CLEP Calculus — Mock Exam

Answer Key

1. C) 3. Direct substitution gives 0/0, so factor: (x² − x − 2)/(x − 2) = (x − 2)(x + 1)/(x − 2) = x + 1 → 3. Distractors: A) 0 stops at the 0/0 form and calls it zero. B) 1 reports the leftover constant from x + 1 instead of evaluating at 2. D) 4 substitutes x = 2 into the wrong surviving factor or double-counts. E) "nonexistent" treats a removable 0/0 as a failure — the hole fills to a clean value. Fix: 0/0 in a rational limit signals a hidden (x − c) factor; cancel, then substitute. [Apply]

2. A) 3x² − 8x + 2. Power rule term by term: d/dx x³ = 3x², d/dx(−4x²) = −8x, d/dx(2x) = 2, d/dx(−7) = 0. Distractors: B) 3x² − 4x + 2 forgets to multiply the −4x² term by its exponent (kept −4 instead of −8). C) x² − 8x + 2 forgets the factor 3 on x³. D) 3x² − 8x − 7 differentiated everything but left the constant −7 in place. E) 3x² − 8x drops the 2x term's derivative of 2. Fix: bring the exponent down as a factor, then reduce it by one; the derivative of a constant is 0, not the constant. [Understand]

3. B) 2x³ − 2x² + 3x + C. Reverse the power rule: ∫6x² = 6x³/3 = 2x³, ∫−4x = −4x²/2 = −2x², ∫3 = 3x. Distractors: A) 12x − 4 + C differentiates instead of integrating. C) 6x³ − 4x² + 3x + C raises each power but forgets to divide by the new exponent. D) 2x³ − 4x² + 3x + C fails to divide the −4x term by 2. E) 2x³ − 2x² + 3 + C integrates 3 as 3 instead of 3x. Fix: to antidifferentiate xⁿ, add 1 to the exponent and divide by the new exponent; a constant integrates to (constant)·x. [Apply]

4. D) 24x(3x² + 1)³. Chain rule: 4(3x² + 1)³ · d/dx(3x² + 1) = 4(3x² + 1)³ · 6x = 24x(3x² + 1)³. Distractors: A) 4(3x² + 1)³ forgets the inner derivative 6x. B) 6x(3x² + 1)³ uses only the inner derivative and drops the outer factor 4. C) 24x(3x² + 1)⁴ multiplies correctly but never reduced the power from 4 to 3. E) 12x(3x² + 1)³ used 3x instead of 6x for the inner derivative. Fix: differentiate the outside (drop the power, subtract one) and multiply by the derivative of the inside. [Apply]

5. C) 8. ∫₀² 3x² dx = [x³]₀² = 2³ − 0³ = 8. Distractors: A) 4 = 2² comes from a botched antiderivative. B) 6 halves the answer as if integrating x² without the 3. D) 12 evaluates the integrand 3x² at x = 2 instead of the antiderivative. E) 16 = 2⁴ over-raises the power. Fix: find the antiderivative first (x³), then subtract its values at the two limits — never plug the limits into the original integrand. [Apply]

6. B) x² e^x (x + 3). Product rule: d/dx(x³ e^x) = 3x² e^x + x³ e^x = x² e^x(x + 3). Distractors: A) 3x² e^x differentiates only x³ and treats e^x as constant (drops the second term). C) x² e^x(x − 3) makes a sign slip in the factoring. D) 3x² + e^x differentiates each factor separately and adds — ignoring the product rule entirely. E) 3x² e^x + x³ drops the e^x on the second term. Fix: (fg)′ = f′g + fg′; here both terms survive and factor to x²e^x(x + 3). [Apply]

7. A) 3/5. Match the argument to the denominator: sin(3x)/(5x) = (3/5)·sin(3x)/(3x) → (3/5)·1 = 3/5. Distractors: B) 1 applies sin(u)/u → 1 without correcting the mismatched coefficients. C) 5/3 inverts the correction factor. D) 0 confuses this with (1 − cos x)/x. E) "nonexistent" misjudges a standard trig limit. Fix: sin(ax)/(bx) → a/b — the sine's coefficient over the denominator's coefficient. [Apply]

