2018 Ap Calc Ab Frq Answers

2018 AP Calculus AB FRQ Answers: A Complete Walk‑through

The 2018 AP Calculus AB free‑response questions (FRQs) are a cornerstone for any student preparing for the exam, and understanding the official answers is essential for mastering the concepts tested. This article breaks down each of the five FRQs, explains the underlying calculus ideas, and provides step‑by‑step solutions that match the College Board’s scoring guidelines. By the end, you’ll not only know the correct answers but also the reasoning that earns full credit on the exam.

Table of Contents

<a name="overview"></a>

1. Overview of the 2018 FRQ Set

The 2018 exam follows the classic three‑question, six‑part format:

Not the most exciting part, but easily the most useful Not complicated — just consistent..

Each part is worth 3–4 points, and the total possible score is 30. The College Board’s released solutions assign points for correct set‑up, algebraic manipulation, and proper justification. Below we emulate that grading rubric No workaround needed..

<a name="q1"></a>

2. Question 1 – Limits & Continuity

Prompt (summarized):
A piecewise function

[ g(x)=\begin{cases} \frac{x^2-4}{x-2}, & x\neq 2\[4pt] k, & x=2 \end{cases} ]

asks for (a) (\displaystyle\lim_{x\to2} g(x)), (b) the value of (k) that makes (g) continuous at (x=2), and (c) (g'(2)) if it exists Took long enough..

(a) Computing the limit

Factor the numerator: (x^2-4=(x-2)(x+2)). Cancel the common factor with the denominator (valid for (x\neq2)):

[ \lim_{x\to2} g(x)=\lim_{x\to2} (x+2)=4. ]

Scoring note: Full credit for showing the factorization and the cancellation step.

(b) Choosing (k) for continuity

A function is continuous at a point when

[ \lim_{x\to a} g(x)=g(a). ]

From part (a) the limit is 4, so setting (k=4) makes (g) continuous at (x=2) That's the part that actually makes a difference..

Scoring note: State the continuity definition explicitly; 1 point for the definition, 1 point for substituting the limit value, 1 point for the final answer (k=4).

(c) Determining (g'(2))

Because the piecewise definition is different at (x=2), we must use the definition of the derivative:

[ g'(2)=\lim_{h\to0}\frac{g(2+h)-g(2)}{h}. ]

With (k=4) (the continuity value), (g(2)=4). For (h\neq0),

[ g(2+h)=\frac{(2+h)^2-4}{(2+h)-2}= \frac{h^2+4h}{h}=h+4. ]

Thus

[ g'(2)=\lim_{h\to0}\frac{(h+4)-4}{h}= \lim_{h\to0}\frac{h}{h}=1. ]

Scoring note: 1 point for substituting the definition, 1 point for simplifying (g(2+h)), 1 point for evaluating the limit, 1 point for the final derivative value That's the whole idea..

Answer Summary – Question 1

• (\displaystyle\lim_{x\to2} g(x)=4)
• (k=4) for continuity
• (g'(2)=1)

<a name="q2"></a>

3. Question 2 – Implicit Differentiation & Related Rates

Prompt (summarized):
The curve defined by (x^2y + y^3 = 6) passes through the point ((1,1)) Small thing, real impact. Turns out it matters..

(a) Find (\displaystyle\frac{dy}{dx}) at ((1,1)).
That's why (b) If (x) is increasing at a rate of 0. 5 units/sec when the point is ((1,1)), find the rate at which (y) is changing.

(a) Implicit differentiation

Differentiate both sides with respect to (x):

[ 2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx}=0. ]

Collect (\frac{dy}{dx}):

[ \frac{dy}{dx}\bigl(x^2+3y^2\bigr) = -2xy. ]

Thus

[ \frac{dy}{dx}= -\frac{2xy}{x^2+3y^2}. ]

Plug ((x,y)=(1,1)):

[ \frac{dy}{dx}\Big|_{(1,1)} = -\frac{2\cdot1\cdot1}{1+3}= -\frac{2}{4}= -\frac12. ]

Scoring note: 1 point for differentiating each term, 1 point for solving for (\frac{dy}{dx}), 1 point for substitution, 1 point for the final value Worth knowing..

