative is -x² cos(x) + 2x sin(x) + 2 cos(x) + C. Applying integration by parts with u = x² and dv = sin(x) dx gives -x² cos(x) plus the integral of 2x cos(x) dx, and a second integration by parts on that remaining integral produces 2x sin(x) + 2 cos(x). Combining these results yields -x² cos(x) + 2x sin(x) + 2 cos(x) + C.
- Evaluate the integral of eˣ sin(x) dx using integration by parts.
- eˣ sin(x) + C
- (1/2) eˣ (sin(x) - cos(x)) + C
- eˣ (sin(x) - cos(x)) + C
- (1/2) eˣ (sin(x) + cos(x)) + C
Correct answer: (1/2) eˣ (sin(x) - cos(x)) + C
The antiderivative is (1/2) eˣ (sin(x) - cos(x)) + C. Applying integration by parts twice reproduces the original integral on the right side, and solving the resulting equation for the integral isolates it. The algebra gives (1/2) eˣ (sin(x) - cos(x)) + C.
- Evaluate the integral of (x + 5) / [x(x - 1)] dx using partial fractions.
- -5 ln |x| + 6 ln |x - 1| + C
- 5 ln |x| - 6 ln |x - 1| + C
- ln |x| + ln |x - 1| + C
- -5 ln |x| - 6 ln |x - 1| + C
Correct answer: -5 ln |x| + 6 ln |x - 1| + C
The antiderivative is -5 ln |x| + 6 ln |x - 1| + C. Decomposing (x + 5)/[x(x - 1)] as A/x + B/(x - 1) and clearing denominators gives x + 5 = A(x - 1) + Bx; substituting x = 0 gives A = -5, and substituting x = 1 gives B = 6. Integrating -5/x + 6/(x - 1) produces -5 ln |x| + 6 ln |x - 1| + C.
- Evaluate the improper integral from 0 to 1 of (1 / √x) dx.
Correct answer: 2
The improper integral equals 2. Because the integrand is unbounded at x = 0, the integral is rewritten as the limit as a approaches 0 from the right of the integral from a to 1; the antiderivative of x⁻¹ᐟ² is 2 √x. Evaluating gives 2 √1 - 2 √a, and as a approaches 0 this approaches 2.
- Evaluate the improper integral from 1 to infinity of (1 / x²) dx.
Correct answer: 1
The improper integral equals 1. Rewriting it as the limit as b approaches infinity of the integral from 1 to b of x⁻² dx, the antiderivative is -1/x, evaluated from 1 to b gives -1/b + 1. As b approaches infinity, -1/b approaches 0, so the value is 1.
- Approximate the integral from 0 to 2 of f using a left Riemann sum with 2 equal subintervals, given f(0) = 4, f(1) = 6, and f(2) = 9.
Correct answer: 10
The left Riemann sum approximation is 10. With 2 equal subintervals on [0, 2], each has width 1, and a left sum uses the left endpoints x = 0 and x = 1. Multiplying f(0) = 4 and f(1) = 6 by the width 1 gives 4 + 6 = 10.
- For a function f that is concave up on [a, b], how does a trapezoidal approximation of the integral of f compare to the exact value?
- It underestimates the integral
- It equals the integral exactly
- It overestimates the integral
- It cannot be determined
Correct answer: It overestimates the integral
A trapezoidal approximation overestimates the integral when f is concave up. The straight chord of each trapezoid lies above a curve that bends upward, so each trapezoid encloses slightly more area than the region under the curve. Summing these too-large pieces gives a total above the exact integral.
- What is the average value of f(x) = 4x³ on the interval [0, 1]?
Correct answer: 1
The average value is 1. The average value of a function on [a, b] equals 1/(b - a) times the integral from a to b, so here it is 1/(1 - 0) times the integral from 0 to 1 of 4x³ dx. The antiderivative x⁴ evaluated from 0 to 1 gives 1 - 0 = 1, and dividing by the width 1 gives 1.
- The Mean Value Theorem for Integrals guarantees that for f(x) = x² on [0, 3], there is a value c in [0, 3] where f(c) equals the average value. What is that average value?
Correct answer: 3
The average value is 3. The average value equals 1/(3 - 0) times the integral from 0 to 3 of x² dx; the antiderivative x³/3 evaluated from 0 to 3 gives 9, and dividing by the width 3 gives 3. The Mean Value Theorem for Integrals then guarantees some c where f(c) = 3.
