AP Calculus AB

AP Calculus AB Practice Test: Practice Test 2

Practice Test 2 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

For $t(x)=\dfrac{2}{1-x}$ , which limit expression represents the behavior of $t(x)$ as $x\to1^-$ ?

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Question 1

For $t(x)=\dfrac{2}{1-x}$ , which limit expression represents the behavior of $t(x)$ as $x\to1^-$ ?

$\displaystyle \lim_{x\to1^-} t(x)=+\infty$ (correct answer)
$\displaystyle \lim_{x\to1^+} t(x)=+\infty$
$\displaystyle \lim_{x\to1} t(x)=-\infty$
$\displaystyle t(1)=+\infty$
$\displaystyle \lim_{x\to-1^-} t(x)=+\infty$

Explanation: For t(x) = 2/(1-x), we analyze the behavior as x approaches 1 from the left (x→1⁻). When x is slightly less than 1, the denominator 1-x is a small positive number. The fraction becomes 2/(small positive) = large positive value. Therefore, lim[x→1⁻] t(x) = +∞ correctly represents this behavior. A common error is misidentifying the sign of (1-x) when x < 1; remember that if x < 1, then 1-x > 0. Another mistake is confusing the direction of approach or using incorrect notation like t(1) = +∞. Notation checklist: Carefully determine the sign of expressions like (a-x), use superscript - for left approach, and verify your sign analysis.

Question 2

A function $b$ is continuous on $[3,11]$ with $b(3)=20$ and $b(11)=4$ . Does the Mean Value Theorem guarantee a $c$ with $b'(c)=-2$ ?

Yes, because $b$ is continuous on $[3,11]$ and differentiable on $(3,11)$ , so some $c$ has $b'(c)=\dfrac{4-20}{11-3}=-2$ . (correct answer)
No, because MVT requires $b$ to be increasing.
Yes, because $b(11)-b(3)=-16$ .
No, because the derivative cannot be negative.
Yes, because $b$ is continuous, so it must have a point where $b'(c)=-2$ .

Explanation: The Mean Value Theorem applies when b is continuous on [3,11] and differentiable on (3,11). Since both conditions are met, MVT guarantees there exists c in (3,11) where b'(c) equals the average rate of change. The average rate of change is (b(11)-b(3))/(11-3) = (4-20)/8 = -16/8 = -2. Therefore, MVT does guarantee a point where b'(c) = -2. Students sometimes incorrectly think MVT requires increasing functions or that negative derivatives cannot occur, but the theorem applies to both increasing and decreasing functions. Derivatives can be positive, negative, or zero.

Question 3

The height of a ball is $h(t)$ . Which represents the ball’s instantaneous vertical velocity at $t=1$ ?

$h(1)-h(0)$
$\dfrac{h(2)-h(0)}{2-0}$
$\dfrac{h(1)-h(0)}{1-0}$
$\displaystyle \lim_{h\to 0}\dfrac{h(1+h)-h(1)}{h}$ (correct answer)
$h(1)$

Explanation: To find the ball's instantaneous vertical velocity at t=1, we need the derivative of height with respect to time at that exact moment. Options A and C both involve the change from t=0 to t=1, with C dividing by the time interval to get average velocity. Option B calculates average velocity over the two-second interval [0,2], while option E just gives the height at t=1. Only option D uses the limit definition, which captures instantaneous velocity by considering what happens as the time interval shrinks to zero around t=1. This limit process ensures we're measuring the rate at exactly t=1, not averaged over any interval. The key identifier: instantaneous rates always involve limits with h→0.

Question 4

What is the correct washer-method integral for the volume when the region between $y=x+2$ and $y=x^2+1$ on $[0,1]$ is revolved about the x-axis?

$\pi\displaystyle\int_{0}^{1}\big((x+2)^2-(x^2+1)^2\big)\,dx$ (correct answer)
$\pi\displaystyle\int_{0}^{1}(x+2)^2\,dx$
$\pi\displaystyle\int_{0}^{1}\big((x^2+1)^2-(x+2)^2\big)\,dx$
$\pi\displaystyle\int_{0}^{1}\big((x+2)-(x^2+1)\big)^2\,dx$
$\pi\displaystyle\int_{0}^{1}\big((x+2)^2+(x^2+1)^2\big)\,dx$

Explanation: This problem requires the washer method to find the volume when revolving around the x-axis. For washers around the x-axis, we need π∫[R²(x) - r²(x)]dx, where R(x) is the outer radius (distance from x-axis to the farther curve) and r(x) is the inner radius (distance from x-axis to the nearer curve). At x = 0: y = x + 2 gives 2, while y = x² + 1 gives 1, so the line y = x + 2 is farther from the x-axis. At x = 1: y = x + 2 gives 3, while y = x² + 1 gives 2, confirming that y = x + 2 remains the outer curve throughout [0,1]. Therefore, the integral is π∫₀¹[(x+2)² - (x²+1)²]dx. Choice D incorrectly squares the difference of the functions rather than taking the difference of squares. Remember: for washers around the x-axis, always use (outer radius)² - (inner radius)², where radii are the y-values of the curves.