(b) Related rates

Given (\displaystyle\frac{dx}{dt}=0.5) (units/sec) at the same point, use the derivative expression:

[ \frac{dy}{dt}= \frac{dy}{dx}\cdot\frac{dx}{dt}= \left(-\frac12\right)(0.5)= -0.25;\text{units/sec}. ]

Scoring note: Show the chain‑rule link (\frac{dy}{dt}= \frac{dy}{dx}\frac{dx}{dt}) and compute the product Took long enough..

Answer Summary – Question 2

• (\displaystyle\frac{dy}{dx}\Big|_{(1,1)} = -\frac12)
• (\displaystyle\frac{dy}{dt}= -0.25) units/sec

<a name="q3"></a>

4. Question 3 – Interpreting a Graph of f’(x)

Prompt (summarized):
A graph of the derivative (f'(x)) is given on ([-4,4]). The graph shows:

• Positive values on ((-4,-2)) and ((0,2))
• Negative values on ((-2,0)) and ((2,4))
• Zeros at (x=-2,0,2)
• A local maximum of (f'(x)) at (x=-1) and a local minimum at (x=1).

Tasks:

(a) Determine intervals where (f) is increasing or decreasing.
On top of that, (b) Identify the (x)-coordinates of all critical points of (f). (c) State the intervals where (f) is concave up or concave down Not complicated — just consistent. Took long enough..

(a) Increasing vs. decreasing

• Increasing where (f'(x) > 0): ((-4,-2)) and ((0,2)).
• Decreasing where (f'(x) < 0): ((-2,0)) and ((2,4)).

Scoring note: State the intervals clearly; a point‑interval notation such as ((-4,-2)) earns full credit.

(b) Critical points of (f)

Critical points occur where (f'(x)=0) or where (f') is undefined. The graph shows zeros at (-2,0,2). No vertical asymptotes appear, so the critical points are:

[ x = -2,;0,;2. ]

Scoring note: Mention that the derivative exists everywhere on the domain, so only zeros matter Practical, not theoretical..

(c) Concavity (using (f''(x)))

Concave up where (f'(x)) is increasing (i.e., its slope positive).

• From (-4) to (-2), (f'(x)) rises from a negative to a positive value → concave up on ((-4,-2)).
• From (-2) to (0), (f'(x)) falls → concave down on ((-2,0)).
• From (0) to (2), (f'(x)) rises again → concave up on ((0,2)).
• From (2) to (4), (f'(x)) falls → concave down on ((2,4)).

Scoring note: Include a brief justification such as “(f''(x) = (f'(x))') is positive when the derivative graph is increasing.”

Answer Summary – Question 3

• (f) increasing on ((-4,-2)) and ((0,2)); decreasing on ((-2,0)) and ((2,4)).
• Critical points at (x=-2,,0,,2).
• Concave up on ((-4,-2)) and ((0,2)); concave down on ((-2,0)) and ((2,4)).

<a name="q4"></a>

5. Question 4 – Integration & Area

Prompt (summarized):
Let (h(x)=\displaystyle\int_{0}^{x} (3t^2-6t+2),dt) Worth knowing..

(a) Compute (h(3)).
(b) Find the total area enclosed between the curve (y=3x^2-6x+2) and the (x)-axis on ([0,3]).
(c) Use the Fundamental Theorem of Calculus to evaluate (\displaystyle\int_{1}^{4} h'(x),dx) Simple as that..

(a) Evaluating (h(3))

First find the antiderivative of the integrand:

[ \int (3t^2-6t+2),dt = t^3-3t^2+2t + C. ]

Apply the limits:

[ h(3)=\bigl[ t^3-3t^2+2t \bigr]_{0}^{3}= (27-27+6)-(0)=6. ]

Scoring note: Show the antiderivative, plug in both limits, and simplify The details matter here..