8. D) (x² + 1)⁴/4 + C. Let u = x² + 1, du = 2x dx; the integral becomes ∫u³ du = u⁴/4 = (x² + 1)⁴/4 + C. Distractors: A) (x² + 1)⁴ + C forgets to divide by 4. B) (x² + 1)⁴/8 + C divides by 8 (mishandling the 2x). C) (x² + 1)³/3 + C never raised the power (integrated u² instead of u³). E) 2x(x² + 1)⁴/4 + C wrongly keeps the 2x after it was absorbed into du. Fix: when du exactly matches the leftover factor, substitute cleanly — the 2x is consumed, not carried. [Apply]

9. E) 2x/(x² + 1). Chain rule for ln u: (ln(x²+1))′ = (1/(x²+1))·(2x) = 2x/(x²+1). Distractors: A) 1/(x² + 1) forgets the inner derivative 2x. B) 1/(2x) differentiates ln of the wrong argument. C) 2x·ln(x² + 1) multiplies by the log instead of dividing by the argument. D) (x² + 1)/(2x) inverts the fraction. Fix: d/dx ln u = u′/u — inner derivative on top, the argument on the bottom. [Apply]

10. B) 1/6. The curves meet where x = x², i.e. x = 0, 1, and x ≥ x² on [0, 1], so area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6. Distractors: A) 1/2 integrates only the top curve x. C) 1/3 integrates only the bottom curve x². D) 1/12 halves the correct answer. E) 1 ignores the subtraction entirely. Fix: area between curves is ∫(top − bottom) over the interval where they enclose a region; find the intersection points first. [Analyze]

11. C) 1/(x + 1)². Quotient rule: [(1)(x + 1) − x(1)]/(x + 1)² = (x + 1 − x)/(x + 1)² = 1/(x + 1)². Distractors: A) 1 divides the derivatives (1/1) instead of using the quotient rule. B) x/(x + 1)² keeps an x in the numerator from a subtraction slip. D) −1/(x + 1)² flips the sign by reversing the numerator order. E) (2x + 1)/(x + 1)² comes from adding rather than subtracting in the numerator. Fix: (f/g)′ = (f′g − fg′)/g² — low·d-high minus high·d-low, order matters for the sign. [Apply]

12. A) sin(x²). By the Fundamental Theorem (Part 1), d/dx ∫₀ˣ sin(t²) dt = sin(x²) — differentiating the accumulation function returns the integrand at the upper limit. Distractors: B) cos(x²) differentiates the integrand instead of just evaluating it. C) 2x cos(x²) needlessly applies the chain rule (the upper limit is plain x). D) sin(x²) − sin(0) treats it like an evaluated definite integral rather than a derivative. E) (1/3)sin(x²) invents an antiderivative constant. Fix: d/dx ∫_a^x f(t) dt = f(x) — no integration and no extra chain factor when the upper limit is simply x. [Understand]

13. D) −3/4. Differentiate implicitly: 2x + 2y·y′ = 0 → y′ = −x/y; at (3, 4), y′ = −3/4. Distractors: A) 3/4 drops the negative sign. B) −4/3 inverts the ratio. C) 4/3 inverts and drops the sign. E) −3/5 uses the hypotenuse 5 (from 3² + 4² = 25) in place of y. Fix: differentiate both sides, attach y′ to every y-term, solve for y′, then substitute the point — here y′ = −x/y. [Apply]

14. E) x⁴ − 1/x + C. ∫4x³ dx = x⁴, and ∫x⁻² dx = x⁻¹/(−1) = −1/x. Distractors: A) 12x² − 2/x³ + C differentiates instead of integrating. B) x⁴ + 1/x + C gets the sign of the second term wrong. C) x⁴ − 2/x³ + C mishandles the negative exponent (used the derivative pattern). D) 4x⁴ − 1/x + C forgets to divide the 4x³ term by the new power 4. Fix: ∫xⁿ dx = xⁿ⁺¹/(n+1); for n = −2 this gives −x⁻¹ = −1/x. [Apply]

