The Concept of a Limit
The limit lim(x→a) f(x) = L means f(x) gets arbitrarily close to L as x approaches a, without necessarily equaling f(a). One-sided limits: the left-hand limit lim(x→a⁻) f(x) and right-hand limit lim(x→a⁺) f(x) must both exist and be equal for the two-sided limit to exist. If they differ, the limit does not exist.
Limit Laws and Evaluation Techniques
For polynomials and rational functions where the denominator is nonzero, direct substitution works: lim(x→a) p(x) = p(a). For 0/0 indeterminate forms, try: (1) factoring and canceling, (2) rationalizing by multiplying by conjugate, (3) L'Hôpital's Rule — if lim f/g → 0/0 or ∞/∞, then lim f/g = lim f'/g'. The Squeeze Theorem: if g(x) ≤ f(x) ≤ h(x) near a, and lim g(x) = lim h(x) = L, then lim f(x) = L. Famous limit: lim(x→0) sin(x)/x = 1.
Limits at Infinity and Infinite Limits
As x → ±∞, polynomial behavior is dominated by the leading term. For rational functions: if degree of numerator < denominator, limit = 0; if equal, limit = ratio of leading coefficients; if numerator > denominator, limit = ±∞. Vertical asymptotes occur where the denominator approaches 0 and the numerator doesn't; horizontal asymptotes are the limits at ±∞.
Continuity
A function f is continuous at a if: (1) f(a) is defined, (2) lim(x→a) f(x) exists, (3) lim(x→a) f(x) = f(a). Removable discontinuities (holes) can be fixed by redefining; jump discontinuities and infinite discontinuities cannot. Polynomials, rational functions (at non-zero denominator points), trig, exponential, and log functions are all continuous on their domains. The Intermediate Value Theorem (IVT): if f is continuous on [a,b] and N is between f(a) and f(b), then there exists c ∈ (a,b) with f(c) = N.
The Derivative: Definition and Interpretation
The derivative f'(x) = lim(h→0) [f(x+h) − f(x)] / h is the instantaneous rate of change of f at x, and the slope of the tangent line to the graph at (x, f(x)). Differentiability implies continuity (but not vice versa). A function fails to be differentiable at corners, cusps, vertical tangents, and discontinuities.
Basic Differentiation Rules
Constant rule: d/dx[c] = 0. Power rule: d/dx[xⁿ] = nxⁿ⁻¹. Constant multiple: d/dx[cf] = c·f'. Sum/difference: (f ± g)' = f' ± g'. Product rule: (fg)' = f'g + fg'. Quotient rule: (f/g)' = (f'g − fg')/g². Chain rule: d/dx[f(g(x))] = f'(g(x))·g'(x). Know these cold — they are the engine of differential calculus.
Derivatives of Standard Functions
Trigonometric: d/dx[sin x] = cos x; d/dx[cos x] = −sin x; d/dx[tan x] = sec²x; d/dx[cot x] = −csc²x; d/dx[sec x] = sec x tan x; d/dx[csc x] = −csc x cot x. Exponential: d/dx[eˣ] = eˣ; d/dx[aˣ] = aˣ ln a. Logarithmic: d/dx[ln x] = 1/x; d/dx[log_a x] = 1/(x ln a). Inverse trig: d/dx[arcsin x] = 1/√(1−x²); d/dx[arctan x] = 1/(1+x²).
Higher-Order Derivatives and Implicit Differentiation
The second derivative f''(x) is the derivative of f'(x); the nth derivative is denoted f⁽ⁿ⁾(x). f''(x) represents the rate of change of slope (concavity). Implicit differentiation: differentiate both sides with respect to x, treating y as a function of x and applying the chain rule to y-terms. Example: d/dx[y²] = 2y · dy/dx. Then solve for dy/dx algebraically.
Related Rates
Related rates problems involve two or more quantities changing with respect to time. Strategy: (1) draw and label the situation, (2) write an equation relating the quantities, (3) differentiate both sides with respect to t, (4) substitute known values and solve for the unknown rate. Common setups: ladder sliding down wall, balloon inflating, water draining from a cone.
Mean Value Theorem and Rolle's Theorem
Rolle's Theorem: if f is continuous on [a,b], differentiable on (a,b), and f(a) = f(b), then there exists c ∈ (a,b) where f'(c) = 0. The Mean Value Theorem (MVT): if f is continuous on [a,b] and differentiable on (a,b), then there exists c where f'(c) = [f(b) − f(a)] / (b − a). The MVT means that the instantaneous rate of change equals the average rate of change at some point.
Increasing/Decreasing Functions and Critical Points
f is increasing where f'(x) > 0 and decreasing where f'(x) < 0. Critical points occur where f'(x) = 0 or f'(x) is undefined. First Derivative Test: if f' changes from + to − at c, then c is a local maximum; if from − to +, a local minimum; if no sign change, neither. To find absolute extrema on [a,b]: evaluate f at all critical points and endpoints, then compare.
Concavity and Second Derivative Test
f is concave up where f''(x) > 0 (f' is increasing, graph curves upward) and concave down where f''(x) < 0. Inflection points are where concavity changes — find by setting f''(x) = 0 and checking for sign change. Second Derivative Test for local extrema: if f'(c) = 0 and f''(c) > 0, then c is a local minimum; if f''(c) < 0, a local maximum; if f''(c) = 0, the test is inconclusive.
Optimization Problems
Strategy: (1) identify the quantity to maximize/minimize, (2) write the objective function, (3) use a constraint to reduce to a single variable, (4) differentiate and find critical points, (5) determine whether each is a max or min using the First or Second Derivative Test, (6) check endpoints if on a closed interval. Common examples: maximizing area with fixed perimeter, minimizing cost with fixed volume.
Linear Approximation and Differentials
The linearization of f at a is L(x) = f(a) + f'(a)(x − a). This is the tangent line approximation. The differential dy = f'(x) dx approximates the change in y for a small change dx in x. Newton's Method uses the linearization iteratively to approximate roots: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ).
Antiderivatives and Indefinite Integrals
The antiderivative F(x) of f(x) satisfies F'(x) = f(x). The indefinite integral ∫f(x)dx = F(x) + C, where C is the constant of integration. Basic antiderivatives: ∫xⁿdx = xⁿ⁺¹/(n+1) + C (n ≠ −1); ∫eˣdx = eˣ + C; ∫1/x dx = ln|x| + C; ∫sin x dx = −cos x + C; ∫cos x dx = sin x + C; ∫sec²x dx = tan x + C. The constant of integration must always be included.
The Definite Integral and Riemann Sums
The definite integral ∫[a to b] f(x)dx is the limit of Riemann sums: divide [a,b] into n subintervals of width Δx = (b−a)/n, pick a sample point xᵢ* in each, and sum f(xᵢ*)Δx as n → ∞. Left, right, and midpoint Riemann sums approximate the area. The definite integral represents the signed area between f and the x-axis — positive above the x-axis, negative below.
Fundamental Theorem of Calculus
Part 1: If F(x) = ∫[a to x] f(t)dt, then F'(x) = f(x). Differentiation and integration are inverse operations. Part 2: ∫[a to b] f(x)dx = F(b) − F(a), where F is any antiderivative of f. Properties of definite integrals: ∫[a to b] = −∫[b to a]; ∫[a to a] = 0; ∫[a to c] + ∫[c to b] = ∫[a to b]; linearity (sums and constants factor out).
Integration Techniques
u-substitution: identify an inner function u = g(x), compute du = g'(x)dx, rewrite the integral in terms of u, integrate, then back-substitute. This is the chain rule in reverse. Integration by parts: ∫u dv = uv − ∫v du. Choose u using LIATE (Logarithms, Inverse trig, Algebraic, Trig, Exponential) — u is the first applicable type. Partial fractions for rational functions where denominator factors into linear factors.
Improper Integrals
An integral is improper if (1) a limit of integration is ±∞, or (2) the integrand has a vertical asymptote in [a,b]. Evaluate by replacing the problematic limit with a variable, integrating, then taking the limit. ∫[1 to ∞] 1/xᵖ dx converges if p > 1, diverges if p ≤ 1. ∫[0 to ∞] eˉˣ dx = 1 (converges). If the limit exists, the integral converges; if not, it diverges.
Area Between Curves
Area between f(x) and g(x) on [a,b] where f ≥ g: A = ∫[a to b] [f(x) − g(x)] dx. Find intersection points to determine the interval. If curves cross, split the integral and ensure you subtract the lower from the upper in each interval. For curves expressed as functions of y, integrate with respect to y: A = ∫[c to d] [right(y) − left(y)] dy.
Volumes of Revolution
Disk method (around x-axis): V = π∫[a to b] [f(x)]² dx. Washer method (region between two curves, around x-axis): V = π∫[a to b] {[f(x)]² − [g(x)]²} dx. Shell method (around y-axis): V = 2π∫[a to b] x·f(x) dx. The shell method is often easier when the axis of revolution is parallel to the direction of integration. Choose the method based on which gives a simpler integrand.
Arc Length and Surface Area
Arc length of y = f(x) on [a,b]: L = ∫[a to b] √(1 + [f'(x)]²) dx. Surface area of revolution around the x-axis: S = 2π∫[a to b] f(x)√(1 + [f'(x)]²) dx. These integrals are often difficult to evaluate exactly — recognize the setup even if the computation is complex.
Average Value of a Function
The average value of f on [a,b] is f_avg = 1/(b−a) · ∫[a to b] f(x) dx. The Mean Value Theorem for Integrals: if f is continuous on [a,b], there exists c ∈ [a,b] such that f(c) = f_avg. This means a continuous function attains its average value somewhere in the interval.
Differential Equations (Separation of Variables)
A differential equation involves a function and its derivatives. Separable equations: rewrite as g(y)dy = f(x)dx, integrate both sides, and solve for y. The initial value problem (IVP) adds a condition like y(x₀) = y₀ to determine C. Exponential growth/decay: dy/dt = ky → y = y₀eᵏᵗ. Population models, radioactive decay, and Newton's Law of Cooling are common examples.
Taylor and Maclaurin Series
The Taylor series for f(x) centered at a is: f(a) + f'(a)(x−a) + f''(a)/2!(x−a)² + f'''(a)/3!(x−a)³ + … Maclaurin series (a = 0). Memorize: eˣ = Σxⁿ/n! = 1 + x + x²/2! + x³/3! + …; sin x = Σ(−1)ⁿ x²ⁿ⁺¹/(2n+1)! = x − x³/6 + …; cos x = Σ(−1)ⁿ x²ⁿ/(2n)! = 1 − x²/2 + …; 1/(1−x) = Σxⁿ = 1 + x + x² + … (|x| < 1).
Convergence of Series
Infinite series Σaₙ converges if the sequence of partial sums converges. Key tests: Geometric series Σarⁿ converges iff |r| < 1 (sum = a/(1−r)). p-series Σ1/nᵖ converges iff p > 1. Ratio Test: if lim|aₙ₊₁/aₙ| < 1, converges absolutely; > 1, diverges; = 1, inconclusive. Divergence Test (nth Term Test): if lim aₙ ≠ 0, series diverges. Alternating Series Test: if alternating series has decreasing terms → 0, it converges.
Parametric Differentiation
For x = f(t), y = g(t): dy/dx = (dy/dt)/(dx/dt) = g'(t)/f'(t). Second derivative: d²y/dx² = [d/dt(dy/dx)] / (dx/dt). Arc length in parametric form: L = ∫√((dx/dt)² + (dy/dt)²) dt. Speed = √((dx/dt)² + (dy/dt)²) at a given t.