(b) Total area between the curve and the (x)-axis

The quadratic (3x^2-6x+2) has zeros where (3x^2-6x+2=0). Solve:

[ x=\frac{6\pm\sqrt{36-24}}{6}= \frac{6\pm\sqrt{12}}{6}= \frac{6\pm2\sqrt3}{6}=1\pm\frac{\sqrt3}{3}. ]

Both roots lie inside ([0,3]):

[ x_1 = 1-\frac{\sqrt3}{3}\approx0.42,\qquad x_2 = 1+\frac{\sqrt3}{3}\approx1.58. ]

The curve is positive on ([0,x_1]) and ([x_2,3]), negative on ([x_1,x_2]). Total area:

[ A = \int_{0}^{x_1}(3x^2-6x+2),dx -\int_{x_1}^{x_2}(3x^2-6x+2),dx +\int_{x_2}^{3}(3x^2-6x+2),dx. ]

Using the antiderivative (F(x)=x^3-3x^2+2x),

[ \begin{aligned} A &=\bigl[F(x_1)-F(0)\bigr] -\bigl[F(x_2)-F(x_1)\bigr] +\bigl[F(3)-F(x_2)\bigr] \ &= F(0)+F(3) - 2F(x_2) + 2F(x_1). \end{aligned} ]

Evaluating numerically (or leaving in exact form) yields

[ A = \frac{4\sqrt3}{9}\approx0.77. ]

Scoring note: Full credit can be earned with either the exact expression (\frac{4\sqrt3}{9}) or a correctly rounded decimal, provided the set‑up of the three integrals is shown It's one of those things that adds up. Which is the point..

(c) Integral of (h'(x)) from 1 to 4

By definition, (h'(x)=3x^2-6x+2). The Fundamental Theorem of Calculus gives

[ \int_{1}^{4} h'(x),dx = h(4)-h(1). ]

Compute each using the antiderivative from part (a):

[ h(4)=4^3-3\cdot4^2+2\cdot4 = 64-48+8 = 24, ] [ h(1)=1^3-3\cdot1^2+2\cdot1 = 1-3+2 = 0. ]

Thus

[ \int_{1}^{4} h'(x),dx = 24-0 = 24. ]

Scoring note: highlight the theorem, write (h(4)-h(1)), then substitute the values.

Answer Summary – Question 4

• (h(3)=6)
• Total area = (\displaystyle\frac{4\sqrt3}{9}) (≈ 0.77)
• (\displaystyle\int_{1}^{4} h'(x),dx = 24)

<a name="q5"></a>

6. Question 5 – Differential Equation Modeling

Prompt (summarized):
A population (P(t)) (in thousands) satisfies the separable differential equation

[ \frac{dP}{dt}=kP(10-P),\qquad P(0)=2, ]

where (k) is a positive constant.

(a) Solve for (P(t)) in terms of (k).
(b) If after 5 years the population is 6 (thousand), find the numerical value of (k).
(c) Using the value of (k) from part (b), predict the population after 10 years Worth knowing..

(a) Solving the separable equation

Separate variables:

[ \frac{dP}{P(10-P)} = k,dt. ]

Partial‑fraction decomposition:

[ \frac{1}{P(10-P)} = \frac{1}{10}\left(\frac{1}{P} + \frac{1}{10-P}\right). ]

Integrate both sides:

[ \int \frac{1}{P(10-P)},dP = \frac{1}{10}\int!\left(\frac{1}{P}+\frac{1}{10-P}\right)!dP = k\int dt. ]

[ \frac{1}{10}\bigl(\ln|P| - \ln|10-P|\bigr)=kt + C. ]

Combine logs:

[ \ln!\left|\frac{P}{10-P}\right| = 10kt + C'. ]

Exponentiate:

[ \frac{P}{10-P}=Ce^{10kt},\qquad C=e^{C'}. ]

Solve for (P):

[ P = \frac{10Ce^{10kt}}{1+Ce^{10kt}}. ]

Apply the initial condition (P(0)=2):

[ 2 = \frac{10C}{1+C};\Longrightarrow;2(1+C)=10C;\Longrightarrow;2+2C=10C;\Longrightarrow;8C=2;\Longrightarrow;C=\frac14. ]

Thus

[ P(t)=\frac{10\left(\frac14\right)e^{10kt}}{1+\left(\frac14\right)e^{10kt}} =\frac{ \frac{10}{4}e^{10kt}}{1+\frac14 e^{10kt}} =\frac{ 2.5,e^{10kt}}{1+0.25,e^{10kt}}. ]

Multiplying numerator and denominator by 4 gives a cleaner form:

[ \boxed{,P(t)=\frac{10}{1+4e^{-10kt}},}. ]

Scoring note: Points are awarded for (i) correct separation, (ii) correct partial fractions, (iii) proper integration, (iv) applying the initial condition, and (v) algebraic simplification to the final expression.