15. A) 1. f′(x) = 3x² − 3 = 0 → x = ±1; f″(x) = 6x, and f″(1) = 6 > 0, so x = 1 is a local minimum. Distractors: B) −1 is the local maximum (f″(−1) = −6 < 0). C) 0 is where f″ = 0 (an inflection point), not an extremum. D) 3 and E) −3 are not critical numbers at all. Fix: set f′ = 0 for critical points, then use the sign of f″: positive means a minimum, negative means a maximum. [Analyze]

16. B) 1/2. This is a standard limit; by L'Hôpital on the 0/0 form, (1 − cos x)/x² → (sin x)/(2x) → 1/2. Distractors: A) 0 confuses this with (1 − cos x)/x (which does go to 0). C) 1 stops after one L'Hôpital step and misreads sin x / (2x). D) 2 inverts the 1/2. E) "nonexistent" misjudges a convergent form. Fix: memorize (1 − cos x)/x² → 1/2; note the single power (1 − cos x)/x → 0 is different. [Apply]

17. C) 8π. Disk method about the x-axis: V = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π·8 = 8π. Distractors: A) 4π forgets to square-then-simplify or halves twice. B) 16π uses [x²/2] incorrectly as x² at 4 without the ½. D) 32π/3 and E) 64π/3 come from integrating x² (as if the radius were x, not √x). Fix: disk volume is π∫[radius]² dx; here (√x)² = x collapses cleanly before integrating. [Apply]

18. D) 20π cm²/s. A = πr², so dA/dt = 2πr·dr/dt = 2π(5)(2) = 20π. Distractors: A) 10π drops the factor 2 from differentiating r². B) 25π uses πr² (the area itself) instead of its rate. C) 4π uses only dr/dt squared or forgets r. E) 100π multiplies area π(5²) by dr/dt without differentiating. Fix: differentiate A = πr² with respect to time: dA/dt = 2πr(dr/dt); substitute the instant's r last. [Apply]

19. A) 3. Average value = (1/(3 − 0))∫₀³ x² dx = (1/3)[x³/3]₀³ = (1/3)(9) = 3. Distractors: B) 9 is ∫₀³ x² without the 1/(b − a) factor. C) 27 is x³ at 3, skipping both the antiderivative division and the averaging. D) 1 over-divides. E) 6 mishandles the 1/3 factor. Fix: average value = [1/(b − a)]∫_a^b f(x) dx; compute the definite integral, then divide by the interval length. [Apply]

20. E) 2. MVT: f′(c) = (f(3) − f(1))/(3 − 1) = (9 − 1)/2 = 4; since f′(x) = 2x, set 2c = 4 → c = 2. Distractors: A) 0 and B) 1 and C) 3 are endpoints or arbitrary points, not the guaranteed value. D) 4 reports the average slope itself instead of solving f′(c) = 4 for c. Fix: compute the secant slope over [a, b], set f′(c) equal to it, and solve for c. [Apply]

21. B) 18. ∫₁⁴ (2x + 1) dx = [x² + x]₁⁴ = (16 + 4) − (1 + 1) = 20 − 2 = 18. Distractors: A) 15 = [x²]₁⁴ forgets to integrate the +1. C) 21 slips in the arithmetic of the bounds. D) 24 evaluates only at the upper limit and ignores the lower. E) 9 halves incorrectly. Fix: antidifferentiate every term (2x → x², 1 → x), then subtract lower-limit value from upper-limit value. [Apply]

22. D) 1 and 3. Velocity v(t) = s′(t) = 3t² − 12t + 9 = 3(t − 1)(t − 3) = 0 at t = 1 and t = 3; both are in t ≥ 0. Distractors: A) 0 only misreads the position, not velocity. B) 1 only and C) 3 only each keep just one root. E) 2 only picks the vertex of the parabola (where speed is minimal, not zero). Fix: "at rest" means velocity = 0; differentiate position, factor, and keep every valid root. [Analyze]