Question 5

A tank’s net flow rate is $r(t)=6-2t$ gal/min for $0\le t\le 4$ . Net change in volume?

$8$ gallons (correct answer)
$0$ gallons
$12$ gallons
$-8$ gallons
$4$ gallons

Explanation: This problem requires accumulation reasoning to find volume change from a linear flow rate. The rate r(t) = 6 - 2t changes linearly from 6 gal/min at t = 0 to -2 gal/min at t = 4, so we integrate ∫₀⁴ (6-2t) dt. This equals [6t - t²]₀⁴ = (24 - 16) - 0 = 8 gallons net increase. The positive inflow initially exceeds the growing outflow, resulting in net gain. Students might choose 12 by evaluating only 6t at t = 4, ignoring the negative -t² term that reduces the accumulation. For linear rates, integrate the entire expression to capture the changing flow dynamics.

Question 6

For $h(x)=\ln\left(\dfrac{(x^2+1)^5}{\sqrt{3x-2}}\right)$ , which rules should be used to find $h'(x)$ ?

Power rule only
Product rule and quotient rule
Chain rule and properties of logarithms to simplify first (correct answer)
Implicit differentiation and power rule
Quotient rule only

Explanation: Selecting the appropriate differentiation procedure often involves recognizing when simplification can make the process more efficient. The function $h(x)=\ln\left(\frac{(x^2+1)^5}{\sqrt{3x-2}}\right)$ appears complex, but logarithm properties allow us to simplify before differentiating. Using the properties $\ln(a/b) = \ln(a) - \ln(b)$ and $\ln(a^n) = n\ln(a)$ , we can rewrite this as $h(x) = 5\ln(x^2+1) - \frac{1}{2}\ln(3x-2)$ . Now we can differentiate using the chain rule on each logarithmic term, which is much simpler than applying the quotient rule to the original expression. While the quotient rule could technically work on the argument inside the logarithm, it would create unnecessary complexity. The recognition strategy is to always check if logarithm properties can simplify your expression before differentiating—this often transforms a complex quotient or product into a sum or difference that's easier to handle.

Question 7

If $R(x)=\int_{1}^{x}\frac{e^{t}}{t}\,dt$ for $x>0$ , what is $R'(x)$ ?

$\displaystyle \int_{1}^{x}\frac{e^{t}}{t}\,dt$
$\dfrac{e^{x}}{x}$ (correct answer)
$\dfrac{e^{x}x-e^{x}}{x^2}$
$\displaystyle \frac{e^{t}}{t}\big|_{t=1}$
$\displaystyle \int_{1}^{x}\left(\frac{e^{t}t-e^{t}}{t^2}\right)dt$

Explanation: This problem applies the Fundamental Theorem of Calculus Part 1, where the derivative of an accumulation function equals the integrand evaluated at the upper limit of integration. Given $R(x) = \int_{1}^{x}\frac{e^t}{t}\,dt$ , we find $R'(x) = \frac{e^x}{x}$ by replacing t with x in the integrand. The process is direct evaluation, not integration or differentiation of the integrand. Choice C ( $\frac{e^x \cdot x - e^x}{x^2}$ ) shows the derivative of $\frac{e^x}{x}$ using the quotient rule, which would be incorrect since FTC Part 1 requires evaluation, not differentiation. To apply FTC Part 1 correctly, identify that x appears as the upper limit and evaluate the integrand at x.

Question 8

Let $P(x) = ax^3 + bx^2 + cx + d$ where $a$ , $b$ , $c$ , and $d$ are constants. If $\frac{d}{dx}[P(x) + Q(x)] = 6x^2 - 4x + 7$ and $Q'(x) = 2x^2 + x - 3$ , what is $P'(x)$ ?

