Question 1 of 25
A tank’s salt amount satisfies . If , which is ?
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AP Calculus AB
Practice Test 10 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.
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Question 1 of 25
A tank’s salt amount S satisfies dtdS=0.03S. If S(0)=50, which is S(t)?
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A tank’s salt amount S satisfies dtdS=0.03S. If S(0)=50, which is S(t)?
Explanation: This problem involves solving an exponential growth differential equation with a specific initial condition. The equation dS/dt = 0.03S indicates the salt amount grows at a rate proportional to its current value, leading to exponential growth. The general solution is S(t) = Ce^(0.03t), and applying the initial condition S(0) = 50 gives us C = 50, so S(t) = 50e^(0.03t). The positive coefficient (0.03) ensures the salt amount increases over time. Choice B (50 + 0.03t) represents linear growth where salt increases at a constant rate of 0.03 units per time, not the accelerating growth described by our differential equation. When solving dy/dt = ky with y(0) = y₀, always substitute to find C = y₀, giving y(t) = y₀e^(kt).
If f′′(x)=3−x for all x, where is f concave up and where is it concave down?
Explanation: This question requires determining concavity from a linear second derivative function. Given f''(x) = 3 - x, we need to find where f''(x) > 0 (concave up) and where f''(x) < 0 (concave down). Setting f''(x) = 0 gives 3 - x = 0, so x = 3 is our critical point. For x < 3, we have 3 - x > 0, so f is concave up on (-∞,3); for x > 3, we have 3 - x < 0, so f is concave down on (3,∞). Students often make the error of reversing the inequality when solving 3 - x > 0, incorrectly concluding x > 3. The key strategy is to remember that f''(x) > 0 means concave up, and always test the sign of f'' on each side of critical points.
If f′′(x)<0 for x<0 and f′′(x)>0 for 0<x<8, where is f concave up?
Explanation: This problem provides direct information about the sign of f''(x) and asks where f is concave up. A function is concave up wherever its second derivative is positive. We're told that f''(x) < 0 for x < 0, which means f is concave down on (-∞, 0), and f''(x) > 0 for 0 < x < 8, which means f is concave up on (0, 8). Therefore, f is concave up on the interval (0, 8). Choice E incorrectly relates concavity to the monotonic behavior of f'(x) in a reversed manner. The fundamental rule: f is concave up where f''(x) > 0.
A continuous function g on [−4,2] has g(−4)=0 and g(2)=5; which statement must be true?
Explanation: This question applies the Extreme Value Theorem to determine what must be guaranteed for a continuous function on a closed interval with specific endpoint values. Since g is continuous on the closed interval [−4,2], the EVT unconditionally guarantees that g must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The given endpoint values g(−4)=0 and g(2)=5 provide information about the function but don't determine where the absolute extrema occur, as the maximum or minimum could occur at interior points. We cannot conclude specific extrema locations without knowing the function's behavior throughout the entire interval. Choice A incorrectly assumes the minimum occurs at x=−4, and choice C incorrectly assumes the maximum occurs at x=2. The fundamental EVT guarantee: continuous function on closed interval ensures both absolute maximum and minimum exist somewhere.
Given f′′(x)>0 for x<1 and f′′(x)<0 for x>1, where is f concave up and concave down?
Explanation: This question tests your understanding of how the second derivative determines concavity. When f''(x) > 0, the function is concave up (shaped like a U), and when f''(x) < 0, the function is concave down (shaped like ∩). Since f''(x) > 0 for x < 1, the function is concave up on (-∞, 1), and since f''(x) < 0 for x > 1, the function is concave down on (1, ∞). Choice B incorrectly reverses these intervals, likely confusing the sign of the second derivative with concavity direction. Remember: positive second derivative means concave up, negative second derivative means concave down.
For h(x)={x+2,x2,x<1x≥1, what type of discontinuity does h have at x=1?