23. C) 100e². The solution of dy/dt = ky is y = y₀e^{kt} = 100e^{0.5t}; at t = 4, y = 100e^{0.5·4} = 100e². Distractors: A) 100e^{0.2} multiplies 0.5·4 wrong (used 0.05·4). B) 200 treats the growth as linear doubling. D) 50e² halves the initial amount. E) 100 + e² adds instead of multiplying by the exponential. Fix: exponential growth is y₀e^{kt}; put the full exponent k·t = 0.5·4 = 2 in place before evaluating. [Apply]

24. A) 2x cos(x²). Chain rule: d/dx sin(x²) = cos(x²)·(2x) = 2x cos(x²). Distractors: B) cos(x²) forgets the inner derivative 2x. C) 2x cos(2x) differentiates the inner argument to 2x inside the cosine too (double error). D) −2x cos(x²) wrongly attaches the minus sign from cos's own derivative. E) cos(2x) mishandles both the argument and the inner factor. Fix: d/dx sin(u) = cos(u)·u′; the argument of cosine stays x², and you multiply by u′ = 2x. [Apply]

25. E) e − 1. Let u = sin x, du = cos x dx; limits x: 0 → π/2 give u: 0 → 1, so ∫₀¹ e^u du = [e^u]₀¹ = e − 1. Distractors: A) e forgets to subtract the lower-limit value e⁰ = 1. B) 1 evaluates e^u at 0 only or mishandles the antiderivative. C) e^{π/2} − 1 fails to change the limits after substituting. D) 0 cancels incorrectly. Fix: with u-substitution in a definite integral, convert the limits to u-values and evaluate directly — don't keep the original x-bounds. [Apply]

26. D) x > 1. f″(x) = 6x − 6 > 0 ⇔ x > 1, so the graph is concave up for x > 1. Distractors: A) x < 1 is where it is concave down. B) x > 0 uses f′ sign logic, not f″. C) all x ignores that concavity changes at the inflection point x = 1. E) x < 0 is unrelated. Fix: concavity is governed by the sign of f″; solve f″(x) > 0 for concave-up intervals. [Analyze]

27. B) 3/2. Numerator and denominator have equal degree (2), so the limit is the ratio of leading coefficients: 3/2. Distractors: A) 0 applies the "denominator wins" rule for the wrong degree comparison. C) 2/3 inverts the ratio. D) 1 assumes the limit of any rational function at infinity is 1. E) ∞ applies the "numerator wins" rule incorrectly. Fix: for x → ∞ of a rational function, compare degrees — equal degrees give the ratio of leading coefficients. [Apply]

28. C) tan x + C. d/dx tan x = sec²x, so ∫sec²x dx = tan x + C. Distractors: A) sec x tan x + C is the antiderivative pattern for sec x tan x (that would integrate to sec x). B) 2 sec²x tan x + C differentiates sec²x instead of antidifferentiating. D) −cot x + C is ∫csc²x dx, the wrong squared function. E) sec x + C confuses the reciprocal-function antiderivatives. Fix: read integration as reversed differentiation — since (tan x)′ = sec²x, the antiderivative of sec²x is tan x. [Understand]

29. A) 1250. Let x be each side perpendicular to the barn and y the side parallel to it; fencing gives 2x + y = 100, so y = 100 − 2x and A = x(100 − 2x) = 100x − 2x². A′ = 100 − 4x = 0 → x = 25, y = 50, A = 25·50 = 1250. Distractors: B) 2500 forgets the wall saves one side (uses a full-perimeter model). C) 625 = 25² uses x² alone. D) 1000 stops before optimizing. E) 5000 mis-sets the constraint. Fix: write area as one variable using the fencing constraint, maximize with A′ = 0, and confirm it's a max; the free wall side is not fenced. [Analyze]

30. E) 32/3. y = 4 − x² meets the x-axis at x = ±2, so area = ∫₋₂² (4 − x²) dx = 2∫₀² (4 − x²) dx = 2[4x − x³/3]₀² = 2(8 − 8/3) = 2(16/3) = 32/3. Distractors: A) 8 uses only 4x at 2 without the x³/3 term. B) 16 forgets the symmetry factor issues and the cubic term. C) 16/3 computes only the half-region ∫₀² and forgets to double. D) 8/3 keeps only the x³/3 piece. Fix: find where the curve crosses the axis for the limits, integrate (top − bottom), and use symmetry carefully (double the half-integral). [Apply]