$8x^2 - 3x + 10$
$4x^2 - 5x + 4$
$8x^2 - 3x + 4$
$4x^2 - 5x + 10$ (correct answer)

Explanation: When you encounter problems involving derivatives of sums, remember that the derivative of a sum equals the sum of derivatives: $\frac{d}{dx}[P(x) + Q(x)] = P'(x) + Q'(x)$ . This fundamental property is key to solving this problem. Since you know that $\frac{d}{dx}[P(x) + Q(x)] = 6x^2 - 4x + 7$ and $Q'(x) = 2x^2 + x - 3$ , you can substitute into the sum rule: $P'(x) + Q'(x) = 6x^2 - 4x + 7$ . Therefore, $P'(x) + (2x^2 + x - 3) = 6x^2 - 4x + 7$ . Solving for $P'(x)$ : $P'(x) = (6x^2 - 4x + 7) - (2x^2 + x - 3) = 6x^2 - 4x + 7 - 2x^2 - x + 3 = 4x^2 - 5x + 10$ . This confirms answer choice D. Choice A ( $8x^2 - 3x + 10$ ) results from incorrectly adding $P'(x) + Q'(x)$ instead of subtracting $Q'(x)$ . Choice B ( $4x^2 - 5x + 4$ ) shows correct coefficient work for the $x^2$ and $x$ terms but an error in the constant term ( $7 - (-3) = 10$ , not $4$ ). Choice C ( $8x^2 - 3x + 4$ ) combines both mistakes: adding instead of subtracting the polynomial terms AND miscalculating the constant. Always remember: when you have $f'(x) + g'(x) = h'(x)$ and need to find one derivative, isolate it by subtracting the known derivative from both sides. Double-check your arithmetic, especially with signs when subtracting polynomials.

Question 9

A position function has velocity $v(t)=2-t$ for $0\le t\le 4$ . What is the net displacement?

$0$ (correct answer)
$4$
$-4$
$2$
$-2$

Explanation: This problem uses accumulation reasoning to find net displacement from a linear velocity function. The velocity v(t) = 2 - t decreases from 2 to -2 over [0,4], changing from positive to negative at t = 2. We integrate ∫₀⁴ (2-t) dt = [2t - t²/2]₀⁴ = (8 - 8) - 0 = 0. The equal areas above and below the t-axis (where v > 0 and v < 0) cancel exactly. Students might choose 2 by evaluating the initial velocity, but accumulation requires integration over the entire interval. When velocity functions cross zero symmetrically, forward and backward displacements can cancel completely.

Question 10

Find $\lim\limits_{h\to 0}\dfrac{(5+h)^2-25}{h}$ , representing a difference quotient at $x=5$ .

$0$
$25$
$10$ (correct answer)
$5$
$\dfrac{1}{10}$

Explanation: This limit represents a difference quotient, a fundamental calculus concept, and initially gives 0/0. Expand the numerator: (5+h)² - 25 = 25 + 10h + h² - 25 = 10h + h². Factor out h to get h(10 + h). The expression becomes h(10 + h)/h, which simplifies to (10 + h) after canceling h. As h approaches 0, the limit equals 10 + 0 = 10. A common error is expanding incorrectly or not factoring out the common h. This demonstrates the essential technique for difference quotients: expand, factor out h, cancel, and substitute.

Question 11

Let $g(x)=-5x^3+4x^6-9x$ . What is $g'(x)$ ?

$-15x^2+24x^5-9$ (correct answer)
$-15x^2+4x^5-9$
$-5x^2+24x^5-9$
$-15x^3+24x^6-9x$
$15x^2+24x^5-9$

Explanation: To find g'(x) for g(x) = -5x^3 + 4x^6 - 9x, we apply the power rule to each term. The power rule gives us: -5x^3 becomes -5·3x^2 = -15x^2, 4x^6 becomes 4·6x^5 = 24x^5, and -9x becomes -9·1x^0 = -9. Therefore, g'(x) = -15x^2 + 24x^5 - 9. Choice B shows 4x^5 instead of 24x^5, which comes from forgetting to multiply 4 by 6 when differentiating 4x^6. Remember that the power rule requires multiplying the coefficient by the exponent before reducing the exponent.

Question 12

Let f and g be functions such that $\lim_{x \to 2} g(x) = 3$ . If f is continuous at $x=3$ and $f(3) = 5$ , what is $\lim_{x \to 2} f(g(x))$ ?

$2$
$3$
$5$ (correct answer)
$8$

Explanation: Because f is continuous at $\lim_{x \to 2} g(x) = 3$ , we can use the property for the limit of a composite function: $\lim_{x \to 2} f(g(x)) = f(\lim_{x \to 2} g(x)) = f(3) = 5$ . Distractor B is the limit of the inner function, not the composite function. Distractor D is the sum of the input and output values ( $3+5$ ).

