CalcIQ — Mock Exam 1: Mid-Course Diagnostic (Units 1 through early 4–5, Lessons 1–20)

All multiple-choice answers and all free-response values below were independently recomputed (and verified symbolically) before keying. No scratch work is keyed — only verified final answers with one-line rationales.

Section I, Part A — Answers

A1. (A). Factor: (x² − 9)/(x − 3) = (x − 3)(x + 3)/(x − 3) = x + 3 → 6. Distractor (B) assumes the 0/0 form is automatically undefined; it is a removable discontinuity, so the limit exists.

A2. (D). sin(5x)/(3x) = (5/3)·[sin(5x)/(5x)] → (5/3)(1) = 5/3. Distractor (C) forgets the inside-coefficient scaling; (B) inverts the ratio.

A3. (A). Ratio of leading coefficients of equal-degree polynomials: 3/5. (B) mistakes equal degree for numerator-dominant; (D) mis-divides.

A4. (A). Left value at x = 1 is 1² = 1; right value is 2(1) + k = 2 + k. Continuity needs 2 + k = 1, so k = −1. (C) solves 2 + k = ... = 1 with a sign slip.

A5. (C). This is the limit definition of f'(3) for f(x) = x²; f'(x) = 2x, so f'(3) = 6. Direct expansion: [(9 + 6h + h²) − 9]/h = 6 + h → 6. (D) reports f(3) = 9 instead of f'(3).

A6. (D). Power rule term by term: 12x³ − 10x; the constant 7 differentiates to 0. (C) keeps the constant.

A7. (D). Product rule: f'(x) = 2x(x³ − 2x) + (x² + 1)(3x² − 2). At x = 2: 2(2)(8 − 4) + (5)(12 − 2) = 4·4 + 5·10 = 16 + 50 = 66. (C) uses only the first product term; (A) multiplies the derivatives.

A8. (C). Quotient rule: [2(x + 3) − (2x − 1)(1)]/(x + 3)² = (2x + 6 − 2x + 1)/(x + 3)² = 7/(x + 3)². (A) drops a sign; (B) divides derivatives.

A9. (A). d/dx[tan x] = sec²x; sec²(π/4) = (√2)² = 2. (C) reports sec(π/4) = √2 (un-squared); (D) confuses with tan(π/4).

A10. (D). Chain rule: 4(3x² + 1)³ · (6x) = 24x(3x² + 1)³. (C) is the classic missing-chain-factor error.

A11. (D). Implicitly 2x + 2y·y' = 0 ⇒ y' = −x/y = −3/4. (C) inverts to −y/x; (B) drops the sign.

A12. (A). (f⁻¹)'(4) = 1/f'(1). Here f'(x) = 3x² + 2, so f'(1) = 5 and g'(4) = 1/5. (B) reports f'(1) instead of its reciprocal.

A13. (B). f'(x) = 2x/(x² + 1) (chain rule on ln); at x = 1, 2/(1 + 1) = 1. (A) forgets the chain factor 2x (uses 1/(x²+1)); (C) reports the function value ln 2.

A14. (C). f' = 5x⁴, f'' = 20x³, f''' = 60x². (A) stops at the second derivative; (D) over-differentiates to the fourth.

A15. (A). A corner: the one-sided derivatives are −1 (left) and +1 (right); they disagree, so f'(2) does not exist even though f is continuous. (D) is false — f is continuous; (C) describes a cusp/vertical-tangent failure, which is not what happens here.

Section I, Part B — Answers

B1. (C). f'(x) = 2xeˣ + x²eˣ = (x² + 2x)eˣ; f'(1) = 3e ≈ 8.155. (A) is just e; (B) uses only the 2xeˣ term (2e ≈ 5.437).

B2. (C). Differentiate x² + y² = 25: 2x·x' + 2y·y' = 0. At x = 3, y = 4, with x' = 2: y' = −(x/y)x' = −(3/4)(2) = −3/2. The top slides down at 3/2 ft/sec. (D) forgets to divide by y; (B) swaps x and y.

B3. (B). f(x) = √x, f'(x) = 1/(2√x), f'(4) = 1/4. Tangent-line estimate: L(4.1) = 2 + (1/4)(0.1) = 2.025. (D) uses slope 1; the true value √4.1 ≈ 2.0248, confirming the linear estimate is slightly high, as expected for a concave-down function.

B4. (B). 0/0 form; L'Hôpital: lim (2e^{2x})/(3cos 3x) = 2/3. (Equivalently (e^{2x}−1)/(2x) → 1 and sin(3x)/(3x) → 1, giving (2x)/(3x) = 2/3.) (D) uses the wrong derivative ratio.

B5. (D). v(t) = 3(t − 1)(t − 3); a(t) = 6t − 12 = 6(t − 2). Speeding up ⇔ v and a have the same sign. On (1, 2): v < 0 and a < 0 (same sign) ⇒ speeding up. On (2, 3): v < 0, a > 0 (opposite) ⇒ slowing down. (C) confuses "v decreasing" with "speed increasing."