31. C) 2. The form is 0/0; L'Hôpital gives (2e^{2x})/(cos x) → (2·1)/1 = 2. Distractors: A) 0 stops at the 0/0 form. B) 1 differentiates e^{2x} as e^{2x} (forgets the chain factor 2). D) 1/2 inverts the result. E) "nonexistent" misjudges a convergent 0/0. Fix: confirm 0/0, then differentiate top and bottom separately; d/dx e^{2x} = 2e^{2x} carries the chain factor. [Apply]

32. B) ln(x² + 1) + C. With u = x² + 1, du = 2x dx, the integral is ∫du/u = ln|u| = ln(x² + 1) + C (and x² + 1 > 0). Distractors: A) 2 arctan x + C is the antiderivative of 2/(x² + 1), a different integrand. C) 2/(x² + 1) + C differentiates instead of integrating. D) (1/2)ln(x² + 1) + C wrongly divides by 2, as if du weren't already 2x dx. E) ln|2x| + C integrates the numerator's argument instead of the denominator. Fix: when the numerator is exactly the denominator's derivative, the integral is ln|denominator| — no extra constant factor. [Apply]

33. D) 72π cm³/s. V = (4/3)πr³, so dV/dt = 4πr²·dr/dt = 4π(3²)(2) = 4π(9)(2) = 72π. Distractors: A) 36π drops the factor 2 (dr/dt) or halves. B) 18π uses 2πr (a circle formula). C) 24π uses 4πr linearly instead of 4πr². E) 144π doubles the correct result. Fix: differentiate V = (4/3)πr³ to dV/dt = 4πr²(dr/dt); square the radius before multiplying. [Apply]

34. A) 9 gallons. Total accumulated = ∫₀³ 2t dt = [t²]₀³ = 9. Distractors: B) 6 evaluates the rate 2t at t = 3 instead of integrating. C) 18 = 2·9 forgets to divide by the new power (antiderivative of 2t is t², not 2t²). D) 3 integrates 2t as 2 giving 2t → 6... or just reads the endpoint time. E) 12 mishandles the arithmetic. Fix: a total amount from a rate is the definite integral of the rate over the time interval: ∫2t dt = t². [Apply]

35. E) increasing and concave down. f′ > 0 means the function is increasing; f″ < 0 means it is concave down. Distractors: A) increasing, concave up flips the sign of f″. B) decreasing, concave down flips the sign of f′. C) decreasing, concave up flips both. D) invents inflection points, which occur only where f″ changes sign, not throughout. Fix: first derivative controls increasing/decreasing; second derivative controls concavity — read the two signs independently. [Evaluate]

36. B) (e² − 1)/2. ∫₀¹ e^{2x} dx = [(1/2)e^{2x}]₀¹ = (1/2)(e² − 1). Distractors: A) e² − 1 forgets the 1/2 from ∫e^{2x} = (1/2)e^{2x}. C) 2(e² − 1) multiplies by 2 instead of dividing. D) e²/2 forgets to subtract the lower-limit value (1/2)e⁰ = 1/2. E) e − 1 integrates e^x instead of e^{2x}. Fix: ∫e^{ax} dx = (1/a)e^{ax}; here a = 2, so divide by 2, then evaluate the bounds. [Apply]

37. D) y = 6x − 9. y′ = 2x, so the slope at x = 3 is 6; point-slope through (3, 9): y − 9 = 6(x − 3) → y = 6x − 9. Distractors: A) y = 6x + 9 uses the wrong sign on the intercept. B) y = 6x forgets the point entirely (line through the origin). C) y = 9x − 18 uses the function value 9 as the slope. E) y = 3x − 9 uses x = 3 as the slope instead of f′(3) = 6. Fix: tangent slope is f′ at the point; then use point-slope form with the actual point of tangency. [Apply]