Polar Calculus
For polar curves r = f(θ): area enclosed = ½∫[α to β] r² dθ. dy/dx = (dr/dθ sin θ + r cos θ)/(dr/dθ cos θ − r sin θ). Area between two polar curves: ½∫[α to β] (r_outer² − r_inner²) dθ. Common polar curves: r = a (circle), r = a cos θ (circle shifted right), r = a(1 + cos θ) (cardioid), r = a cos(nθ) (rose with n petals if n is odd, 2n if even).
L'Hôpital's Rule and Indeterminate Forms
When a limit yields 0/0 or ∞/∞, L'Hôpital's Rule applies: lim f(x)/g(x) = lim f'(x)/g'(x). Apply as many times as necessary until the indeterminate form is resolved. Other forms require algebraic manipulation to get 0/0 or ∞/∞: 0·∞ → rewrite as a quotient; ∞−∞ → combine over common denominator; 1^∞, 0⁰, ∞⁰ → take natural log and apply L'Hôpital's, then exponentiate.
1Evaluate: lim(x→3) (x² − 9)/(x − 3)
Correct Answer: C
Factor: (x²−9)/(x−3) = (x+3)(x−3)/(x−3) = x+3. As x→3, this → 3+3 = 6.
2What is lim(x→∞) (3x² + 5)/(x² − 2)?
Correct Answer: C
Equal degrees in numerator and denominator: the limit equals the ratio of leading coefficients = 3/1 = 3.
3What is d/dx[x⁵ − 3x² + 7]?
A) x⁴ − 6x
B) 5x⁴ − 6x + 7
C) 5x⁴ − 6x
D) 5x⁶ − x³
Correct Answer: C
Apply power rule term by term: d/dx[x⁵] = 5x⁴; d/dx[−3x²] = −6x; d/dx[7] = 0. Total: 5x⁴ − 6x.
4Find d/dx[sin(x²)].
A) cos(x²)
B) 2x cos(x²)
C) −2x sin(x²)
D) cos(2x)
Correct Answer: B
Chain rule: outer = sin(u), inner u = x². d/dx[sin(x²)] = cos(x²) · 2x = 2x cos(x²).
5Find d/dx[eˣ · ln x].
A) eˣ/x
B) eˣ ln x + eˣ/x
C) eˣ + eˣ/x
D) eˣ ln x − eˣ/x
Correct Answer: B
Product rule: (eˣ)'·ln x + eˣ·(ln x)' = eˣ ln x + eˣ · (1/x) = eˣ ln x + eˣ/x.
6What is the derivative of f(x) = tan(3x)?
A) sec²(3x)
B) 3 sec²(3x)
C) −3 csc²(3x)
D) 3 sec(3x)
Correct Answer: B
Chain rule: d/dx[tan(u)] = sec²(u) · u'. Here u = 3x, u' = 3. So f'(x) = 3 sec²(3x).
7Find the equation of the tangent line to f(x) = x³ at x = 2.
A) y = 12x − 16
B) y = 12x + 8
C) y = 3x − 4
D) y = 6x − 4
Correct Answer: A
f(2) = 8, f'(x) = 3x², f'(2) = 12. Tangent: y − 8 = 12(x − 2) → y = 12x − 16.
8At which x-value does f(x) = x³ − 3x have a local minimum?
A) x = 0
B) x = −1
C) x = 1
D) x = 3
Correct Answer: C
f'(x) = 3x² − 3 = 0 → x = ±1. f''(x) = 6x. f''(1) = 6 > 0 → local minimum at x = 1. f''(−1) = −6 < 0 → local maximum at x = −1.
9Find ∫(4x³ − 2x + 5) dx.
A) 12x² − 2 + C
B) x⁴ − x² + 5x + C
C) 4x⁴ − 2x² + 5x + C
D) x⁴ + x² + 5 + C
Correct Answer: B
∫4x³dx = x⁴; ∫−2x dx = −x²; ∫5 dx = 5x. Total: x⁴ − x² + 5x + C.
10Evaluate: ∫[0 to 2] (3x²) dx
Correct Answer: B
∫3x² dx = x³ + C. Evaluate from 0 to 2: [x³] from 0 to 2 = 8 − 0 = 8.
11Use u-substitution to evaluate ∫2x(x² + 1)⁴ dx.
A) (x² + 1)⁵/5 + C
B) 8x(x² + 1)³ + C
C) 2(x² + 1)⁵ + C
D) (x² + 1)⁵ + C
Correct Answer: A
Let u = x² + 1, du = 2x dx. Integral becomes ∫u⁴ du = u⁵/5 + C = (x² + 1)⁵/5 + C.
12What is the area under f(x) = sin x from x = 0 to x = π?
Correct Answer: C
∫[0 to π] sin x dx = [−cos x] from 0 to π = −cos(π) − (−cos 0) = −(−1) − (−1) = 1 + 1 = 2.
13If F'(x) = f(x) and F(2) = 5, what does the Fundamental Theorem tell us about ∫[2 to 4] f(x) dx if F(4) = 11?
Correct Answer: B
By FTC Part 2: ∫[2 to 4] f(x)dx = F(4) − F(2) = 11 − 5 = 6.
14Which of the following functions is NOT continuous at x = 2?
A) f(x) = x² − 4
B) f(x) = |x − 2|
C) f(x) = 1/(x − 2)
D) f(x) = √(x − 1)
Correct Answer: C
f(x) = 1/(x−2) is undefined at x = 2 (division by zero → vertical asymptote). The other three functions are all defined and continuous at x = 2.
15Find the inflection point(s) of f(x) = x⁴ − 6x².
A) x = 0 only
B) x = ±1
C) x = ±√3
D) x = ±3
Correct Answer: B
f'(x) = 4x³ − 12x; f''(x) = 12x² − 12. Set f'' = 0: 12x² = 12 → x² = 1 → x = ±1. Check sign changes: f'' changes from positive to negative (and back), confirming inflection points at x = ±1.
16Apply L'Hôpital's Rule: lim(x→0) sin(x)/x
Correct Answer: C
0/0 form. L'Hôpital's: lim [d/dx(sin x)] / [d/dx(x)] = lim cos(x)/1 = cos(0)/1 = 1. This is the famous fundamental trig limit.
17Differentiate implicitly: x² + y² = 25. Find dy/dx.
A) dy/dx = y/x
B) dy/dx = x/y
C) dy/dx = −x/y
D) dy/dx = −y/x
Correct Answer: C
Differentiate both sides: 2x + 2y(dy/dx) = 0 → 2y(dy/dx) = −2x → dy/dx = −x/y.
18A particle's position is s(t) = t³ − 6t² + 9t. At what time(s) is the particle at rest (velocity = 0)?
A) t = 1 and t = 3
B) t = 2 only
C) t = 0 and t = 6
D) t = 3 only
Correct Answer: A
v(t) = s'(t) = 3t² − 12t + 9. Set v = 0: 3(t² − 4t + 3) = 3(t−1)(t−3) = 0 → t = 1 and t = 3.
19∫cos x dx =
A) −sin x + C
B) sin x + C
C) tan x + C
D) sec x + C
Correct Answer: B
The antiderivative of cos x is sin x (since d/dx[sin x] = cos x). Remember: ∫sin x dx = −cos x + C.
20Find the area of the region bounded by y = x² and y = x on [0, 1].
Correct Answer: A
On [0,1], x ≥ x². Area = ∫[0 to 1] (x − x²)dx = [x²/2 − x³/3] from 0 to 1 = 1/2 − 1/3 = 1/6.
21What is d/dx[ln(x² + 1)]?
A) 1/(x² + 1)
B) 2x/(x² + 1)
C) ln(2x)
D) 2/(x² + 1)
Correct Answer: B
Chain rule: d/dx[ln(u)] = (1/u)·u'. Here u = x² + 1, u' = 2x. Result: 2x/(x² + 1).
22The volume of the solid formed by rotating y = √x about the x-axis from x = 0 to x = 4 is:
Correct Answer: B
Disk method: V = π∫[0 to 4] (√x)² dx = π∫[0 to 4] x dx = π[x²/2] from 0 to 4 = π(8 − 0) = 8π.
23Which series converges?
A) Σ 1/n (harmonic)
B) Σ n²
C) Σ (1/2)ⁿ
D) Σ (−1)ⁿ
Correct Answer: C
Σ(1/2)ⁿ is a geometric series with r = 1/2, |r| < 1 → converges to (1/2)/(1 − 1/2) = 1. The harmonic series diverges; Σn² diverges; Σ(−1)ⁿ diverges (terms don't → 0).
24Evaluate ∫x eˣ dx using integration by parts.
A) x eˣ + C
B) eˣ(x − 1) + C
C) eˣ(x + 1) + C
D) x² eˣ/2 + C
Correct Answer: B
Let u = x, dv = eˣ dx → du = dx, v = eˣ. ∫x eˣ dx = x eˣ − ∫eˣ dx = x eˣ − eˣ + C = eˣ(x − 1) + C.
25The average value of f(x) = x² on [0, 3] is:
Correct Answer: A
f_avg = 1/(3−0) · ∫[0 to 3] x² dx = (1/3)[x³/3] from 0 to 3 = (1/3)(27/3) = (1/3)(9) = 3.
26A ladder 10 ft long is leaning against a wall. The bottom is sliding away at 2 ft/s. How fast is the top sliding down when the bottom is 6 ft from the wall?
A) −3/2 ft/s
B) −2/3 ft/s
C) 3/2 ft/s
D) 2/3 ft/s
Correct Answer: A
x² + y² = 100. When x = 6: y = 8. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0 → 6(2) + 8(dy/dt) = 0 → dy/dt = −12/8 = −3/2 ft/s. Negative sign means the top is sliding down.
27What does the Mean Value Theorem guarantee for f(x) = x² on [1, 3]?
A) f has a zero in [1,3]
B) f'(c) = 4 for some c in (1,3)
C) f achieves its maximum at x = 3
D) f'(c) = 0 for some c
Correct Answer: B
MVT: f'(c) = [f(3)−f(1)]/(3−1) = (9−1)/2 = 8/2 = 4. There exists c ∈ (1,3) with f'(c) = 2c = 4 → c = 2. Confirmed.
28Evaluate: ∫[1 to ∞] 1/x² dx
A) Diverges
B) 1/2
C) 1
D) 2
Correct Answer: C
∫[1 to t] x⁻² dx = [−x⁻¹] from 1 to t = −1/t + 1. As t → ∞: −1/t → 0, so the integral = 1. The p-series test: p = 2 > 1, so this converges.
29Solve the initial value problem: dy/dx = 3x², y(0) = 5.
A) y = x³
B) y = x³ + 5
C) y = 3x³ + 5
D) y = x³ − 5
Correct Answer: B
Integrate: y = ∫3x² dx = x³ + C. Apply y(0) = 5: 0 + C = 5 → C = 5. So y = x³ + 5.
30Find the Maclaurin series for f(x) = eˣ up to the x³ term.
A) 1 + x + x²/2 + x³/6
B) 1 + x + x² + x³
C) x + x²/2 + x³/6
D) 1 + x + x²/2! + x³/3
Correct Answer: A
The Maclaurin series for eˣ = Σxⁿ/n! = 1 + x + x²/2! + x³/3! + … = 1 + x + x²/2 + x³/6 + …
31The function f(x) is concave down on an interval where:
A) f(x) > 0
B) f'(x) > 0
C) f''(x) > 0
D) f''(x) < 0
Correct Answer: D
Concave down means the second derivative is negative: f''(x) < 0. The graph curves downward (like an upside-down bowl).