Question 13

A formula uses $f(x)=\arctan(\tfrac{3}{2}x)$ . What is $f'(x)$ ?

$\dfrac{3/2}{1+\left(\tfrac{3}{2}x\right)^2}$ (correct answer)
$\dfrac{3}{1+\left(\tfrac{3}{2}x\right)^2}$
$\dfrac{1}{1+\left(\tfrac{3}{2}x\right)^2}$
$\tfrac{3}{2}\sec^2\!\left(\tfrac{3}{2}x\right)$
$-\dfrac{3/2}{1+\left(\tfrac{3}{2}x\right)^2}$

Explanation: This problem tests the skill of differentiating inverse trigonometric functions, specifically the arctan function. Derivative of arctan(u) with u as (3/2)x is 1 over 1 plus u squared times 3/2. Results in 3/2 over 1 plus ((3/2)x) squared. Increases magnitude from the 3/2 factor. Choice B uses 3 instead, doubling incorrectly. Pattern: Multiply standard arctan derivative by the fractional coefficient.

Question 14

Find the washer-method integral when the region between $x=\tfrac{2}{y}+5$ and $x=\tfrac{y}{2}+2$ on $[1,2]$ is revolved about the y-axis.

$\pi\int_{1}^{2}\left[\left(\tfrac{y}{2}+2\right)^2-\left(\tfrac{2}{y}+5\right)^2\right]dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$ (correct answer)
$\pi\int_{1}^{2}\left(\tfrac{2}{y}+5\right)^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
$\pi\int_{1}^{2}\left[\left(\tfrac{2}{y}+5\right)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$

Explanation: The washer method is used to find the volume of a solid of revolution with a hole, formed by revolving a region between two curves around an axis. In this case, revolving around the y-axis, the radii are the x-values of the curves. To determine which is the outer and inner radius, compare the two functions: 2/y + 5 > y/2 + 2 for y in [1,2], so outer radius is 2/y + 5 and inner is y/2 + 2. The volume integral is then π times the integral from 1 to 2 of [(2/y + 5)^2 - (y/2 + 2)^2] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, not correctly forming the washer area. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up π ∫ (outer^2 - inner^2) d(variable perpendicular to axis), with limits where the region exists.

Question 15

Let $t(x)=\begin{cases}\frac{\sin x}{x},&x\ne0\\1,&x=0\end{cases}$ . What type of discontinuity occurs at $x=0$ ?

No discontinuity (continuous) (correct answer)
Jump discontinuity
Infinite discontinuity
Removable discontinuity
Oscillating discontinuity

Explanation: This function has no discontinuity at x = 0. The limit of sin(x)/x as x approaches 0 is a fundamental limit equal to 1, and t(0) is defined as 1. Since lim(x→0) sin(x)/x = 1 = t(0), the function is continuous at x = 0. This is a classic example where a rational expression that's undefined at a point can be made continuous by defining the function value to equal the limit. Students often think this is a removable discontinuity because sin(x)/x is undefined at x = 0, but since t(0) = 1 matches the limit, there's no discontinuity. To determine continuity: check that the limit exists and equals the function value at the point.

Question 16

For a waveform, evaluate $\int \sec^2(5x-2)\,dx$ to find the phase accumulation.

$\tan(5x-2)+C$
$\tfrac{1}{5}\tan(5x-2)+C$ (correct answer)
$5\tan(5x-2)+C$
$\sec^2(5x-2)+C$
$\tfrac{1}{5}\tan x + C$

Explanation: This integral requires u-substitution because we have a composite function inside the secant squared. Let u = 5x - 2, so du = 5 dx, which means dx = ⅕ du. The integral becomes ∫ sec²(u) · ⅕ du = ⅕ ∫ sec²(u) du = ⅕ tan(u) + C = ⅕ tan(5x-2) + C. Choice A (tan(5x-2) + C) forgets the crucial factor of ⅕ that comes from the chain rule in reverse. When integrating trigonometric functions with linear arguments, always divide by the coefficient of x in the argument.

Question 17

A continuous profit function $P(x)$ satisfies $P(1)=-5$ and $P(3)=7$ . What must exist?

A break-even point $c\in(1,3)$ with $P(c)=0$ . (correct answer)
A point $c\in(1,3)$ with $P(c)=10$ .
A point $c\in(0,1)$ with $P(c)=0$ .
A point $c\in(1,3)$ with $P(c)=-6$ .
No conclusion can be drawn because $P$ could jump between $x=1$ and $x=3$ .