Explanation: The function h(x) has a jump discontinuity at x=1. From the left, limx→1−h(x)=limx→1−(x+2)=3. From the right, limx→1+h(x)=limx→1+(x2)=1. At x=1, we use the second formula since 1≥1, so h(1)=12=1. The left and right limits exist but are different (3 vs 1), which is the defining characteristic of a jump discontinuity. Students might think this is removable since h(1) equals one of the limits, but removable discontinuities require both one-sided limits to be equal. A jump discontinuity occurs when both one-sided limits exist as finite values but are not equal to each other.
For γ(x)={x2+1,0,x=−2x=−2, what type of discontinuity is at x=−2?
Explanation: This function has a removable discontinuity at x = -2. Since γ(x) = x²+1 for x ≠ -2, the limit as x approaches -2 is lim(x→-2) (x²+1) = 5, but γ(-2) = 0. Since the limit exists but doesn't equal the function value (5 ≠ 0), this creates a removable discontinuity. The discontinuity could be 'removed' by redefining γ(-2) = 5. Students might think this is continuous because x²+1 is a polynomial, but the piecewise definition changes the value at x = -2. To identify removable discontinuities: check if polynomial limits differ from redefined function values at specific points.
If t is differentiable and one-to-one with t(1)=2 and t′(1)=−6, what is (t−1)′(2)?
Explanation: This question requires finding the derivative of an inverse function. The formula (t−1)′(b)=t′(a)1 applies when t(a)=b. Given t(1)=2, we know t−1(2)=1, and we need (t−1)′(2). Using the formula with a=1 and b=2: (t−1)′(2)=t′(1)1=−61=−61. Choice A (−6) represents t′(1) without reciprocation, a common mistake. Remember: the derivative of an inverse function equals one divided by the original derivative at the corresponding input.
A particle’s velocity is modeled by v(t)=sec(3t). What is v′(t)?
Explanation: This problem requires differentiating the secant function, one of the reciprocal trigonometric functions. The derivative of sec(u) is sec(u)tan(u) times the derivative of u. Here, u = 3t, so by the chain rule, we get v'(t) = sec(3t)tan(3t) · 3 = 3sec(3t)tan(3t). A common error is forgetting the chain rule factor of 3, which would give just sec(3t)tan(3t) (choice B). Remember that derivatives of reciprocal trig functions always involve products: sec and tan go together, as do csc and cot.
A function satisfies dxdy=−4xy. What is the general solution for y(x)?
Explanation: This problem requires solving a differential equation using separation of variables. With dy/dx = -4 x y, separate as dy/y = -4 x dx, assuming y ≠ 0. Integrate for ln|y| = -2 x^2 + C. Solve to y = C e^{-2 x^2}. The option y = e^{-2 x^2} + C is wrong as it adds C outside. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.
The velocity of a runner is v(t)=15t2+6t−1. Which is an antiderivative of v(t)?
Explanation: This question tests the skill of finding antiderivatives by reversing the power rule for differentiation. To find the antiderivative of v(t) = 15t² + 6t - 1, increase the exponent of each term by one and divide by the new exponent, resulting in (15/3)t³ + (6/2)t² - t + C. This process reverses differentiation because differentiating 5t³ + 3t² - t + C yields 15t² + 6t - 1, matching the original velocity. The constant C accounts for the fact that differentiation eliminates constants, so antiderivatives differ by a constant. A tempting distractor like choice D fails because it incorrectly uses a coefficient of 15 instead of 5 for the t³ term. Always remember that integration is the reverse of differentiation, so check your antiderivative by differentiating it back to see if you get the original function.
The temperature of a cooling drink is T(t) degrees Celsius after t minutes. Interpret T′(5).
Explanation: This question involves interpreting the derivative in a temperature context. Since T(t) represents temperature in degrees Celsius after t minutes, T'(5) represents the instantaneous rate of change of temperature with respect to time at exactly t = 5 minutes. This tells us how fast the temperature is changing at that specific moment, measured in degrees Celsius per minute. Students often confuse this with the actual temperature T(5) or the average rate of change over 5 minutes. The derivative T'(5) specifically captures the instantaneous rate at t = 5, not an average or total value. When interpreting derivatives, remember they represent instantaneous rates of change with units of output per input at a specific input value.