32A rancher wants to fence a rectangular area of 200 sq ft. What dimensions minimize the total fencing used?
A) 10 ft × 20 ft
B) 14.14 ft × 14.14 ft (i.e., √200 × √200)
C) 5 ft × 40 ft
D) 25 ft × 8 ft
Correct Answer: B
Minimize P = 2x + 2y subject to xy = 200. Substituting y = 200/x: P = 2x + 400/x. P' = 2 − 400/x² = 0 → x² = 200 → x = √200 ≈ 14.14. Square shape minimizes perimeter for a given area.
33What is lim(x→0) (1 − cos x)/x²?
Correct Answer: C
0/0 form. Apply L'Hôpital's twice: (1−cos x)/x² → sin x/2x → cos x/2. As x→0: cos(0)/2 = 1/2.
34Which test definitively shows Σ 1/n² converges?
A) Divergence Test
B) p-Series Test (p = 2 > 1)
C) Geometric Series (r = 1/n)
D) Integral Test only
Correct Answer: B
Σ1/nᵖ is a p-series; it converges when p > 1. Here p = 2 > 1, so the series converges. The integral test also works, but the p-series test is the most direct.
35Find d/dx[arctan(x)].
A) 1/√(1−x²)
B) −1/(1+x²)
C) 1/(1+x²)
D) sec²(x)
Correct Answer: C
The derivative of arctan(x) is 1/(1+x²). This is a standard formula to memorize. d/dx[arcsin x] = 1/√(1−x²); d/dx[arctan x] = 1/(1+x²).
36Solve the differential equation dy/dx = 2y with y(0) = 3.
A) y = 3e²ˣ
B) y = 2e³ˣ
C) y = e²ˣ + 3
D) y = 3 + 2x
Correct Answer: A
Separate: dy/y = 2dx → ln|y| = 2x + C → y = Ce²ˣ. Apply y(0) = 3: C = 3. So y = 3e²ˣ.
37What is the linearization of f(x) = √x at x = 9?
A) L(x) = 3 + (x − 9)/6
B) L(x) = 3 + (x − 9)/3
C) L(x) = 9 + (x − 9)/6
D) L(x) = 3 + x/6
Correct Answer: A
f(9) = 3; f'(x) = 1/(2√x); f'(9) = 1/6. L(x) = f(9) + f'(9)(x−9) = 3 + (x−9)/6.
38Compute ∫[0 to 1] x/(x² + 1) dx.
A) ln(2)/2
B) ln(2)
C) π/4
D) 1/2
Correct Answer: A
Let u = x²+1, du = 2x dx. Integral becomes ½∫[1 to 2] 1/u du = ½[ln u] from 1 to 2 = ½(ln 2 − 0) = (ln 2)/2.
39The acceleration of a particle is a(t) = 6t. If v(0) = 2, what is v(3)?
Correct Answer: C
v(t) = ∫a(t) dt = ∫6t dt = 3t² + C. v(0) = 2: C = 2. v(t) = 3t² + 2. v(3) = 3(9) + 2 = 27 + 2 = 29. Wait, that gives 29. Actually v(3) = 27 + 2 = 29. The correct answer is B) 29.
40Which of the following is the correct statement of the Intermediate Value Theorem?
A) If f'(c) = 0, then c is a local extremum
B) If f is continuous on [a,b] and N is between f(a) and f(b), then f(c) = N for some c ∈ (a,b)
C) If f is differentiable on (a,b), then f'(c) = [f(b)−f(a)]/(b−a)
D) If f is increasing, then f' > 0 everywhere
Correct Answer: B
This is the IVT: a continuous function on a closed interval takes on every value between its endpoint values. Option C describes the Mean Value Theorem.
41Evaluate: ∫ sec²(x) dx
A) −csc²(x) + C
B) sec(x) tan(x) + C
C) tan(x) + C
D) 2 sec(x) + C
Correct Answer: C
d/dx[tan x] = sec²x, so ∫sec²x dx = tan x + C. This is a standard antiderivative to memorize.
42What is d²y/dx² for y = x⁴?
Correct Answer: B
First derivative: dy/dx = 4x³. Second derivative: d²y/dx² = 12x².
43Find the absolute maximum of f(x) = −x² + 4x on [0, 5].
Correct Answer: A
f'(x) = −2x + 4 = 0 → x = 2. Evaluate: f(0) = 0, f(2) = 4, f(5) = −25 + 20 = −5. Maximum value is 4 at x = 2.
44The Ratio Test for convergence of Σaₙ says if lim|aₙ₊₁/aₙ| = L, then the series:
A) Converges if L > 1, diverges if L < 1
B) Converges if L < 1, diverges if L > 1
C) Always converges if L = 1
D) Diverges for all L
Correct Answer: B
Ratio Test: if L < 1 → absolute convergence; L > 1 (or ∞) → divergence; L = 1 → inconclusive (need another test). Used most often for factorial and exponential series.
45If a function's derivative is positive on (2, 5) and negative on (5, 8), then at x = 5:
A) f has a local minimum
B) f has an inflection point
C) f has a local maximum
D) f is discontinuous
Correct Answer: C
By the First Derivative Test: if f' changes from positive to negative at x = 5, then f has a local maximum at x = 5 (function increases then decreases).
46∫ 1/x dx = ?
A) x⁰ + C
B) −1/x² + C
C) ln|x| + C
D) ln(x²) + C
Correct Answer: C
The power rule ∫xⁿdx = xⁿ⁺¹/(n+1) fails at n = −1. Instead, ∫(1/x)dx = ln|x| + C. The absolute value is needed because ln is only defined for positive x, but 1/x is defined for all x ≠ 0.
47Find d/dx[(x² + 1)⁵⁰].
A) 50(x² + 1)⁴⁹
B) 100x(x² + 1)⁴⁹
C) 50x(x² + 1)⁴⁹
D) 50(2x)⁴⁹
Correct Answer: B
Chain rule: outer = u⁵⁰, inner = x² + 1. d/dx = 50(x² + 1)⁴⁹ · 2x = 100x(x² + 1)⁴⁹.
48What does the shell method formula V = 2π∫[a to b] x·f(x) dx compute?
A) Area under the curve from a to b
B) Volume of revolution around the x-axis
C) Volume of revolution around the y-axis
D) Surface area of revolution
Correct Answer: C
The shell method V = 2π∫x·f(x)dx computes the volume of the solid formed when the region under y = f(x) is revolved around the y-axis. Each shell has radius x, height f(x), and thickness dx.
49What is the first-degree Taylor polynomial (linearization) of f(x) = sin x at x = 0?
A) f(x) ≈ 1
B) f(x) ≈ x − x³/6
C) f(x) ≈ x
D) f(x) ≈ 1 + x
Correct Answer: C
f(0) = 0, f'(x) = cos x, f'(0) = 1. First-degree (linearization): L(x) = 0 + 1·(x−0) = x. That's why sin x ≈ x for small x — the famous small-angle approximation.
50Evaluate: d/dx[∫[0 to x] √(t² + 1) dt]
A) √(x⁴ + 1)
B) √(x² + 1)
C) ∫[0 to x] 2t/(√(t²+1)) dt
D) x√(x² + 1)
Correct Answer: B
By the Fundamental Theorem of Calculus, Part 1: d/dx[∫[0 to x] f(t) dt] = f(x). Simply replace t with x in the integrand: √(x² + 1).
51Evaluate: lim(x→4) (√x − 2)/(x − 4)
A) 0
B) 1/4
C) 1/2
D) Undefined
Correct Answer: B
Direct substitution gives 0/0. Rationalize the numerator: multiply by (√x + 2)/(√x + 2) to get (x − 4)/[(x − 4)(√x + 2)]. Cancel (x − 4): 1/(√x + 2). As x → 4: 1/(2 + 2) = 1/4.
52Evaluate: lim(x→0) (1/x − 1/sin x)
Correct Answer: A
Combine fractions: (sin x − x)/(x sin x). Near x = 0, sin x ≈ x − x³/6, so sin x − x ≈ −x³/6 and x sin x ≈ x². Thus the expression ≈ (−x³/6)/x² = −x/6 → 0. The limit is 0.
53For f(x) = { x² + 1, x < 2; ax − 1, x ≥ 2 } to be continuous at x = 2, what must a equal?
Correct Answer: D
Left-hand limit: lim(x→2⁻) = 2² + 1 = 5. Right-hand limit: lim(x→2⁺) = 2a − 1. For continuity: 2a − 1 = 5 → 2a = 6 → a = 3. With a = 3, f(2) = 3(2)−1 = 5 ✓. Actually a = 3 (not 5/2). The answer is A.
54The Intermediate Value Theorem guarantees a root of f(x) = x³ − 2x − 5 on which interval?
A) [0, 1]
B) [1, 2]
C) [2, 3]
D) [3, 4]
Correct Answer: C
f(2) = 8 − 4 − 5 = −1 < 0 and f(3) = 27 − 6 − 5 = 16 > 0. Since f is continuous and changes sign on [2, 3], by the IVT there exists c ∈ (2, 3) with f(c) = 0. This is the interval containing the actual root x ≈ 2.094.
55Find dy/dx by implicit differentiation: x² + y² = 25
A) −x/y
B) x/y
C) −y/x
D) 2x + 2y
Correct Answer: A
Differentiate both sides with respect to x: 2x + 2y(dy/dx) = 0. Solve for dy/dx: 2y(dy/dx) = −2x → dy/dx = −x/y. This gives the slope of the circle x² + y² = 25 at any point (x, y) on it.
56A sphere's radius is increasing at 2 cm/s. How fast is the volume increasing when r = 3 cm?
A) 36π cm³/s
B) 72π cm³/s
C) 24π cm³/s
D) 54π cm³/s
Correct Answer: B
V = (4/3)πr³ → dV/dt = 4πr²·(dr/dt). At r = 3, dr/dt = 2: dV/dt = 4π(9)(2) = 72π cm³/s. The chain rule connects the rate of change of volume to the rate of change of radius.
57A 10-foot ladder leans against a wall. Its base slides away at 1 ft/s. How fast is the top sliding down when the base is 6 ft from the wall?
A) −3/4 ft/s
B) −4/3 ft/s
C) 3/4 ft/s
D) −1 ft/s
Correct Answer: A
x² + y² = 100. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0. When x=6: y=√(100−36)=8. With dx/dt=1: 2(6)(1) + 2(8)(dy/dt) = 0 → 12 + 16(dy/dt) = 0 → dy/dt = −12/16 = −3/4 ft/s. Negative means the top is sliding down.
58Apply the Mean Value Theorem to f(x) = x² on [1, 3]. Find the value c guaranteed by the MVT.
A) c = 2
B) c = √3
C) c = 1.5
D) c = √5
Correct Answer: A
MVT: f'(c) = [f(3)−f(1)]/(3−1) = (9−1)/2 = 4. f'(x) = 2x, so 2c = 4 → c = 2. Check: c = 2 ∈ (1, 3) ✓. The MVT guarantees at least one point where the instantaneous rate equals the average rate over the interval.
59Use Newton's method with x₀ = 2 to find x₁ for f(x) = x² − 5.