38. C) √(1 + x³). Fundamental Theorem (Part 1): d/dx ∫₁ˣ √(1 + t³) dt = √(1 + x³) — plug the upper limit x into the integrand. Distractors: A) √(1 + t³) leaves the dummy variable t instead of substituting x. B) (1 + x³)^{3/2} integrates or mis-powers the integrand. D) 1/[2√(1 + x³)] differentiates the square root instead of just evaluating it. E) √(1 + x³) − √2 subtracts a lower-limit term, treating it like an evaluated definite integral. Fix: the derivative of ∫_a^x f(t) dt is simply f(x) — replace t with x, nothing more. [Understand]

39. A) g is continuous everywhere but not differentiable at x = 2. |x − 2| is a continuous V-shape whose corner at x = 2 gives left slope −1 and right slope +1; the mismatch means no derivative there. Distractors: B) claims a discontinuity, but the graph connects. C) claims differentiability everywhere, ignoring the corner. D) denies continuity, which holds. E) is self-contradictory — differentiability requires continuity. Fix: a corner is continuous but not differentiable; differentiability implies continuity, never the reverse. [Evaluate]

40. E) y = x² + 1. Integrate dy/dx = 2x to y = x² + C; the condition y(0) = 1 gives C = 1, so y = x² + 1. Distractors: A) y = 2 ignores the variable dependence. B) y = x² drops the initial condition (would give y(0) = 0). C) y = 2x + 1 fails to antidifferentiate 2x. D) y = x² − 1 uses the wrong sign for C. Fix: solve the differential equation by integrating, then use the initial condition to pin down the constant C. [Apply]

41. B) −2. Acceleration a(t) = v′(t) = 2t − 4, so a(1) = 2(1) − 4 = −2. Distractors: A) 0 reports v(1) = 1 − 4 + 3 = 0 (velocity, not acceleration). C) 2 keeps only the 2t term. D) −4 uses the constant term of v without differentiating. E) −1 slips in the arithmetic. Fix: acceleration is the derivative of velocity; differentiate v(t) first, then substitute the time. [Apply]

42. C) 0. The form is ∞/∞; L'Hôpital gives (1/x)/1 = 1/x → 0 as x → ∞. Distractors: A) ∞ assumes the numerator dominates, but x grows faster than ln x. B) 1 treats it as equal-rate growth. D) e injects the base of the log erroneously. E) "nonexistent" misjudges a convergent limit. Fix: x outgrows ln x, so (ln x)/x → 0; L'Hôpital confirms it in one step. [Apply]

43. D) −4 sin(2x). f′(x) = 2cos(2x), then f″(x) = 2·(−sin(2x))·2 = −4 sin(2x). Distractors: A) −sin(2x) forgets both chain factors of 2. B) −2 sin(2x) applies the chain factor only once. C) 4 sin(2x) drops the negative sign from differentiating cosine. E) −4 cos(2x) differentiates one step too few (keeps cosine). Fix: each derivative of sin(2x)/cos(2x) multiplies by the inner factor 2, so two derivatives contribute 2·2 = 4, with a sign flip from cos → −sin. [Apply]

44. E) lim_{h→0} [f(2 + h) − f(2)]/h. This is the limit definition of the derivative at x = 2. Distractors: A) omits the /h denominator. B) adds instead of subtracting the function values. C) drops the limit, leaving just a difference quotient (an average rate, not the derivative). D) uses x → 0 with a mismatched expression. Fix: f′(a) = lim_{h→0} [f(a + h) − f(a)]/h — the difference quotient with the limit as h → 0. [Understand]

Scaled Score Estimator

CLEP reports a scaled score from 20 to 80; ACE recommends credit at 50, which corresponds to roughly half the questions correct (~22/44 on this mock). The table below approximates raw correct → scaled score.

For an odd raw score, interpolate between the neighboring rows. Disclaimer: CLEP's exact raw-to-scaled conversion is proprietary and varies slightly by form; this table is a transparent linear approximation for self-assessment only, not an official predictor.