CSE1200 -- Exam Patterns

1. Power Series: Interval / Radius of Convergence Very High

END 2019 O1a-b • 2020 O9a • 2023 O1a • 2024 O1a • 2025 O1a | RES 2019 O2a-b • 2023 O1a • 2024 O1a-c • 2025 O1a

Given a power series ∑ c_n(x − a)ⁿ, find the radius of convergence R, the interval of convergence, and determine behavior at the boundary points.

Methods to Solve

Ratio Test: Compute L = lim |c_n+1/c_n|. Then R = 1/L. Works best when c_n involves factorials, exponentials, or products.
Root Test: Compute L = lim |c_n|^1/n. Then R = 1/L. Best for c_n = (something)ⁿ.
Boundary check: At x = a ± R, substitute and test convergence separately using alternating series test, p-series, comparison test, or limit comparison test.
Handle non-standard forms: If the series has x²ⁿ or x³ⁿ⁺¹, substitute u = x² or u = x³ to reduce to standard form, then convert back.

Example (END 2024 O1a): Find interval of convergence of ∑ (−1)ⁿ/πⁿ · (n+1)/n² · (x+1)ⁿ. Ratio test gives R = π, center a = −1. Check x = −1 − π (diverges by comparison) and x = −1 + π (converges by alternating series test). Interval: (−1−π, −1+π].

2. Critical Points + Second Derivative Test (Multivariable) Very High

END 2019 O2 • 2020 O10 • 2023 O2a • 2024 O2a • 2025 O2a | RES 2019 O3b-c • 2023 O2a • 2024 O2a • 2025 O2a

Find all critical points of f(x,y) by solving ∇f = 0, then classify each as local max, local min, or saddle point using the second derivative test (Hessian determinant).

Methods to Solve

Find critical points: Solve the system f_x = 0 and f_y = 0 simultaneously.
Compute the Hessian determinant: D = f_xx f_yy − (f_xy)² at each critical point.
Classify: If D > 0 and f_xx > 0 ⇒ local min. If D > 0 and f_xx < 0 ⇒ local max. If D < 0 ⇒ saddle point. If D = 0 ⇒ test is inconclusive.
When D = 0: Use other arguments (e.g., show f values along different paths through the point differ).
Restricted domains: Only consider critical points that lie in the interior of the given domain D.

Example (END 2020 O10): f(x,y) = 4x³ + 6x²y + y³ − 3y². Solve f_x = 12x² + 12xy = 0 and f_y = 6x² + 3y² − 6y = 0. Critical points: (0,0), (0,2), (−1,1). At (0,2): D = −144 < 0 ⇒ saddle.

3. Double Integrals (Setup, Evaluate, Interchange Order) Very High

END 2019 S4 • 2020 O11 • 2023 O3a • 2024 O3 • 2025 O3 | RES 2019 S8 • 2023 O3 • 2024 O3 • 2025 O3

Compute a double integral &iint;_D f(x,y) dA over a given region D. Often requires sketching the region, choosing the correct order of integration (dx dy vs dy dx), and sometimes interchanging the order to make the integral tractable.

Methods to Solve

Sketch the region: Draw the boundary curves and identify intersection points to understand the domain D.
Choose integration order: Pick the order (dy dx or dx dy) that gives simpler limits. If one order leads to an integral you cannot evaluate (e.g., ∫ cos(x²) dx), switch order.
Set up limits: For dy dx: outer limits are x = a to x = b, inner limits are y = g₁(x) to y = g₂(x). For dx dy: reverse roles.
Interchange order: Sketch the region, re-describe boundaries in the other variable, and rewrite the integral.
Split regions: If the region boundary changes form, split into sub-regions and sum the integrals.
Evaluate: Integrate the inner integral first, treating the outer variable as a constant, then integrate the result.

Example (END 2023 O3a): Compute &iint;_T cos(x²) dA over the triangle with vertices (0,0), (2,0), (2,2). Integrating dy dx: ∫₀² ∫₀^x cos(x²) dy dx = ∫₀² x cos(x²) dx = sin(4)/2.