B6. (B). Secant slope on [0, 2] is (8 − 0)/2 = 4; f'(x) = 3x² = 4 ⇒ c = 2/√3 ≈ 1.155 (in (0, 2); the negative root is rejected). (A) guesses the midpoint of slopes; (C) is √2.

B7. (C). f'(x) = 2x·cos(x²); f'(1) = 2cos(1) ≈ 2(0.5403) ≈ 1.081. (A) is cos 1 (missing the 2x chain factor); (B) is sin 1.

B8. (B). Symmetric difference quotient: g'(2.0) ≈ [g(2.1) − g(1.9)]/(2.1 − 1.9) = (17.45 − 12.43)/0.2 = 5.02/0.2 = 25.1. (A) uses the backward quotient (14.78 − 12.43)/0.1 = 23.5; (C) uses the forward quotient (17.45 − 14.78)/0.1 = 26.7.

Section II — Model Solutions & Rubrics

FRQ 1 — Particle motion (calculator) · 9 points

Throughout, x(t) = 10e^{−t/4} sin t and v(t) = x'(t) = 10e^{−t/4}[cos t − (1/4) sin t] (equivalently (5/2)e^{−t/4}(4cos t − sin t)).

(a) v(1) = x'(1) ≈ 2.570 cm/sec (TI-84: nDeriv(10e^(−X/4)sin(X), X, 1)). Since v(1) > 0, the particle is moving in the positive direction at t = 1. (2 pts: 1 for v(1) ≈ 2.570 with units; 1 for "positive direction, because v(1) > 0.")

(b) a(1) = x''(1) ≈ −8.248 cm/sec² (numerical second derivative, or nDeriv of v). At t = 1, v(1) ≈ 2.570 > 0 and a(1) ≈ −8.248 < 0. Because the velocity and acceleration have opposite signs, the speed is decreasing at t = 1. (3 pts: 1 for a(1) ≈ −8.248; 1 for comparing signs of v(1) and a(1); 1 for the correct conclusion "speed decreasing" with the sign-based justification.)

(c) The particle changes direction where v changes sign. Solving v(t) = 0 numerically on (0, 8) (TI-84: graph v(t) and use 2nd → CALC → zero, or Solver) gives the first zero at t ≈ 1.326. For t just below 1.326, v > 0; for t just above, v < 0. Because v changes from positive to negative at t ≈ 1.326, the particle changes direction there (from moving in the positive direction to moving in the negative direction). (2 pts: 1 for t ≈ 1.326; 1 for justifying the direction change via the sign change of v.)

(d) x(1) = 10e^{−1/4} sin 1 ≈ 6.553 cm. At t = 1 second, the particle is located about 6.553 cm to the right of the origin, moving in the positive direction (v(1) ≈ 2.570 cm/sec) but slowing down (a(1) < 0). (2 pts: 1 for x(1) ≈ 6.553 with units; 1 for a contextual sentence consistent with parts (a)–(b).)

FRQ 2 — Implicit differentiation (no calculator) · 9 points

(a) Differentiate x² − xy + y² = 3 implicitly with respect to x, using the product rule on xy:

2x − (y + x·y') + 2y·y' = 0
2x − y − x·y' + 2y·y' = 0
y'(2y − x) = y − 2x
y' = (y − 2x)/(2y − x) = (2x − y)/(x − 2y).

At (1, −1): y' = (2(1) − (−1))/((1) − 2(−1)) = 3/3 = 1. Tangent line: y − (−1) = 1·(x − 1), i.e. y = x − 2. (3 pts: 1 for correct implicit differentiation including the product-rule term on xy; 1 for the slope = 1 at (1, −1); 1 for the tangent-line equation y = x − 2.)

(b) The tangent is horizontal where dy/dx = 0, i.e. where the numerator 2x − y = 0 (and the denominator ≠ 0), so y = 2x. Substituting into the curve:

x² − x(2x) + (2x)² = 3 ⇒ x² − 2x² + 4x² = 3 ⇒ 3x² = 3 ⇒ x = ±1.

This gives the points (1, 2) and (−1, −2). At each, 2x − y = 0 while x − 2y ≠ 0 (1 − 4 = −3 and −1 + 4 = 3), so dy/dx = 0 and the tangent is horizontal. (3 pts: 1 for setting the numerator 2x − y = 0; 1 for solving with the curve to get x = ±1; 1 for both points (1, 2) and (−1, −2) with the denominator check.)

(c) A vertical tangent occurs where dy/dx is undefined because the denominator x − 2y = 0 (and the numerator ≠ 0), so x = 2y. Substituting:

(2y)² − (2y)y + y² = 3 ⇒ 4y² − 2y² + y² = 3 ⇒ 3y² = 3 ⇒ y = ±1.

This gives the points (2, 1) and (−2, −1). At each the numerator 2x − y is nonzero (4 − 1 = 3 and −4 + 1 = −3), so dy/dx → ±∞, indicating a vertical tangent. (3 pts: 1 for setting the denominator x − 2y = 0; 1 for solving with the curve to get y = ±1; 1 for both points (2, 1) and (−2, −1) with the explanation that a zero denominator and nonzero numerator make the slope undefined → vertical tangent.)