A particle has v(2)<0 and a(2)>0; which statement about direction at t=2 is correct?
Explanation: This straight-line motion problem focuses on directional information from velocity and acceleration signs. Since v(2)<0, the particle has negative velocity, meaning it's moving left at t=2. The sign of acceleration a(2)>0 tells us about speed changes but doesn't determine the current direction of motion. Only the velocity sign determines direction: negative velocity means moving left. Students might think positive acceleration affects the current direction, but direction depends solely on velocity sign. The key strategy is that velocity sign determines direction (positive = right, negative = left).
A continuous function g on [0,6] has critical points x=1,5. If g(0)=3,g(1)=4,g(5)=2,g(6)=5, where is the absolute maximum?
Explanation: This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=1, but g(1)=4 is less than g(6)=5, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
If an antiderivative of g is G(x)=sinx+x2, find ∫0πg(x)dx.
Explanation: This problem applies the Fundamental Theorem of Calculus Part 2, where we use a known antiderivative to evaluate a definite integral. Since G(x) is an antiderivative of g(x), we have G'(x) = g(x), which means ∫[0 to π] g(x)dx = G(π) - G(0). We evaluate G(π) = sin(π) + π² = 0 + π² = π² and G(0) = sin(0) + 0² = 0 + 0 = 0. Therefore, ∫[0 to π] g(x)dx = G(π) - G(0) = π² - 0 = π². Option A represents the common error of subtracting in the wrong order (lower minus upper), which would give a negative result when the integral should be positive. Remember that FTC Part 2 always evaluates as F(upper) - F(lower), maintaining the direction of integration.
A car’s distance from home is d(t). Which expression is the instantaneous speed at t=6 hours?
Explanation: Instantaneous speed is the magnitude of instantaneous velocity, which requires the derivative of the distance function. Choice D uses the limit definition of the derivative, capturing the instantaneous rate at exactly t=6. Choices A, B, and E all calculate average speeds over various intervals (1 hour, 4 hours, and 4 hours respectively). Choice C gives the total distance traveled in 6 hours, not a rate. Many students think that d(6)-d(5) represents instantaneous speed because it uses a small interval, but this is still the average speed over that hour. The key principle: instantaneous rates require limits as the time interval approaches zero, while any calculation over a fixed interval gives an average rate.
The region between y=6−x2 and y=2x+1 on 0≤x≤1 is rotated about the x-axis; which washer integral is correct?
Explanation: This problem requires the washer method for rotation about the x-axis with a parabola and linear function. We must identify which curve is farther from the x-axis on [0,1]. At x = 0: y = 6 - 0² = 6, while y = 2(0) + 1 = 1, so the parabola is above. At x = 1: y = 6 - 1² = 5, while y = 2(1) + 1 = 3, confirming y = 6 - x² remains above throughout. For rotation about the x-axis, the outer radius is (6 - x²) and the inner radius is (2x + 1), giving the washer integral π∫₀¹[(6 - x²)² - (2x + 1)²]dx. Option B incorrectly places (2x + 1)² first, reversing the subtraction and producing a negative result. Remember the washer formula structure: π∫[R²(x) - r²(x)]dx where R is the outer (farther) radius.
A slope field for dxdy=y−x is shown. Which statement about the solution through (0,1) is true?
Explanation: This question requires using slope field reasoning to analyze solution behavior. For dy/dx = y - x at point (0,1), the slope is 1 - 0 = 1, so the curve is initially increasing. Moving slightly right from (0,1) to small positive x values, the slopes remain positive and even increase since y > x in that region, indicating the curve is concave up. Nearby points show slopes that increase as we move right and up, confirming concave up behavior. Choice A fails because the solution increases rather than decreases initially. To read slope fields effectively, first evaluate the slope at the given point, then examine how slopes change in the immediate neighborhood to determine concavity.