4. Complex Roots (zⁿ = c) Very High

END 2019 S3 • 2020 S2 • 2023 S2 • 2024 O5a • 2025 S3 | RES 2019 S7b • 2023 S6 • 2024 O5a • 2025 O5a

Find all complex solutions to zⁿ = c, where c is a complex number. Express solutions in polar or a + bi form.

Methods to Solve

Convert c to polar form: Write c = r e^iθ where r = |c| and θ = arg(c).
Apply the root formula: z_k = r^1/n e^{i(θ + 2πk)/n} for k = 0, 1, ..., n−1.
Convert to a + bi: Use z_k = r^1/n(cos(φ_k) + i sin(φ_k)) where φ_k = (θ + 2πk)/n.
Sketch: The n roots are equally spaced on a circle of radius r^1/n, separated by angle 2π/n.
Special cases: For equations like (z − a)ⁿ = c, solve wⁿ = c first then shift: z = w + a.

Example (END 2025 S3): Solve (z−2)³ = −8. Set w = z−2, solve w³ = −8 = 8e^iπ. w_k = 2e^i(π+2πk)/3. Then z_k = w_k + 2 gives z = 1 ± i√3 and z = 0.

5. Linearization / Linear Approximation Very High

MID 2020 Q7-8 • 2023 Q9 • 2025 Q9 | END 2019 S5c • 2020 S6a • 2025 O1b | RES 2019 S5 • 2023 S1

Find the linearization L(x) of a function near a point, or use it to approximate a value. For multivariable functions, find the tangent plane.

Methods to Solve

Single variable: L(x) = f(a) + f'(a)(x − a). Choose a where f(a) and f'(a) are easy to compute.
Multivariable: L(x,y) = f(a,b) + f_x(a,b)(x−a) + f_y(a,b)(y−b).
Error estimation (differentials): |Δf| ≈ |f'(a)| · |Δx|. The absolute error in f is approximately |f'| times the error in x.
Common approximations: sin(x) ≈ x, cos(x) ≈ 1, e^x ≈ 1 + x, ln(1+x) ≈ x, (1+x)ⁿ ≈ 1 + nx, all near x = 0.

Example (MID 2023 Q9): Approximate ln(1.1). Use f(x) = ln(x) near a = 1: L(x) = ln(1) + (1/1)(x−1) = x − 1. So ln(1.1) ≈ 0.1.

6. Directional Derivative Very High

END 2019 S5d • 2020 S6b-c • 2023 S3 • 2024 S2b • 2025 S1 | RES 2019 S6b • 2023 S4 • 2024 S2b

Compute the directional derivative D_uf(a,b) = ∇f(a,b) · u, the rate of change of f in a given direction u. May also ask for the direction of steepest ascent/descent or the maximum/minimum directional derivative.

Methods to Solve

Compute the gradient: ∇f = (f_x, f_y). Evaluate at the given point.
Normalize the direction: If u is not a unit vector, compute u/|u|.
Dot product: D_uf = ∇f · û (the unit vector).
Maximum rate of change: max D_uf = |∇f|, achieved in the direction u = ∇f/|∇f|.
Minimum rate of change: min D_uf = −|∇f|, achieved in direction u = −∇f/|∇f| (steepest descent).

Example (END 2023 S3): f(x,y) = x cos(πy) − xy at P = (1,1). ∇f = (cos(πy) − y, −πx sin(πy) − x) = (−2, −1). Steepest descent direction: (2,1)/√5. Directional derivative in that direction: −√5.

7. Improper Integrals Very High

MID 2020 Q11 • 2023 Q5, Q12 • 2024 Q7b • 2025 Q8 | END 2025 S4 | RES 2019 O1 • 2023 S5 • 2025 S6b

Evaluate integrals with infinite limits (Type 1) or integrands with singularities (Type 2). Determine convergence or divergence, and if convergent, find the value.