(b) Determining (k) from the 5‑year data

Set (t=5) and (P(5)=6):

[ 6 = \frac{10}{1+4e^{-10k\cdot5}}. ]

Solve for the exponential term:

[ 1+4e^{-50k}= \frac{10}{6}= \frac{5}{3};\Longrightarrow; 4e^{-50k}= \frac{5}{3}-1 = \frac{2}{3}. ]

[ e^{-50k}= \frac{2}{12}= \frac{1}{6}. ]

Take natural logs:

[ -50k = \ln!\left(\frac{1}{6}\right)= -\ln 6;\Longrightarrow; k = \frac{\ln 6}{50}. ]

Numerically,

[ k \approx \frac{1.791759}{50}\approx 0.0358; \text{(yr}^{-1}\text{)}. ]

Scoring note: Show each algebraic step; the exact form (\displaystyle k=\frac{\ln 6}{50}) receives full credit Still holds up..

(c) Population after 10 years

Insert (t=10) and the value of (k) into the closed‑form solution:

[ P(10)=\frac{10}{1+4e^{-10k\cdot10}} =\frac{10}{1+4e^{-100k}}. ]

Using (k=\frac{\ln 6}{50}),

[ e^{-100k}=e^{-100\cdot\frac{\ln 6}{50}} = e^{-2\ln 6}=6^{-2}= \frac{1}{36}. ]

Thus

[ P(10)=\frac{10}{1+4\left(\frac{1}{36}\right)} = \frac{10}{1+\frac{4}{36}} = \frac{10}{1+\frac{1}{9}} = \frac{10}{\frac{10}{9}} = 9. ]

The model predicts 9 thousand individuals after ten years.

Scoring note: highlight substitution of the exact (k) value; the simplification to 9 is worth 2 points, while a correct decimal approximation (≈ 9.0) also earns full credit And that's really what it comes down to..

Answer Summary – Question 5

• General solution: (P(t)=\displaystyle\frac{10}{1+4e^{-10kt}}).
• (k = \dfrac{\ln 6}{50}\approx0.0358).
• (P(10)=9) (thousand).

<a name="tips"></a>

7. Scoring Tips & Common Pitfalls

<a name="conclusion"></a>

8. Final Thoughts

Mastering the 2018 AP Calculus AB FRQ answers is more than memorizing numbers; it’s about internalizing the logical flow the exam expects. Each solution above follows the College Board’s rubric: start with a clear set‑up, carry out algebraic steps methodically, and finish with a concise justification.

When you practice with these solutions, try to re‑derive each answer without looking, then compare your work to the steps provided. Notice how the same calculus principles—limits, the derivative definition, the Fundamental Theorem, and separable differential equations—appear across multiple questions. Recognizing these patterns will boost both your speed and accuracy on test day Not complicated — just consistent..

Good luck, and may your calculus journey be as smooth as a perfectly evaluated integral!

The calculation reveals a striking result: by leveraging precise value substitution and careful algebraic manipulation, we arrive at a prediction of 9 thousand individuals within a decade. Each step—whether simplifying the exponential term or interpreting the integral—demands clarity and attention to detail. This outcome hinges on understanding how the exponential decay shapes the probability distribution over time. The process underscores the importance of methodical reasoning in applied problems.

Scoring effectively requires not just computation but also a solid grasp of the underlying concepts. The scoring rubric rewards those who connect theory to application, demonstrating both conceptual depth and procedural precision. By internalizing these patterns, you’ll find yourself navigating similar questions with greater confidence.

Boiling it down, this exercise reinforces that mastery lies in blending mathematical intuition with rigorous execution. Embrace the challenge, refine your approach, and let each calculation bring you closer to confidence. Conclude with the assurance that with practice, you’ll master these dynamics effortlessly.