A) x₁ = 9/4
B) x₁ = 7/4
C) x₁ = 2.25
D) x₁ = 5/2
Correct Answer: A
Newton's method: xₙ₊₁ = xₙ − f(xₙ)/f'(xₙ). f(2) = 4−5 = −1; f'(x) = 2x → f'(2) = 4. x₁ = 2 − (−1)/4 = 2 + 1/4 = 9/4. This approximates √5; the exact value is ≈ 2.236. Note: options A and C are equivalent (9/4 = 2.25).
60For f(x) = 2x³ − 9x² + 12x − 4, use the First Derivative Test to classify x = 1 and x = 2.
A) x = 1 is a local min, x = 2 is a local max
B) x = 1 is a local max, x = 2 is a local min
C) Both are inflection points
D) x = 1 is local max, x = 2 is neither
Correct Answer: B
f'(x) = 6x² − 18x + 12 = 6(x²−3x+2) = 6(x−1)(x−2). Sign chart: x<1: (+)(−)(−)=+>0; 12: (+)(+)(+)=>0. f' goes +→−at x=1 (local max) and −→+at x=2 (local min).
61A cylindrical can must hold 500 cm³. To minimize the total surface area, what ratio of height to radius should be used?
A) h/r = 1
B) h/r = 2
C) h/r = π
D) h/r = 4
Correct Answer: B
SA = 2πr² + 2πrh; volume constraint: V = πr²h → h = V/(πr²). Substituting and minimizing: dSA/dr = 0 gives 4πr = 2πV/πr² → 4r = 2h/r → h = 2r → h/r = 2. The optimal can has height equal to its diameter.
62Evaluate: ∫ tan x dx
A) sec²x + C
B) ln|cos x| + C
C) ln|sec x| + C
D) −ln|cos x| + C
Correct Answer: C
∫ tan x dx = ∫ sin x/cos x dx. Let u = cos x, du = −sin x dx: −∫(1/u)du = −ln|u| + C = −ln|cos x| + C = ln|sec x| + C (since −ln|cos x| = ln|1/cos x| = ln|sec x|). Both C and D represent the same function.
63Evaluate: ∫ x·eˣ dx using integration by parts.
A) xeˣ + eˣ + C
B) xeˣ − eˣ + C
C) x²eˣ/2 + C
D) eˣ(x−1) + C
Correct Answer: B
Let u = x, dv = eˣ dx → du = dx, v = eˣ. IBP formula: ∫u dv = uv − ∫v du = xeˣ − ∫eˣ dx = xeˣ − eˣ + C = eˣ(x−1) + C. Options B and D are equivalent. Verify by differentiating: d/dx[xeˣ − eˣ] = eˣ + xeˣ − eˣ = xeˣ ✓.
64Evaluate: ∫ sin²x dx
A) −cos²x/2 + C
B) x/2 − sin(2x)/4 + C
C) sin(2x)/4 + C
D) −cos(2x)/2 + C
Correct Answer: B
Use half-angle identity: sin²x = (1 − cos 2x)/2. Then ∫sin²x dx = ∫(1 − cos 2x)/2 dx = x/2 − sin(2x)/4 + C. This technique applies whenever even powers of sin or cos appear.
65To evaluate ∫√(9 − x²) dx, what trigonometric substitution is used?
A) x = 3tan θ
B) x = 3sin θ
C) x = 3sec θ
D) x = 9sin θ
Correct Answer: B
For integrands of the form √(a² − x²), use x = a sin θ. Here a = 3: x = 3sin θ, dx = 3cos θ dθ, and √(9−x²) = √(9−9sin²θ) = 3cos θ. The three trig substitution patterns: √(a²−x²) → x=asinθ; √(a²+x²) → x=atanθ; √(x²−a²) → x=asecθ.
66Set up (but do not evaluate) the partial fraction decomposition for: (3x + 5)/[(x−1)(x+2)²]
A) A/(x−1) + B/(x+2)²
B) A/(x−1) + B/(x+2) + C/(x+2)²
C) (Ax+B)/(x−1) + C/(x+2)²
D) A/(x−1)(x+2)²
Correct Answer: B
When a linear factor is repeated, include a term for each power up to the multiplicity. For (x+2)²: include B/(x+2) and C/(x+2)². For the non-repeated linear factor (x−1): include A/(x−1). Correct setup: A/(x−1) + B/(x+2) + C/(x+2)².
67Does ∫[1 to ∞] 1/x² dx converge or diverge?
A) Diverges to ∞
B) Converges to 1
C) Converges to 2
D) Converges to 1/2
Correct Answer: B
∫[1 to ∞] x⁻² dx = lim(b→∞) [−1/x] from 1 to b = lim(b→∞) (−1/b + 1) = 0 + 1 = 1. The integral converges to 1. For ∫[1 to ∞] 1/xᵖ dx: converges when p > 1, diverges when p ≤ 1. Here p = 2 > 1 → converges.
68Find the arc length of f(x) = x^(3/2) from x = 0 to x = 4.
A) ∫[0 to 4] √(1 + (9x/4)) dx
B) ∫[0 to 4] √(1 + (3√x/2)²) dx
C) ∫[0 to 4] (1 + 9x/4) dx
D) 4² + (4^(3/2))²
Correct Answer: B
Arc length formula: L = ∫√(1 + [f'(x)]²) dx. f'(x) = (3/2)x^(1/2) = 3√x/2. [f'(x)]² = 9x/4. L = ∫[0 to 4]√(1 + (3√x/2)²) dx = ∫[0 to 4]√(1 + 9x/4) dx. Options A and B are equivalent; B shows the derivative explicitly.
69Find the volume generated by rotating y = x² from x = 0 to x = 2 about the x-axis using the disk method.
A) 32π/5
B) 16π/5
C) 8π
D) 4π
Correct Answer: A
V = π∫[0 to 2] [f(x)]² dx = π∫[0 to 2] x⁴ dx = π[x⁵/5] from 0 to 2 = π·32/5 = 32π/5. The disk method formula V = π∫[a to b] [f(x)]² dx rotates the region under the curve about the x-axis.
70Using the shell method, find the volume when the region bounded by y = x², y = 0, x = 2 is rotated about the y-axis.
A) 4π
B) 8π
C) 16π/3
D) 8π/3
Correct Answer: B
Shell method (rotation about y-axis): V = 2π∫[0 to 2] x·f(x) dx = 2π∫[0 to 2] x·x² dx = 2π∫[0 to 2] x³ dx = 2π[x⁴/4] from 0 to 2 = 2π·4 = 8π. The shell method formula is V = 2π∫(radius)(height) dx.
71Solve the separable differential equation: dy/dx = 2xy, with y(0) = 3.
A) y = 3e^(x²)
B) y = e^(x²) + 2
C) y = 3e^(2x)
D) y = 3x² + 3
Correct Answer: A
Separate: dy/y = 2x dx. Integrate: ln|y| = x² + C → y = Ae^(x²). Apply IC y(0) = 3: 3 = Ae⁰ = A. So y = 3e^(x²). This is a classic separable ODE; the technique is to isolate y-terms on one side and x-terms on the other before integrating.
72A slope field for dy/dx = x − y has a slope of 0 along which line?
A) y = 0
B) x = 0
C) y = x
D) y = −x
Correct Answer: C
Slope = 0 where dy/dx = x − y = 0 → y = x. Along the line y = x, all slope field segments are horizontal (slope zero). The solution curves approach the line y = x + 1 as the equilibrium line of this linear ODE.
73Apply Euler's method with h = 0.1 to dy/dx = y, y(0) = 1. Find y(0.1).
A) 1.1
B) 1.105
C) e^0.1
D) 1.01
Correct Answer: A
Euler's method: yₙ₊₁ = yₙ + h·f(xₙ, yₙ). At (x₀, y₀) = (0, 1): y₁ = 1 + 0.1·f(0,1) = 1 + 0.1·1 = 1.1. The exact solution is y = eˣ, so the exact value y(0.1) = e^0.1 ≈ 1.105. Euler's method gives the approximate value 1.1.
74Apply the Integral Test to determine if Σ(n=1 to ∞) 1/n² converges.
A) Diverges, since ∫[1 to ∞] 1/x² dx diverges
B) Converges, since ∫[1 to ∞] 1/x² dx = 1 (converges)
C) Cannot apply; 1/x² is not decreasing
D) Converges to exactly 1
Correct Answer: B
f(x) = 1/x² is positive, continuous, and decreasing for x ≥ 1. ∫[1 to ∞] 1/x² dx = 1 (converges). By the Integral Test, Σ 1/n² also converges. (The exact sum is π²/6, but the test only establishes convergence, not the sum.)
75For the alternating series Σ(n=1 to ∞) (−1)^(n+1)/n, what does the Alternating Series Test conclude?
A) Diverges
B) Converges; error ≤ 1/6 if summing first 5 terms
C) Converges; error ≤ 1/(n+1) when stopping at nth term
D) Converges absolutely
Correct Answer: C
The AST requires: (1) bₙ = 1/n > 0, (2) bₙ is decreasing, (3) lim bₙ = 0. All satisfied → converges. Error bound: |error| ≤ bₙ₊₁ (the first omitted term). This series converges to ln 2, but only conditionally (the series of absolute values, the harmonic series, diverges).
76Find the radius of convergence of Σ(n=0 to ∞) xⁿ/n! using the Ratio Test.
A) R = 0
B) R = 1
C) R = e
D) R = ∞
Correct Answer: D
Ratio test: |aₙ₊₁/aₙ| = |x^(n+1)/(n+1)!| · |n!/xⁿ| = |x|/(n+1) → 0 as n → ∞, for any finite x. Since the limit is 0 < 1 for all x, the series converges everywhere: R = ∞. This series is the Maclaurin series for eˣ.
77If f(x) = Σ(n=0 to ∞) cₙxⁿ, what is f'(x)?
A) Σ(n=0 to ∞) n·cₙxⁿ
B) Σ(n=1 to ∞) n·cₙx^(n−1)
C) Σ(n=0 to ∞) cₙx^(n+1)/(n+1)
D) Σ(n=0 to ∞) cₙ·eˣ
Correct Answer: B
Power series can be differentiated term by term within the radius of convergence: f'(x) = Σ(n=1 to ∞) n·cₙx^(n−1). Note the index starts at n=1 (the n=0 term, c₀, is a constant whose derivative is 0). This is the same rule as differentiating a polynomial, extended to infinite series.
78The Maclaurin series for eˣ is Σ xⁿ/n!. What is the Maclaurin series for e^(−x²)?
A) Σ (−1)ⁿx^(2n)/n!
B) Σ x^(2n)/n!
C) Σ (−x²)ⁿ/n!
D) −Σ xⁿ/n!
Correct Answer: A
Substitute (−x²) in place of x in the series for eˣ: e^(−x²) = Σ(n=0 to ∞) (−x²)ⁿ/n! = Σ (−1)ⁿ(x²)ⁿ/n! = Σ (−1)ⁿx^(2n)/n!. Options A and C are equivalent; A displays the coefficients explicitly. This series has R = ∞.
79Find the washer method volume when the region between y = √x and y = x is rotated about the x-axis.
A) π/6
B) π/3
C) π/2
D) 2π/3
Correct Answer: B
Intersection: √x = x → x = 0, 1. Outer radius R = √x, inner radius r = x (on [0,1], √x ≥ x). V = π∫[0 to 1] [(√x)² − x²] dx = π∫[0 to 1] (x − x²) dx = π[x²/2 − x³/3] from 0 to 1 = π(1/2 − 1/3) = π/6. Wait: that gives π/6, so answer is A.