FRQ 3 — MVT, IVT, and the derivative limit (no calculator) · 9 points

(a) Average rate of change on [0, 2]:

[f(2) − f(0)]/(2 − 0) = [(8 − 6 + 1) − (0 − 0 + 1)]/2 = (3 − 1)/2 = 2/2 = 1.

The average rate of change of f on [0, 2] is 1. (2 pts: 1 for f(2) = 3 and f(0) = 1; 1 for the value 1.)

(b) f is a polynomial, so it is continuous on the closed interval [0, 2] and differentiable on the open interval (0, 2); the hypotheses of the Mean Value Theorem are satisfied. Therefore there exists at least one c in (0, 2) with f'(c) = [f(2) − f(0)]/(2 − 0) = 1. Now f'(x) = 3x² − 3, so

3c² − 3 = 1 ⇒ 3c² = 4 ⇒ c² = 4/3 ⇒ c = ±2/√3.

Only c = 2/√3 = 2√3/3 ≈ 1.155 lies in (0, 2); c = −2/√3 is rejected. By the MVT, c = 2√3/3. (3 pts: 1 for stating the hypotheses (continuous on [0,2], differentiable on (0,2)); 1 for the equation 3c² − 3 = 1; 1 for c = 2√3/3, rejecting the negative root.)

(c) f is a polynomial, hence continuous on the closed interval [0, 1]. We have f(0) = 1 > 0 and f(1) = 1 − 3 + 1 = −1 < 0, so 0 lies between f(1) = −1 and f(0) = 1. By the Intermediate Value Theorem, since f is continuous on [0, 1] and f(1) < 0 < f(0), there exists a value x in (0, 1) with f(x) = 0. Thus f(x) = 0 has a solution in (0, 1). (2 pts: 1 for naming the IVT and citing continuity on [0, 1]; 1 for showing the sign change f(0) = 1 > 0, f(1) = −1 < 0 and concluding a root exists. Citing the MVT here earns no credit — this is a statement about a function value, which is the IVT's job.)

(d) lim_{h→0} [f(1 + h) − f(1)]/h is by definition f'(1). Since f'(x) = 3x² − 3, f'(1) = 3 − 3 = 0. The limit equals 0, and it represents the instantaneous rate of change of f at x = 1, i.e. the slope of the tangent line to y = f(x) at x = 1. (2 pts: 1 for recognizing the limit as f'(1) and evaluating it to 0; 1 for interpreting it as the derivative / instantaneous rate of change / tangent slope at x = 1.)

Score Bands (approximate — for self-assessment only)

This diagnostic is not an official AP score and is intended only to gauge mid-course progress on Lessons 1–20. Weight the two sections equally (Section I = 50%, Section II = 50%):

Section I score   = (MC correct / 23) × 50
Section II score  = (FRQ points / 27) × 50          [9 + 9 + 9 = 27 points]
Composite (0–100) = Section I score + Section II score

| Composite | Approximate band | Interpretation (Lessons 1–20 only) |

|---|---|---|

| 80–100 | 5-ish | Strong command of limits, all derivative rules, and early applications. Ready for Unit 5 analysis. |

| 67–79 | 4-ish | Solid; tighten justification language (state theorem hypotheses) and chain-rule fluency. |

| 52–66 | 3-ish | Passing range; review related rates, implicit differentiation, and MVT/IVT distinctions. |

| 38–51 | 2-ish | Revisit Lessons 6–16 (derivative rules) and 17–20 (applications) before moving on. |

| 0–37 | 1-ish | Re-study the limit and derivative foundations (Lessons 1–11) before continuing. |

Diagnostic focus areas keyed to lessons: missed A1–A5 → Lessons 1–6 (limits, derivative definition); A6–A10 → Lessons 7–12 (rules, chain); A11–A14 → Lessons 13–15 (implicit, inverse, log, higher-order); A15, B1–B8 & all FRQs → Lessons 10, 17–20 (differentiability, motion, related rates, linear approximation, L'Hôpital, MVT/IVT).

CalcIQ · Mock Exam 1 — Mid-Course Diagnostic (after Lesson 20). Covers Lessons 1–20 only. Next: Lesson 21 — First Derivative Test & Intervals.

This exam is independent study material and is not endorsed by or affiliated with the College Board. "AP" and "Advanced Placement" are trademarks of the College Board.

Accuracy note: All 23 multiple-choice answers and all three free-response solutions were independently recomputed and symbolically verified. Selected verified values: A1 = 6; A2 = 5/3; A7 = f'(2) = 66; A12 = 1/5; B1 = 3e ≈ 8.155; B5 speeding up on (1, 2); B8 = 25.1. FRQ 1: v(1) ≈ 2.570, a(1) ≈ −8.248, first direction change t ≈ 1.326, x(1) ≈ 6.553. FRQ 2: slope 1 at (1, −1); horizontal tangents at (1, 2) and (−1, −2); vertical tangents at (2, 1) and (−2, −1). FRQ 3: average rate 1; MVT c = 2√3/3 ≈ 1.155; IVT root in (0, 1); f'(1) = 0. Content is restricted to Lessons 1–20 — no derivative tests, optimization, curve sketching, or integration.