Explanation: The Intermediate Value Theorem (IVT) for continuous P on [1, 3] with P(1) = -5 and P(3) = 7 guarantees values between -5 and 7. Since -5 < 0 < 7, there exists c in [1, 3] with P(c) = 0, and as 0 ≠ -5, 0 ≠ 7, c is in (1, 3). This ensures a break-even point. A common error is thinking discontinuity allows jumping over values, but continuity prevents that. Selecting 10, outside the range, is incorrect. 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).

Question 18

Suppose $S(x)=\int_{3}^{x}\big(\cos t+\dfrac{1}{t^2}\big)\,dt$ . What is $S'(x)$ ?

$\cos x+\dfrac{1}{x^2}$ (correct answer)
$-\sin x-\dfrac{2}{x^3}$
$\int_{3}^{x}\big(\cos t+\dfrac{1}{t^2}\big)\,dt$
$\cos t+\dfrac{1}{t^2}$
$\left[\sin t-\dfrac{1}{t}\right]_{3}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function S(x) is defined as the integral from 3 to x of (cos t + 1/t^2) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative S'(x) is simply the integrand evaluated at t = x, so S'(x) = cos x + 1/x^2. This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 19

Select the washer-method integral for revolving the region between $x=\frac{1}{y+1}+3$ and $x=\tfrac{y}{2}+2$ on $[0,2]$ about the y-axis.

$\pi\int_{0}^{2}\left[\left(\tfrac{y}{2}+2\right)^2-\left(\tfrac{1}{y+1}+3\right)^2\right]dy$
$\pi\int_{0}^{2}\left[\left(\tfrac{1}{y+1}+3\right)^2-\left(\tfrac{y}{2}+2\right)^2\right]dy$ (correct answer)
$\pi\int_{0}^{2}\left(\tfrac{1}{y+1}+3\right)^2dy$
$\pi\int_{0}^{2}\left[\left(\tfrac{1}{y+1}+3\right)-\left(\tfrac{y}{2}+2\right)\right]^2dy$
$\pi\int_{0}^{2}\left[\left(\tfrac{1}{y+1}+3\right)^2+\left(\tfrac{y}{2}+2\right)^2\right]dy$

Explanation: The washer method is used to find the volume of a solid of revolution with a hole, formed by revolving a region between two curves around an axis. In this case, revolving around the y-axis, the radii are the x-values of the curves. To determine which is the outer and inner radius, compare the two functions: 1/(y + 1) + 3 > y/2 + 2 for y in [0,2], so outer radius is 1/(y + 1) + 3 and inner is y/2 + 2. The volume integral is then π times the integral from 0 to 2 of [(1/(y + 1) + 3)^2 - (y/2 + 2)^2] dy. A tempting distractor is choice D, which squares the difference of the radii instead of differencing the squares, incorrectly computing the volume. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up π ∫ (outer^2 - inner^2) d(variable perpendicular to axis), with limits where the region exists.

Question 20

Let $f(x)=\begin{cases}\dfrac{x^2-2x-15}{x-5},&x\ne5\\12,&x=5\end{cases}$ . What value should replace $12$ to remove the discontinuity at $x=5$ ?

$12$
$-3$
$8$ (correct answer)
$5$
$15$

Explanation: f(x) shows removable discontinuity at x=5. Simplifies: (x-5)(x+3)/(x-5) = x+3, limit 5+3=8. Replace 12 with 8, choice C. Confusion: 0/0. As a transferable strategy, always factor rational expressions and cancel common factors to evaluate limits and remove discontinuities.

Question 21

A continuous function $f$ on $[-1,2]$ satisfies $f(-1)=0$ and $f(2)=5$ . What must be true?

There exists $c\in(-1,2)$ such that $f(c)=3$ . (correct answer)
There exists $c\in(-1,2)$ such that $f(c)=-1$ .
There exists $c\in(-1,2)$ such that $f(c)=6$ .
There exists $c\in(2,4)$ such that $f(c)=3$ .
A root in $(-1,2)$ must exist.

Explanation: The Intermediate Value Theorem (IVT) for continuous f on [-1, 2] with f(-1) = 0 and f(2) = 5 guarantees values between 0 and 5. The value 3 is strictly between 0 and 5, so there exists c in [-1, 2] with f(c) = 3. Since 3 ≠ 0 and 3 ≠ 5, c is in (-1, 2). This highlights that IVT can guarantee non-zero values when appropriate. A common error is confusing this with guaranteeing a root, but 0 is already at an endpoint and not necessarily inside. Assuming the interval must be positive is irrelevant to IVT. 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).