80Newton's Law of Cooling states dT/dt = k(T − Tₐ). If a body at 90°C cools in a 20°C room, which expression gives T(t)?
A) T(t) = 70e^(kt) + 20
B) T(t) = 90e^(kt)
C) T(t) = 20 + 70e^(−kt) where k > 0
D) T(t) = 70 − kt + 20
Correct Answer: C
Solving dT/dt = k(T−20) with T(0) = 90: T(t) = 20 + Ce^(kt). At t=0: 90 = 20 + C → C = 70. Since the body cools, T decreases, meaning k < 0. Writing as T(t) = 20 + 70e^(−kt) with k > 0 shows the temperature decaying toward the ambient 20°C.
81Evaluate: ∫ x²eˣ dx (using integration by parts twice)
A) x²eˣ − 2xeˣ + 2eˣ + C
B) x²eˣ + 2xeˣ + C
C) (x²−2x+2)eˣ + C
D) Both A and C
Correct Answer: D
IBP (LIATE: u = x², dv = eˣ dx): x²eˣ − ∫2xeˣ dx. Apply IBP again to ∫2xeˣ dx (u=2x, dv=eˣ): 2xeˣ − 2eˣ. Result: x²eˣ − 2xeˣ + 2eˣ + C = (x²−2x+2)eˣ + C. Options A and C are the same expression factored differently.
82For f(x) = x⁴ − 8x², find the inflection points.
A) x = 0 only
B) x = ±2 only
C) x = 0, ±2
D) x = ±2√3/3
Correct Answer: B
f'(x) = 4x³ − 16x; f''(x) = 12x² − 16. Set f''(x) = 0: 12x² = 16 → x² = 4/3 → x = ±2/√3 = ±2√3/3. Check sign changes: f'' changes sign at these points → inflection points. At x=0: f''(0) = −16 ≠ 0, so x=0 is not an inflection point (it's a local max). The inflection points are at x = ±2√3/3 ≈ ±1.155.
83Find dy/dx by implicit differentiation: sin(xy) = x
A) dy/dx = (1 − y·cos(xy))/(x·cos(xy))
B) dy/dx = cos(xy)
C) dy/dx = 1/(x·cos(xy))
D) dy/dx = y/x
Correct Answer: A
Differentiate both sides: cos(xy)·(y + x·dy/dx) = 1 (chain rule and product rule on xy). Expand: y·cos(xy) + x·cos(xy)·dy/dx = 1. Solve: dy/dx = (1 − y·cos(xy))/(x·cos(xy)).
84Evaluate: ∫ sec x dx
A) tan x + C
B) ln|sec x + tan x| + C
C) sec x tan x + C
D) ln|sec x| + C
Correct Answer: B
The classic result: ∫ sec x dx = ln|sec x + tan x| + C. Derived by multiplying sec x by (sec x + tan x)/(sec x + tan x), giving ∫(sec²x + sec x tan x)/(sec x + tan x) dx. Let u = sec x + tan x, du = (sec x tan x + sec²x) dx, giving ∫du/u = ln|u| + C.
85Use u-substitution to evaluate: ∫ sin x · cos²x dx
A) cos³x/3 + C
B) −cos³x/3 + C
C) sin³x/3 + C
D) −sin²x·cos x/2 + C
Correct Answer: B
Let u = cos x, du = −sin x dx → sin x dx = −du. ∫sin x·cos²x dx = ∫u²·(−du) = −u³/3 + C = −cos³x/3 + C. Verify: d/dx[−cos³x/3] = −3cos²x·(−sin x)/3 = cos²x·sin x ✓.
86Find the interval of convergence of Σ(n=1 to ∞) xⁿ/n. (Radius of convergence R = 1; check endpoints.)
A) (−1, 1)
B) [−1, 1)
C) [−1, 1]
D) (−1, 1]
Correct Answer: B
Ratio test gives R = 1. Check x = 1: Σ 1/n (harmonic series) → diverges. Check x = −1: Σ (−1)ⁿ/n (alternating harmonic series) → converges by AST. Interval of convergence: [−1, 1). This series represents −ln(1−x) for x ∈ [−1, 1).
87Find the area of the region bounded by y = x² and y = 4.
Correct Answer: B
Intersection: x² = 4 → x = ±2. Area = ∫[−2 to 2] (4 − x²) dx = [4x − x³/3] from −2 to 2 = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 32/3. By symmetry: 2∫[0 to 2](4−x²) dx = 2[8 − 8/3] = 32/3.
88Evaluate: lim(x→0) (1 − cos x)/x²
Correct Answer: C
Apply L'Hôpital's Rule (0/0 form): lim sin x/(2x) — still 0/0. Apply again: lim cos x/2 = 1/2. Alternatively, use the Maclaurin series: 1 − cos x ≈ x²/2 for small x, so (1−cosx)/x² ≈ (x²/2)/x² = 1/2.
89∫ sec²x dx = ?
A) 2 sec x tan x + C
B) tan x + C
C) sec x + C
D) sin x/cos²x + C
Correct Answer: B
∫ sec²x dx = tan x + C. This is a standard result from the derivative formula d/dx[tan x] = sec²x, reversed. It is one of the most frequently used integration results in calculus.
90One-sided limits: for f(x) = |x|/x, evaluate lim(x→0⁺) f(x) and lim(x→0⁻) f(x).
A) Both equal 0
B) Both equal 1
C) Left = −1, right = 1
D) Left = 1, right = −1
Correct Answer: C
For x > 0: |x|/x = x/x = 1, so lim(x→0⁺) = 1. For x < 0: |x|/x = −x/x = −1, so lim(x→0⁻) = −1. Since one-sided limits differ, the two-sided limit does not exist. f(x) = |x|/x = sgn(x), the sign function.
91A Norman window (rectangle topped by a semicircle) has a perimeter of 12 ft. To maximize the area, what width x should be chosen?
A) x = 12/(4 + π)
B) x = 12/(2 + π)
C) x = 3
D) x = 6/π
Correct Answer: A
Width = x; semicircle radius = x/2. Perimeter: x + 2h + πx/2 = 12 → h = (12 − x − πx/2)/2. Area = xh + π(x/2)²/2. Substitute h and maximize: dA/dx = 0 yields x = 12/(4 + π) ≈ 1.68 ft. This is a classic constrained optimization problem.
92Evaluate: ∫ 1/√(1 − x²) dx
A) arctan x + C
B) arcsin x + C
C) arcsec x + C
D) −arccos x + C
Correct Answer: B
∫ 1/√(1−x²) dx = arcsin x + C. This is a standard form. Note: −arccos x + C is equivalent (since arcsin x + arccos x = π/2, a constant). Three key inverse trig integrals: ∫1/√(1−x²) dx = arcsin x + C; ∫1/(1+x²) dx = arctan x + C; ∫1/(x√(x²−1)) dx = arcsec x + C.
93Compare: does Σ(n=1 to ∞) 1/(n² + n) converge or diverge?
A) Diverges by comparison with Σ 1/n
B) Converges by comparison with Σ 1/n²
C) Diverges by ratio test
D) Converges, sum = 1
Correct Answer: D
Partial fractions: 1/(n²+n) = 1/[n(n+1)] = 1/n − 1/(n+1). This is a telescoping series: Sₙ = 1 − 1/(n+1) → 1 as n → ∞. The series converges to 1. By comparison: 1/(n²+n) < 1/n², and Σ1/n² converges, confirming convergence.
94Find the Taylor series for f(x) = cos x centered at x = 0 (first four non-zero terms).
A) 1 − x²/2 + x⁴/24 − x⁶/720 + ⋯
B) x − x³/6 + x⁵/120 − ⋯
C) 1 + x²/2 + x⁴/24 + ⋯
D) 1 − x + x²/2 − x³/6 + ⋯
Correct Answer: A
cos x = Σ(n=0 to ∞) (−1)ⁿ x^(2n)/(2n)! = 1 − x²/2! + x⁴/4! − x⁶/6! + ⋯ = 1 − x²/2 + x⁴/24 − x⁶/720 + ⋯. Option B is the Maclaurin series for sin x. This series has infinite radius of convergence.
95Evaluate: ∫ x sin x dx
A) x cos x − sin x + C
B) −x cos x + sin x + C
C) x sin x + cos x + C
D) −x sin x + cos x + C
Correct Answer: B
IBP with u = x, dv = sin x dx → du = dx, v = −cos x. ∫x sin x dx = −x cos x − ∫(−cos x) dx = −x cos x + sin x + C. Verify: d/dx[−x cos x + sin x] = −cos x + x sin x + cos x = x sin x ✓.
96Find d²y/dx² for y = x³ − 6x at the point x = 2.
Correct Answer: A
y = x³ − 6x; y' = 3x² − 6; y'' = 6x. At x = 2: y'' = 6(2) = 12. Since y'' > 0 at x = 2, the function is concave up there. (The critical points are at x = ±√2; x=2 is not a critical point.)
97Evaluate: ∫[0 to 1] x/√(1 + x²) dx
A) √2 − 1
B) (√2 − 1)/2
C) 1
D) √2
Correct Answer: A
Let u = 1 + x², du = 2x dx → x dx = du/2. When x=0: u=1; x=1: u=2. ∫ (1/2)u^(−1/2) du from 1 to 2 = [u^(1/2)] from 1 to 2 = √2 − 1.
98What does it mean for a series to converge absolutely?
A) Σaₙ converges and all terms are positive
B) Σ|aₙ| converges
C) Σaₙ converges but Σ|aₙ| diverges
D) The partial sums are bounded
Correct Answer: B
A series Σaₙ converges absolutely if Σ|aₙ| converges. Absolute convergence implies convergence (but not vice versa). A series that converges but not absolutely is called conditionally convergent. Example: Σ(−1)ⁿ/n converges conditionally (Σ1/n diverges) while Σ(−1)ⁿ/n² converges absolutely (Σ1/n² converges).
99For f(x) = x·ln x, find the local minimum on (0, ∞).
A) (1, 0)
B) (e, e)
C) (1/e, −1/e)
D) (1, 1)
Correct Answer: C
f'(x) = ln x + x·(1/x) = ln x + 1. Set f'(x) = 0: ln x = −1 → x = e⁻¹ = 1/e. f''(x) = 1/x > 0 at x = 1/e → local minimum. f(1/e) = (1/e)·ln(1/e) = (1/e)·(−1) = −1/e. Minimum at (1/e, −1/e).
100Evaluate: ∫ e^(2x) sin x dx (integration by parts, cycling technique)
A) e^(2x)(2sin x − cos x)/5 + C
B) e^(2x)(sin x − 2cos x)/5 + C
C) e^(2x)(2sin x + cos x)/5 + C
D) e^(2x)(sin x + cos x)/2 + C
Correct Answer: A
Let I = ∫e^(2x)sin x dx. IBP: u=sin x, dv=e^(2x)dx → v=e^(2x)/2. I = (e^(2x)sin x)/2 − (1/2)∫e^(2x)cos x dx. Apply IBP again: ∫e^(2x)cos x dx = (e^(2x)cos x)/2 + (1/2)∫e^(2x)sin x dx = (e^(2x)cos x)/2 + I/2. So I = (e^(2x)sin x)/2 − (1/2)[(e^(2x)cos x)/2 + I/2] → I = e^(2x)sin x/2 − e^(2x)cos x/4 − I/4 → 5I/4 = e^(2x)(2sin x − cos x)/4 → I = e^(2x)(2sin x − cos x)/5 + C.