A continuous function f satisfies f(0)=−1 and f(1)=2. Which must be true?
Explanation: The Intermediate Value Theorem (IVT) ensures continuous f with f(0) = -1 and f(1) = 2 takes values between -1 and 2. Since -1 < 0 < 2, there exists c in [0, 1] with f(c) = 0, and as 0 ≠ -1, 0 ≠ 2, c is in (0, 1). This is a root guarantee from opposite signs. A common error is choosing 3, outside the range. Unequal endpoints do not prevent application; they enable it here. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).
If S(x)=∫4xtcostdt, what is S′(x)?
Explanation: This problem applies the Fundamental Theorem of Calculus Part 1, which tells us that differentiating an integral with variable upper limit gives the integrand evaluated at that limit. For S(x)=∫4xtcostdt, taking the derivative yields S′(x)=tcost evaluated at t=x. This gives us S′(x)=xcosx, which is the integrand with t replaced by x. Choice B (xsinx) might attract students who mistakenly think they need to integrate cost to get sint, but we're differentiating the accumulation function, not integrating. To recognize FTC Part 1, look for dxd of an integral—the answer is always the integrand evaluated at x.
A data-fitting step requires ∫x+5x2+10x+30dx. Which is an antiderivative?
Explanation: This problem requires performing polynomial long division as preparatory algebra before integrating the rational function. The algebraic step is necessary because the degree of the numerator is equal to the degree of the denominator plus one, making the rational function improper. To integrate, we must first divide to express it as a polynomial plus a proper fraction. This allows us to integrate the polynomial terms directly and use logarithmic integration for the fractional remainder. A tempting distractor like choice B fails because it uses a coefficient of 1 instead of 5 for the logarithmic term, resulting from an incorrect remainder in the division. Always perform long division or simplify rational functions before integration to ensure accurate antiderivatives.
Let u(x)={x−1x2−1,2,x=1x=1. Is u continuous at x=1, and why?
Explanation: For continuity at x = 1, we need u(1) defined, lim[x→1] u(x) to exist, and these values to match. The function defines u(1) = 2, satisfying condition 1. To find the limit, we simplify the rational expression: lim[x→1] (x² - 1)/(x - 1) = lim[x→1] (x + 1)(x - 1)/(x - 1) = lim[x→1] (x + 1) = 2, so the limit exists and equals 2. Since lim[x→1] u(x) = 2 = u(1), all continuity conditions are satisfied. This shows how removable discontinuities can be "removed" by defining the function value to equal the limit. Continuity checklist: (1) u(1) = 2 ✓, (2) Limit = 2 after factoring ✓, (3) They match ✓.
A continuous function g on [1,7] has critical points x=2,6. If g(1)=4,g(2)=1,g(6)=3,g(7)=0, where is the absolute minimum?
Explanation: This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=2, but g(2)=1 is greater than g(7)=0, so it's not the minimum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.
A function is S(t)=−14t2+7t+2. What is S′(t)?
Explanation: The sum and constant multiple rules under linearity simplify S(t) = -14t^2 + 7t + 2. Derivative of -14t^2 is -28t, of 7t is 7, of +2 is 0. Thus, S'(t) = -28t + 7. A common misuse is forgetting the negative on quadratic terms. Linear coefficients might be mishandled. For quadratics, use linearity to differentiate powers separately, ignoring constants.
For v(x)={1/x,0,x=0x=0, is v continuous at x=0, and why?
Explanation: For continuity at x=0, v(0) must be defined, limit exist, and equal. v(0)=0, but limit of 1/x doesn't exist (diverges to ±∞), so not continuous. The point value can't overcome the limit failure. Common mistake: thinking a defined value implies continuity. Infinite discontinuity here. Checklist: (1) Defined? (2) Limit exists? (3) Equals?