Question 22

A rate is given by $u(x)=\tan\!\left(\ln(x^2+4)\right)$ . What is $u'(x)$ ?

$\sec^2\!\left(\ln(x^2+4)\right)$
$\dfrac{2x}{x^2+4}\sec^2\!\left(\ln(x^2+4)\right)$ (correct answer)
$\dfrac{1}{x^2+4}\sec^2\!\left(\ln(x^2+4)\right)$
$\dfrac{2x}{x^2+4}\tan\!\left(\ln(x^2+4)\right)$
$2x\sec^2\!\left(\ln(x^2+4)\right)$

Explanation: This tangent of a log requires chain rule for trig-log composite. Outer tan(u), u = ln(x^2 + 4), derivative sec^2(u), inner u' = 2x/(x^2 + 4). So u'(x) = (2x/(x^2 + 4)) sec^2(ln(x^2 + 4)). Common omission: forgetting 2x from inner derivative. Confusing tan derivative with sec tan is another error. Recognize trig of log polynomial and chain fully. This pattern aids in complex nests, building derivative skills.

Question 23

If $f(x)=(2x^3-1)^2$ , what is the second derivative $f''(x)$ ?

$f''(x)=24x(2x^3-1)+72x^4$ (correct answer)
$f''(x)=24x(2x^3-1)$
$f''(x)=12x^2(2x^3-1)$
$f''(x)=24x(2x^3-1)-72x^4$
$f''(x)=48x^2(2x^3-1)$

Explanation: Chain rule on squares yields higher derivatives via product rule. The first is f'(x) = 12x^2 (2x^3 - 1). The second is f''(x) = 24x (2x^3 - 1) + 72x^4, expanding properly. A common stopping error is not applying product rule for the second. Simplify by factoring if possible. A transferable strategy is to treat as composition and differentiate step-by-step, verifying with expansion.

Question 24

Find the volume setup: base bounded by $y=x^2$ and $y=4$ , semicircular cross sections perpendicular to the $x$ -axis.

$\displaystyle \int_{-2}^{2} \pi\big(4-x^2\big)^2\,dx$
$\displaystyle \int_{-2}^{2} \frac{\pi}{8}\big(4-x^2\big)^2\,dx$ (correct answer)
$\displaystyle \int_{0}^{4} \frac{\pi}{8}\big(\sqrt{y}-(-\sqrt{y})\big)^2\,dy$
$\displaystyle \int_{-2}^{2} \frac12\big(4-x^2\big)^2\,dx$
$\displaystyle \int_{-2}^{2} \frac{\pi}{2}\big(4-x^2\big)\,dx$

Explanation: This problem requires finding the volume of a solid with non-rectangular cross-sections, specifically semicircles perpendicular to the x-axis. The base region is bounded by y = x² and y = 4, which intersect when x² = 4, giving x = ±2. At each x-value, the cross-section has diameter equal to the vertical distance between the curves: 4 - x². For a semicircle with diameter d, the area is (π/8)d², so each cross-section has area (π/8)(4 - x²)². The volume integral is ∫_{-2}^{2} (π/8)(4 - x²)² dx. A common error would be using the full circle formula πr² = (π/4)d², which gives choice A. The key strategy is to identify the cross-section shape, find its area formula in terms of the base dimension, then integrate over the appropriate bounds.

Question 25

For $Y(x)=\begin{cases}\frac{x^2-49}{x-7},&x\ne7\\15,&x=7\end{cases}$ , what type of discontinuity occurs at $x=7$ ?

Infinite discontinuity
Removable discontinuity (correct answer)
No discontinuity (continuous)
Jump discontinuity
Corner discontinuity

Explanation: This function has a removable discontinuity at x = 7. When x ≠ 7, we can factor the numerator as (x-7)(x+7), so Y(x) = (x-7)(x+7)/(x-7) = x+7 for x ≠ 7. The limit as x approaches 7 is lim(x→7) (x+7) = 14, but Y(7) = 15. Since the limit exists but doesn't equal the function value (14 ≠ 15), this creates a removable discontinuity. The discontinuity could be 'removed' by redefining Y(7) = 14. Students often don't recognize the difference of squares factorization in the numerator. To identify removable discontinuities: factor completely and check if limits differ from function values.

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