101Use the Squeeze Theorem to evaluate: lim(x→0) x²·sin(1/x)
A) 1
B) Does not exist
C) 0
D) ∞
Correct Answer: C
Since −1 ≤ sin(1/x) ≤ 1 for all x ≠ 0, we have −x² ≤ x²sin(1/x) ≤ x². Both −x² and x² approach 0 as x→0, so by the Squeeze Theorem, lim(x→0) x²sin(1/x) = 0. The key is that the factor x² "squeezes" the oscillating function sin(1/x) to zero.
102For f(x) = { x² + 1 if x < 2; ax + b if x ≥ 2 }, find values of a and b that make f continuous and differentiable at x = 2.
A) a = 4, b = −3
B) a = 2, b = 1
C) a = 4, b = 1
D) a = 2, b = −3
Correct Answer: A
For continuity: lim(x→2⁻) = 4 + 1 = 5 must equal f(2) = 2a + b → 2a + b = 5. For differentiability: left derivative = 2x|_(x=2) = 4; right derivative = a. So a = 4. Then 2(4) + b = 5 → b = −3. Verify: f(2⁺) = 4(2) − 3 = 5 = f(2⁻) ✓, and derivatives match ✓.
103Find dy/dx using implicit differentiation: x³ + y³ = 6xy
A) (6y − 3x²)/(3y² − 6x)
B) (2y − x²)/(y² − 2x)
C) (y − x²)/(y² − x)
D) (3x² − 6y)/(6x − 3y²)
Correct Answer: B
Differentiate both sides with respect to x: 3x² + 3y²(dy/dx) = 6y + 6x(dy/dx). Collect dy/dx terms: 3y²(dy/dx) − 6x(dy/dx) = 6y − 3x². Factor: dy/dx(3y² − 6x) = 6y − 3x² = 3(2y − x²). So dy/dx = 3(2y − x²) / [3(y² − 2x)] = (2y − x²)/(y² − 2x).
104A 10-foot ladder leans against a wall. The bottom slides away at 2 ft/sec. How fast is the top sliding down when the bottom is 6 feet from the wall?
A) −3/2 ft/sec
B) −8/3 ft/sec
C) −3/4 ft/sec
D) −2/3 ft/sec
Correct Answer: A
Let x = distance from wall, y = height. x² + y² = 100. When x = 6: y = √(100 − 36) = 8. Differentiate: 2x(dx/dt) + 2y(dy/dt) = 0 → x(dx/dt) + y(dy/dt) = 0. Substitute: 6(2) + 8(dy/dt) = 0 → 8(dy/dt) = −12 → dy/dt = −3/2 ft/sec. Negative means sliding down.
105Use logarithmic differentiation to find y' for y = (x²+1)³ · √(x−1) / (2x+3)⁴
A) y · [6x/(x²+1) + 1/(2(x−1)) − 8/(2x+3)]
B) y · [3/(x²+1) + 1/(x−1) − 4/(2x+3)]
C) y · [6x/(x²+1) − 1/(2(x−1)) + 8/(2x+3)]
D) y · [2x/(x²+1) + 1/(2(x−1)) − 4/(2x+3)]
Correct Answer: A
Take ln: ln y = 3ln(x²+1) + (1/2)ln(x−1) − 4ln(2x+3). Differentiate: y'/y = 3·(2x)/(x²+1) + (1/2)·1/(x−1) − 4·2/(2x+3) = 6x/(x²+1) + 1/(2(x−1)) − 8/(2x+3). Multiply both sides by y: y' = y·[6x/(x²+1) + 1/(2(x−1)) − 8/(2x+3)].
106Find the third derivative of f(x) = x⁴ − 3x³ + 5x − 2.
A) 24x − 18
B) 12x² − 18x + 5
C) 4x³ − 9x²
D) 24
Correct Answer: A
f(x) = x⁴ − 3x³ + 5x − 2. f'(x) = 4x³ − 9x² + 5. f''(x) = 12x² − 18x. f'''(x) = 24x − 18. Each differentiation reduces the degree by 1. The constant 5 vanishes at first derivative, and the remaining terms are differentiated normally.
107Find the linear approximation L(x) of f(x) = √x at a = 25, and use it to approximate √26.
A) L(x) = 5 + (1/10)(x − 25); √26 ≈ 5.1
B) L(x) = 5 + (1/5)(x − 25); √26 ≈ 5.2
C) L(x) = 5 + (1/25)(x − 25); √26 ≈ 5.04
D) L(x) = 5 + (x − 25); √26 ≈ 6
Correct Answer: A
f(x) = √x = x^(1/2), so f'(x) = 1/(2√x). At a = 25: f(25) = 5, f'(25) = 1/(2·5) = 1/10. Linear approximation: L(x) = f(a) + f'(a)(x−a) = 5 + (1/10)(x − 25). For x = 26: L(26) = 5 + (1/10)(1) = 5.1. (Actual value: √26 ≈ 5.0990.)
108Verify the Mean Value Theorem for f(x) = x³ − x on [0, 2]. Find the value(s) of c guaranteed by the theorem.
A) c = 2/√3
B) c = 1
C) c = √3
D) c = 4/3
Correct Answer: A
f is continuous on [0,2] and differentiable on (0,2). f(0) = 0, f(2) = 8 − 2 = 6. MVT slope = (6−0)/(2−0) = 3. f'(x) = 3x² − 1. Set 3c² − 1 = 3 → 3c² = 4 → c² = 4/3 → c = 2/√3 = 2√3/3 ≈ 1.155. Since 0 < 2/√3 < 2, the value exists in (0,2) as guaranteed.
109Find the absolute extrema of f(x) = x³ − 6x² + 9x + 1 on [0, 5].
A) Absolute max: f(5) = 21; Absolute min: f(1) = 5 (wait — check f(3))
B) Absolute max: f(5) = 21; Absolute min: f(3) = 1
C) Absolute max: f(5) = 21; Absolute min: f(0) = 1
D) Absolute max: f(1) = 5; Absolute min: f(3) = 1
Correct Answer: B
f'(x) = 3x² − 12x + 9 = 3(x² − 4x + 3) = 3(x−1)(x−3). Critical points: x = 1, 3. Evaluate at critical points and endpoints: f(0) = 1, f(1) = 1 − 6 + 9 + 1 = 5, f(3) = 27 − 54 + 27 + 1 = 1, f(5) = 125 − 150 + 45 + 1 = 21. Absolute maximum = 21 at x = 5; Absolute minimum = 1 at x = 3 (and x = 0).
110Determine the intervals of concavity and inflection points of f(x) = x⁴ − 4x³.
A) Concave up on (−∞,0) and (2,∞); concave down on (0,2); inflection points at x=0,2
B) Concave up on (2,∞) only; concave down on (−∞,2); inflection point at x=2
C) Concave up everywhere; no inflection points
D) Concave up on (−∞,0) only; inflection point at x=0
Correct Answer: A
f'(x) = 4x³ − 12x². f''(x) = 12x² − 24x = 12x(x − 2). f''(x) = 0 at x = 0 and x = 2. Sign analysis: x < 0: f''(x) = 12(neg)(neg) > 0 (concave up); 0 < x < 2: f''(x) = 12(pos)(neg) < 0 (concave down); x > 2: f''(x) = 12(pos)(pos) > 0 (concave up). Inflection points at x = 0 and x = 2 where concavity changes.
111A farmer has 600 feet of fence and wants to enclose a rectangular area, then divide it in half with a fence parallel to one side. What dimensions maximize the enclosed area?
A) 100 ft × 150 ft
B) 150 ft × 100 ft (same thing — wrong answer)
C) 100 ft × 200 ft
D) 75 ft × 150 ft
Correct Answer: A
Let x = width, y = length. The dividing fence is parallel to the width, so total fence: 3x + 2y = 600 → y = (600 − 3x)/2. Area: A = xy = x(600 − 3x)/2 = 300x − (3/2)x². A'(x) = 300 − 3x = 0 → x = 100. y = (600 − 300)/2 = 150. Dimensions: 100 ft × 150 ft. A''(x) = −3 < 0 confirms maximum.
112Evaluate: ∫₀¹ x·e^(x²) dx using u-substitution.
A) (e − 1)/2
B) e/2
C) (e + 1)/2
D) e − 1
Correct Answer: A
Let u = x², then du = 2x dx → x dx = du/2. Change limits: x=0 → u=0; x=1 → u=1. ∫₀¹ x·e^(x²) dx = ∫₀¹ eᵘ·(du/2) = (1/2)[eᵘ]₀¹ = (1/2)(e¹ − e⁰) = (1/2)(e − 1) = (e−1)/2.
113Evaluate: ∫ x²·ln x dx using integration by parts.
A) (x³/3)·ln x − x³/9 + C
B) (x³/3)·ln x + x³/9 + C
C) x³·ln x − x³/3 + C
D) (x²/2)·ln x − x²/4 + C
Correct Answer: A
LIATE: u = ln x (Logarithm), dv = x² dx → du = 1/x dx, v = x³/3. IBP: ∫u dv = uv − ∫v du = (x³/3)·ln x − ∫(x³/3)·(1/x) dx = (x³/3)·ln x − (1/3)∫x² dx = (x³/3)·ln x − (1/3)·(x³/3) + C = (x³/3)·ln x − x³/9 + C.
114Evaluate: ∫ sin³x dx
A) −cos x + cos³x/3 + C
B) cos x − cos³x/3 + C
C) −sin²x·cos x/3 + C
D) −cos³x/3 + C
Correct Answer: A
Write sin³x = sin²x · sin x = (1 − cos²x)·sin x. Let u = cos x, du = −sin x dx. ∫(1 − cos²x)·sin x dx = −∫(1 − u²) du = −(u − u³/3) + C = −cos x + cos³x/3 + C. This technique works whenever sin or cos appears to an odd power: peel off one factor and use the Pythagorean identity.
115Use partial fractions to evaluate: ∫ (3x + 1)/[(x − 1)(x + 2)] dx
A) (4/3)ln|x−1| + (5/3)ln|x+2| + C
B) ln|x−1| + 2ln|x+2| + C
C) (4/3)ln|x−1| − (5/3)ln|x+2| + C
D) 2ln|x−1| + ln|x+2| + C
Correct Answer: A
Write (3x+1)/[(x−1)(x+2)] = A/(x−1) + B/(x+2). Multiply both sides by (x−1)(x+2): 3x+1 = A(x+2) + B(x−1). Set x=1: 4 = 3A → A = 4/3. Set x=−2: −5 = −3B → B = 5/3. So ∫[A/(x−1) + B/(x+2)] dx = (4/3)ln|x−1| + (5/3)ln|x+2| + C.
116Find the area between the curves y = x² and y = 4x − x² on [0, 2].
Correct Answer: B
Find which is on top: 4x − x² − x² = 4x − 2x² = 2x(2−x) ≥ 0 on [0,2], so y = 4x − x² is on top. Area = ∫₀² [(4x − x²) − x²] dx = ∫₀² (4x − 2x²) dx = [2x² − 2x³/3]₀² = [8 − 16/3] − 0 = 24/3 − 16/3 = 8/3.
117Find the volume of the solid formed by revolving y = √x from x = 0 to x = 4 around the x-axis using the disk method.
Correct Answer: B
Disk method: V = π∫₀⁴ [f(x)]² dx = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx = π[x²/2]₀⁴ = π·(16/2 − 0) = 8π. Each cross-section is a disk with radius r = √x and area πr² = πx. Summing all disks from x=0 to x=4 gives volume 8π.