- Evaluate the integral of (6x) / (x² + 1) dx using substitution.
- 6 ln(x² + 1) + C
- 3 ln(x² + 1) + C
- 6 arctan(x) + C
- ln(x² + 1) + C
Correct answer: 3 ln(x² + 1) + C
The antiderivative is 3 ln(x² + 1) + C. Letting u = x² + 1 gives du = 2x dx, so 6x dx = 3 du, and the integral becomes 3 times the integral of (1/u) du = 3 ln |u|. Since x² + 1 is always positive, substituting back gives 3 ln(x² + 1) + C.
- Evaluate the definite integral from 1 to e of (1/x) dx.
Correct answer: 1
The definite integral equals 1. An antiderivative of 1/x is ln(x), and evaluating from 1 to e gives ln(e) - ln(1) = 1 - 0 = 1. Computing the antiderivative at the upper limit minus its value at the lower limit yields 1.
- Find the derivative with respect to x of the integral from x to 5 of cos(t²) dt.
- cos(x²)
- -cos(x²)
- cos(25)
- -cos(25)
Correct answer: -cos(x²)
The derivative is -cos(x²). When the variable x is the lower limit, reversing the limits introduces a negative sign, so the derivative equals the negative of the integrand evaluated at x. Replacing t with x in cos(t²) and applying the sign gives -cos(x²).
- Evaluate the integral of tan(x) dx.
- sec ²(x) + C
- ln |cos(x)| + C
- -ln |cos(x)| + C
- ln |sec(x) + tan(x)| + C
Correct answer: -ln |cos(x)| + C
The antiderivative is -ln |cos(x)| + C. Writing tan(x) as sin(x)/cos(x) and letting u = cos(x) gives du = -sin(x) dx, so the integral becomes the negative of the integral of (1/u) du = -ln |u|. Substituting back gives -ln |cos(x)| + C.
- Evaluate the integral of x cos(x) dx using integration by parts.
- X sin(x) + cos(x) + C
- X sin(x) - cos(x) + C
- -x sin(x) + cos(x) + C
- X cos(x) + sin(x) + C
Correct answer: X sin(x) + cos(x) + C
The antiderivative is x sin(x) + cos(x) + C. Choosing u = x and dv = cos(x) dx gives du = dx and v = sin(x), so the formula uv minus the integral of v du produces x sin(x) minus the integral of sin(x) dx. Since the integral of sin(x) is -cos(x), this becomes x sin(x) + cos(x) + C.
- A bacteria culture changes at a rate of R(t) = 6 - 2t thousand cells per hour for t in [0, 5]. What is the net change in the culture from t = 0 to t = 5?
- 5 thousand cells
- -5 thousand cells
- 30 thousand cells
- 10 thousand cells
Correct answer: 5 thousand cells
The net change is 5 thousand cells. Net change equals the definite integral of the rate, so it is the integral from 0 to 5 of (6 - 2t) dt. An antiderivative is 6t - t², and evaluating from 0 to 5 gives 30 - 25 = 5 thousand cells.
- Evaluate the integral of 5 dx where 5 is a constant.
Correct answer: 5x + C
The antiderivative is 5x + C. The antiderivative of a constant k is kx, because differentiating kx returns the constant k. Applying this to the constant 5 gives 5x + C.
- Approximate the integral from 1 to 7 of f using a right Riemann sum with 3 equal subintervals, given f(1) = 2, f(3) = 5, f(5) = 8, and f(7) = 4.
Correct answer: 34
The right Riemann sum approximation is 34. With 3 equal subintervals on [1, 7], each has width 2, and a right sum uses the right endpoints x = 3, 5, and 7. Multiplying f(3) = 5, f(5) = 8, and f(7) = 4 by the width 2 gives 2(5 + 8 + 4) = 2(17) = 34.
- Evaluate the definite integral from 0 to 2 of (4x - 3) dx.
Correct answer: 2
The definite integral equals 2. An antiderivative of 4x - 3 is 2x² - 3x, and evaluating from 0 to 2 gives [2(4) - 6] - 0 = 8 - 6 = 2. Computing the antiderivative at the upper limit minus its value at the lower limit yields 2.