118Find the volume when the region bounded by y = x², y = 0, and x = 2 is revolved around the y-axis using the shell method.
A) 4π
B) 8π
C) 16π/5
D) 32π/5
Correct Answer: B
Shell method (revolving around y-axis): V = 2π∫ x·f(x) dx = 2π∫₀² x·x² dx = 2π∫₀² x³ dx = 2π[x⁴/4]₀² = 2π·(16/4) = 2π·4 = 8π. Each cylindrical shell has radius x, height x², and thickness dx. The factor 2πx is the circumference of each shell.
119Find the volume of the solid generated by revolving the region between y = x² and y = 4 around y = 4 using the washer method.
A) 256π/15
B) 128π/5
C) 64π/3
D) 512π/15
Correct Answer: D
Intersection: x² = 4 → x = ±2. Axis y=4. Outer radius R = 4 − x² (distance from y=4 to y=x²... wait: the region is between y=x² and y=4, rotating around y=4). R = 4 − x² (outer, from y=4 to y=x²... wait: washer has R = 0 (no hole since curve y=4 is the axis), r = 4 − x². V = π∫₋₂² (4 − x²)² dx = 2π∫₀² (16 − 8x² + x⁴) dx = 2π[16x − 8x³/3 + x⁵/5]₀² = 2π[32 − 64/3 + 32/5] = 2π·256/15 = 512π/15.
120Find the average value of f(x) = sin x on [0, π].
Correct Answer: B
Average value = (1/(b−a))·∫ₐᵇ f(x) dx = (1/(π−0))·∫₀π sin x dx = (1/π)·[−cos x]₀π = (1/π)·(−cos π + cos 0) = (1/π)·(1 + 1) = 2/π ≈ 0.637. This means a horizontal line at height 2/π has the same area under it on [0,π] as the curve sin x.
121Set up (do not evaluate) the arc length of y = x² from x = 0 to x = 2.
A) ∫₀² √(1 + 4x²) dx
B) ∫₀² √(1 + 2x) dx
C) ∫₀² (1 + 4x²) dx
D) ∫₀² √(1 + x⁴) dx
Correct Answer: A
Arc length formula: L = ∫ₐᵇ √(1 + [f'(x)]²) dx. For y = x²: dy/dx = 2x, so [f'(x)]² = 4x². Therefore L = ∫₀² √(1 + 4x²) dx. This integral requires trigonometric substitution (x = (1/2)tan θ) to evaluate in closed form, but the setup only requires the formula with the correct derivative.
122Solve the separable differential equation: dy/dx = y·cos x, with initial condition y(0) = 3.
A) y = 3·e^(sin x)
B) y = 3·sin x + 3
C) y = e^(sin x) + 2
D) y = 3·cos x
Correct Answer: A
Separate: dy/y = cos x dx. Integrate both sides: ln|y| = sin x + C. Exponentiate: |y| = e^(sin x + C) = Ae^(sin x). Apply IC y(0) = 3: 3 = A·e^(sin 0) = A·e⁰ = A. So y = 3·e^(sin x). Verify: dy/dx = 3·e^(sin x)·cos x = y·cos x ✓.
123A population grows at a rate proportional to its size. If the initial population is 500 and doubles in 10 years, what is the population after 25 years?
A) 500·2^(2.5) ≈ 2828
B) 500·e^(25) ≈ enormous
C) 1000·25/10 = 2500
D) 500·2^(2) = 2000
Correct Answer: A
P(t) = P₀·e^(kt) = 500·e^(kt). Doubling condition: P(10) = 1000 → 500e^(10k) = 1000 → e^(10k) = 2 → k = ln2/10. P(25) = 500·e^(25·ln2/10) = 500·e^((5/2)ln2) = 500·2^(5/2) = 500·4√2 = 500·2^(2.5) ≈ 500·5.657 ≈ 2828.
124A logistic growth model has P' = 0.3P(1 − P/100). At what population size is the growth rate maximized?
A) P = 100
B) P = 50
C) P = 30
D) P = 75
Correct Answer: B
For logistic growth P' = rP(1 − P/K), the growth rate P' is a downward-opening parabola in P. It's maximized when P = K/2. Here K = 100 (carrying capacity), so maximum growth rate occurs at P = 50. At P = 0 or P = 100, growth rate = 0. The maximum growth rate is 0.3·50·(1 − 50/100) = 0.3·50·0.5 = 7.5 individuals per time unit.
125Determine whether the series Σ(n=1 to ∞) 1/n^(3/2) converges or diverges.
A) Diverges by p-series test (p = 3/2 > 1)
B) Converges by p-series test (p = 3/2 > 1)
C) Diverges by comparison with harmonic series
D) Converges only conditionally
Correct Answer: B
The p-series Σ1/nᵖ converges if and only if p > 1. Here p = 3/2 > 1, so the series converges. This is an absolute (unconditional) convergence — all terms are positive. Compare: Σ1/n (p=1) is the harmonic series which diverges; Σ1/n² (p=2) converges to π²/6. The p-series test is one of the most fundamental convergence tests.
126Apply the ratio test to Σ(n=0 to ∞) n!/3ⁿ. What does the test conclude?
A) Converges (L = 1/3 < 1)
B) Diverges (L = ∞ > 1)
C) Inconclusive (L = 1)
D) Converges (L = 0 < 1)
Correct Answer: B
aₙ = n!/3ⁿ. aₙ₊₁/aₙ = [(n+1)!/3^(n+1)] / [n!/3ⁿ] = (n+1)!/n! · 3ⁿ/3^(n+1) = (n+1)/3. As n→∞, (n+1)/3 → ∞. Since L = ∞ > 1, the series diverges by the ratio test. Factorials grow faster than exponentials — n! ≫ 3ⁿ for large n, so terms aₙ → ∞.
127Apply the ratio test to Σ(n=0 to ∞) xⁿ/n! for any fixed x. Find the radius of convergence.
A) R = 1
B) R = e
C) R = 0
D) R = ∞
Correct Answer: D
aₙ = xⁿ/n!. |aₙ₊₁/aₙ| = |x^(n+1)/(n+1)!| · |n!/xⁿ| = |x|/(n+1) → 0 as n→∞ for any fixed x. Since L = 0 < 1 for all x, the series converges for all real (and complex) x. R = ∞. This series is the Maclaurin series for eˣ: Σxⁿ/n! = eˣ, confirming it converges everywhere.
128Find the Maclaurin series for f(x) = cos x and use it to find the first three nonzero terms.
A) 1 − x²/2! + x⁴/4! − ...
B) x − x³/3! + x⁵/5! − ...
C) 1 + x + x²/2! + x³/3! + ...
D) 1 − x + x²/2 − x³/6 + ...
Correct Answer: A
For cos x: f(0)=1, f'(0)=−sin(0)=0, f''(0)=−cos(0)=−1, f'''(0)=sin(0)=0, f⁽⁴⁾(0)=cos(0)=1. Maclaurin: f(x) = Σ f⁽ⁿ⁾(0)xⁿ/n! = 1 − x²/2! + x⁴/4! − x⁶/6! + ... = Σ(n=0 to ∞) (−1)ⁿx^(2n)/(2n)!. The three nonzero terms are 1, −x²/2, +x⁴/24. Note: Option B is the Maclaurin series for sin x.
129Find the 3rd-degree Taylor polynomial for f(x) = ln x centered at a = 1.
A) (x−1) − (x−1)²/2 + (x−1)³/3
B) (x−1) − (x−1)²/4 + (x−1)³/9
C) ln(1) + (x−1) + (x−1)²/2 + (x−1)³/6
D) (x−1) + (x−1)²/2 + (x−1)³/3
Correct Answer: A
f(x)=ln x: f(1)=0. f'(x)=1/x: f'(1)=1. f''(x)=−1/x²: f''(1)=−1. f'''(x)=2/x³: f'''(1)=2. Taylor polynomial T₃(x) = f(1) + f'(1)(x−1) + f''(1)(x−1)²/2! + f'''(1)(x−1)³/3! = 0 + (x−1) + (−1)(x−1)²/2 + 2(x−1)³/6 = (x−1) − (x−1)²/2 + (x−1)³/3.
130Use the Maclaurin series for 1/(1−x) to find the series for 1/(1+x²), and identify the radius of convergence.
A) Σ(−1)ⁿx^(2n), |x| < 1, R = 1
B) Σxⁿ, |x| < 1, R = 1
C) Σ(−1)ⁿx^n, |x| < 1, R = 1
D) Σ x^(2n), |x| < 1, R = 1
Correct Answer: A
Known: 1/(1−u) = Σuⁿ for |u| < 1. Substitute u = −x²: 1/(1−(−x²)) = 1/(1+x²) = Σ(−x²)ⁿ = Σ(−1)ⁿx^(2n) = 1 − x² + x⁴ − x⁶ + ... Convergence: |−x²| < 1 → x² < 1 → |x| < 1. Note: Integrating this gives arctan x = Σ(−1)ⁿx^(2n+1)/(2n+1), the Leibniz formula.
131Evaluate lim(x→0) (sin x − x + x³/6) / x⁵ using Maclaurin series.
Correct Answer: B
sin x = x − x³/3! + x⁵/5! − ... = x − x³/6 + x⁵/120 − ... So sin x − x + x³/6 = x⁵/120 − x⁷/5040 + ... Dividing by x⁵: (sin x − x + x³/6)/x⁵ = 1/120 − x²/5040 + ... As x→0, limit = 1/120 = 1/5!. L'Hôpital's rule would require applying it 5 times; Maclaurin series is far more efficient.
132Find dy/dx for the parametric equations x = t² − 1, y = t³ − 3t.
A) (3t² − 3)/(2t)
B) (3t)/(2)
C) 2t/(3t² − 3)
D) (t³ − 3t)/(t² − 1)
Correct Answer: A
dy/dx = (dy/dt)/(dx/dt). dx/dt = 2t; dy/dt = 3t² − 3. So dy/dx = (3t² − 3)/(2t). Note the curve has horizontal tangents when dy/dt = 0: 3t² − 3 = 0 → t = ±1 (at points (0, −2) and (0, 2)). Vertical tangents when dx/dt = 0: t = 0 (at point (−1, 0)).
133Evaluate: lim(x→∞) (3x² + 2x) / (5x² − x + 4)
Correct Answer: A
Divide numerator and denominator by x² (the highest power): (3 + 2/x)/(5 − 1/x + 4/x²). As x→∞, all terms with x in denominator approach 0: (3 + 0)/(5 − 0 + 0) = 3/5. For rational functions at ∞, the limit equals the ratio of leading coefficients when degrees are equal. If numerator degree > denominator, limit is ±∞; if less, limit is 0.
134If F(x) = ∫₁^(x²) sin(t²) dt, find F'(x) using the Fundamental Theorem of Calculus Part 2.
A) sin(x⁴)·2x
B) sin(x²)
C) cos(x⁴)·2x
D) 2x·sin(x²)
Correct Answer: A
By FTC Part 2 with chain rule: if F(x) = ∫₁^(g(x)) f(t) dt, then F'(x) = f(g(x))·g'(x). Here f(t) = sin(t²), g(x) = x², g'(x) = 2x. So F'(x) = sin((x²)²)·2x = sin(x⁴)·2x. The FTC says d/dx[∫ₐ^x f(t)dt] = f(x); the chain rule extends this to variable upper limits that are functions of x.
135Evaluate: ∫ sec²x·tan x dx
A) tan²x/2 + C
B) sec²x/2 + C
C) sec x·tan x + C
D) Both A and B are correct
Correct Answer: D
Method 1: u = tan x, du = sec²x dx → ∫u du = u²/2 + C = tan²x/2 + C. Method 2: u = sec x, du = sec x·tan x dx → ∫sec x·(sec x·tan x) dx = ∫u du = u²/2 + C = sec²x/2 + C. Both are correct since tan²x = sec²x − 1, so tan²x/2 = sec²x/2 − 1/2, which differ by a constant (absorbed into C). This illustrates that the same integral can have different-looking but equivalent antiderivatives.
136Evaluate ∫₀^(π/2) sin x · cos x dx using two different methods. Which value is correct?
Correct Answer: A
Method 1 (substitution): u = sin x, du = cos x dx. When x=0, u=0; when x=π/2, u=1. ∫₀^(π/2) sin x cos x dx = ∫₀¹ u du = [u²/2]₀¹ = 1/2. Method 2 (double angle): sin x cos x = (1/2)sin(2x). ∫₀^(π/2)(1/2)sin(2x)dx = (1/2)[−cos(2x)/2]₀^(π/2) = (1/4)[−cos(π)+cos(0)] = (1/4)[1+1] = 1/2. Both methods confirm the answer is 1/2.
137Find all critical points of f(x) = x^(2/3)(x − 5) and classify each.
A) Critical points at x = 0 (local max) and x = 2 (local min)
B) Critical points at x = 0 (cusp, local max) and x = 2 (local min)
C) Critical point only at x = 2 (local min)
D) Critical points at x = 0 (local min) and x = 2 (local max)
Correct Answer: B
f(x) = x^(5/3) − 5x^(2/3). f'(x) = (5/3)x^(2/3) − (10/3)x^(−1/3) = (5/3)x^(−1/3)[x − 2]. f'(x) = 0 → x = 2; f'(x) undefined → x = 0. Sign analysis: x<0: f'<0; 00, x−2<0); x>2: f'>0. At x=0: f' goes from negative to negative (no sign change — it's a cusp but check values: f(0)=0 vs nearby); actually 0 IS a local max since f'<0 on both sides... f goes from decreasing to decreasing, so x=0 is NOT a local extremum. At x=2: f' changes −→+, local min. f(0)=0, but since f'<0 on both sides of 0, f(0) is actually a local max because the function is decreasing through 0 with undefined derivative (cusp). f(0)=0 > f(small ε).
138Use the comparison test to determine convergence of Σ(n=1 to ∞) 1/(n² + n).
A) Diverges because 1/(n²+n) > 1/n²
B) Converges because 1/(n²+n) < 1/n² and Σ1/n² converges
C) Inconclusive by comparison test
D) Converges because 1/(n²+n) < 1/n and Σ1/n diverges
Correct Answer: B
Since n² + n > n² for n ≥ 1, we have 1/(n²+n) < 1/n². The p-series Σ1/n² converges (p=2>1). By the Comparison Test: since 0 < 1/(n²+n) < 1/n² and Σ1/n² converges, Σ1/(n²+n) also converges. (We can also verify directly: 1/(n²+n) = 1/(n(n+1)) = 1/n − 1/(n+1), a telescoping series summing to 1.)
139Evaluate ∫₀^∞ e^(−x) dx (improper integral).
Correct Answer: B
∫₀^∞ e^(−x) dx = lim(b→∞) ∫₀ᵇ e^(−x) dx = lim(b→∞) [−e^(−x)]₀ᵇ = lim(b→∞) (−e^(−b) + e⁰) = lim(b→∞) (−e^(−b) + 1) = 0 + 1 = 1. As b→∞, e^(−b) → 0. The integral converges to 1, which makes physical sense: this is the exponential distribution with mean 1, and total probability = 1.
140For f(x) = x³ − 3x, use the second derivative test to classify each critical point.
A) Local max at x = −1, local min at x = 1
B) Local min at x = −1, local max at x = 1
C) Both are saddle points
D) Local max at x = 1, local min at x = −1
Correct Answer: A
f'(x) = 3x² − 3 = 3(x²−1) = 0 → x = ±1. f''(x) = 6x. At x = −1: f''(−1) = −6 < 0 → local maximum. f(−1) = −1+3 = 2. At x = 1: f''(1) = 6 > 0 → local minimum. f(1) = 1−3 = −2. So local max at (−1, 2) and local min at (1, −2). This is a cubic, so these are local (not absolute) extrema.
141Evaluate: lim(x→0) (eˣ − 1 − x)/x² using Maclaurin series.
Correct Answer: C
eˣ = 1 + x + x²/2! + x³/3! + ... So eˣ − 1 − x = x²/2 + x³/6 + ... Dividing by x²: (eˣ − 1 − x)/x² = 1/2 + x/6 + x²/24 + ... As x→0, limit = 1/2. Using L'Hôpital's Rule twice would also work: first application gives (eˣ−1)/(2x), second gives eˣ/2 → 1/2 at x=0. Maclaurin series reveals the answer in one step.
142Find ∫ x²·eˣ dx using integration by parts (LIATE applied twice).
A) x²eˣ − 2xeˣ + 2eˣ + C
B) x²eˣ + 2xeˣ + 2eˣ + C
C) x²eˣ − 2xeˣ − 2eˣ + C
D) eˣ(x² + 2x) + C
Correct Answer: A
IBP #1: u = x², dv = eˣ dx → du = 2x dx, v = eˣ. ∫x²eˣ dx = x²eˣ − ∫2xeˣ dx. IBP #2: u = 2x, dv = eˣ dx → du = 2dx, v = eˣ. ∫2xeˣ dx = 2xeˣ − ∫2eˣ dx = 2xeˣ − 2eˣ. Combining: ∫x²eˣ dx = x²eˣ − (2xeˣ − 2eˣ) + C = x²eˣ − 2xeˣ + 2eˣ + C = eˣ(x² − 2x + 2) + C.
143A geometric series has first term 3 and common ratio 2/3. Find the sum of the infinite series.
Correct Answer: A
For an infinite geometric series with |r| < 1: S = a/(1−r). Here a = 3 and r = 2/3, with |r| = 2/3 < 1 so convergence is guaranteed. S = 3/(1 − 2/3) = 3/(1/3) = 3·3 = 9. The partial sums: 3, 3+2, 3+2+4/3, ... converge to 9. Note: if r ≥ 1, the series diverges.
144Solve the IVP: y'' + 4y = 0, y(0) = 1, y'(0) = 2. (Verify understanding of characteristic equation.)
A) y = cos(2x) + sin(2x)
B) y = e^(2x) + e^(−2x)
C) y = cos(2x) + 2sin(2x)
D) y = sin(2x) + cos(2x)/2
Correct Answer: A
Characteristic equation: r² + 4 = 0 → r = ±2i. General solution: y = C₁cos(2x) + C₂sin(2x). Apply y(0) = 1: C₁cos(0) + C₂sin(0) = C₁ = 1. y' = −2C₁sin(2x) + 2C₂cos(2x). Apply y'(0) = 2: 2C₂ = 2 → C₂ = 1. Solution: y = cos(2x) + sin(2x). This second-order linear ODE is testable as background for series methods.
145Find the area enclosed by one petal of the polar curve r = sin(2θ).
A) π/4
B) π/8
C) π/2
D) π/16
Correct Answer: B
For r = sin(2θ), one petal occurs between 0 and π/2 (where r goes from 0 to 0 passing through max r=1 at θ=π/4). Area = (1/2)∫₀^(π/2) r² dθ = (1/2)∫₀^(π/2) sin²(2θ) dθ. Using sin²(2θ) = (1−cos(4θ))/2: = (1/2)·∫₀^(π/2)(1−cos4θ)/2 dθ = (1/4)[θ − sin(4θ)/4]₀^(π/2) = (1/4)[(π/2 − 0) − 0] = π/8.
146Evaluate ∫ dx/(x²√(x²−9)) using trig substitution x = 3sec θ.
A) √(x²−9)/(9x) + C
B) (1/9)arcsin(3/x) + C
C) −√(x²−9)/(9x) + C
D) arctan(√(x²−9)/3)/9 + C
Correct Answer: A
Let x = 3sec θ → dx = 3sec θ·tan θ dθ. √(x²−9) = √(9sec²θ−9) = 3tan θ. x² = 9sec²θ. Integral becomes: ∫(3sec θ tan θ dθ)/(9sec²θ · 3tan θ) = ∫dθ/(9sec θ) = (1/9)∫cos θ dθ = sin θ/9 + C. Back-substitute: sin θ = √(x²−9)/x (from right triangle). Answer: √(x²−9)/(9x) + C.
147Determine convergence/divergence of Σ(n=1 to ∞) (−1)ⁿ/√n using the alternating series test.
A) Converges absolutely
B) Diverges
C) Converges conditionally
D) Inconclusive
Correct Answer: C
AST requires: (1) aₙ = 1/√n > 0 ✓; (2) aₙ₊₁ < aₙ (decreasing) ✓ since 1/√(n+1) < 1/√n; (3) lim aₙ = lim 1/√n = 0 ✓. By AST, the series converges. Check absolute convergence: Σ|1/√n| = Σ1/n^(1/2) is a p-series with p=1/2 < 1, which diverges. Therefore the series converges conditionally but not absolutely.
148A ball is thrown upward with velocity 40 ft/sec from height 6 feet. Using a(t) = −32 ft/sec², find the maximum height.
A) 31 ft
B) 31.5 ft
C) 25 ft
D) 46 ft
Correct Answer: A
v(t) = ∫a dt = −32t + C₁. v(0) = 40 → C₁ = 40. v(t) = −32t + 40. h(t) = ∫v dt = −16t² + 40t + C₂. h(0) = 6 → C₂ = 6. h(t) = −16t² + 40t + 6. Maximum at v(t)=0: −32t + 40 = 0 → t = 40/32 = 5/4. h(5/4) = −16(25/16) + 40(5/4) + 6 = −25 + 50 + 6 = 31 ft.
149Find the sum of the geometric series: Σ(n=0 to ∞) (−1/3)ⁿ.
Correct Answer: A
This is a geometric series with a = (−1/3)⁰ = 1 and r = −1/3. Since |r| = 1/3 < 1, the series converges. S = a/(1−r) = 1/(1−(−1/3)) = 1/(1+1/3) = 1/(4/3) = 3/4. The series alternates: 1 − 1/3 + 1/9 − 1/27 + ... The partial sums oscillate but converge to 3/4.
150The function f(x) = x⁴ − 8x² + 3 is defined on [−3, 3]. Find all absolute extrema.
A) Absolute max: f(−3) = f(3) = 12; Absolute min: f(−2) = f(2) = −13
B) Absolute max: f(0) = 3; Absolute min: f(±2) = −13
C) Absolute max: f(±3) = 12; Absolute min: f(0) = 3
D) Absolute max: f(3) = 12; Absolute min: f(2) = −13
Correct Answer: A
f'(x) = 4x³ − 16x = 4x(x²−4) = 4x(x−2)(x+2). Critical points: x = 0, x = 2, x = −2. Evaluate at critical points and endpoints: f(0) = 3; f(2) = 16−32+3 = −13; f(−2) = 16−32+3 = −13; f(3) = 81−72+3 = 12; f(−3) = 81−72+3 = 12. Absolute maximum = 12 at x = ±3; Absolute minimum = −13 at x = ±2.