Methods to Solve

Type 1 (infinite limits): Replace ∞ with a limit variable t: ∫_a^∞ f(x) dx = lim_t→∞ ∫_a^t f(x) dx.
Type 2 (singularity): If f has a singularity at c ∈ [a,b], split: ∫_a^b = lim_ε→0⁺ ∫_a^c−ε + lim_δ→0⁺ ∫_c+δ^b.
Integration by parts: Often needed for integrands like ln(x)/xⁿ. Apply IBP then evaluate the limit.
Comparison test: If |f(x)| ≤ g(x) and ∫ g converges, then ∫ f converges absolutely.
p-test: ∫₁^∞ 1/x^p dx converges iff p > 1. ∫₀¹ 1/x^p dx converges iff p < 1.
Identify all singularities: Check for singularities at every point where the integrand is undefined, not just the endpoints.

Example (MID 2020 Q11): ∫₁^∞ ln(x)/x³ dx. IBP: u = ln(x), dv = x⁻³dx. Get [−ln(x)/(2x²)]₁^∞ + ∫ 1/(2x³) dx = 0 + 1/4 = 1/4.

8. Recursive Sequences (Monotonicity, Boundedness, Limit) Very High

MID 2020 Q12 • 2023 Q11 • 2024 Q8 • 2025 Q10 | END 2019 O3 | RES 2019 O4 • 2023 S2

Given a sequence defined by a₀ = c and a_n+1 = f(a_n), prove it converges (usually via monotone convergence theorem) and find the limit.

Methods to Solve

Prove monotonicity: Show a_n+1 ≥ a_n (increasing) or a_n+1 ≤ a_n (decreasing) by induction.
Prove boundedness: Show a_n ≤ M (if increasing) or a_n ≥ m (if decreasing) by induction.
Monotone Convergence Theorem: A bounded monotone sequence converges.
Find the limit: Assume lim a_n = L, then L = f(L). Solve the fixed-point equation. Discard solutions outside the range of the sequence.
Alternative: show divergence. Sometimes the sequence diverges -- prove by contradiction (assume L exists, get contradiction from L = f(L)).

Example (MID 2025 Q10): a₀ = −2, a_n+1 = √(2a_n + 8). Show increasing: a_n+1 ≥ a_n. Show bounded: a_n ≤ 4. Fixed point: L = √(2L+8) ⇒ L² = 2L+8 ⇒ L = 4 (discard L = −2). Answer: 4.

9. Implicit Differentiation Very High

MID 2020 Q6 • 2023 Q10 • 2024 Q4 • 2025 Q13 | RES 2019 S4 • 2023 O5 • 2025 S3

Given an implicit equation F(x,y) = 0, find dy/dx or the slope of the tangent line at a given point. Sometimes find points where the tangent is horizontal or vertical (extremal points on the curve).

Methods to Solve

Differentiate implicitly: Differentiate both sides with respect to x, treating y as a function of x. Apply the chain rule: d/dx[f(y)] = f'(y) · dy/dx.
Solve for dy/dx: Collect all dy/dx terms on one side, factor, and isolate.
Horizontal tangent: Set dy/dx = 0, i.e., set the numerator of dy/dx to zero while the denominator is nonzero.
Vertical tangent / rightmost point: Set dx/dy = 0 (equivalently, the denominator of dy/dx is zero while the numerator is nonzero).
Evaluate: Substitute the given point (a,b) into the expression for dy/dx.

Example (MID 2020 Q6): 2x² − x = y³ − 7y. Differentiate: 4x − 1 = (3y² − 7) dy/dx. At (2,−1): dy/dx = (8−1)/(3−7) = 7/(−4) = −7/4.

10. Limits (0/0 Indeterminate Form) High

MID 2020 Q3 • 2023 Q2-3 • 2024 Q2, Q13 • 2025 Q1 | RES 2019 S3a • 2025 S1

Compute limits where direct substitution gives 0/0. Requires algebraic manipulation or L'Hôpital's rule to resolve the indeterminate form.

Methods to Solve

Factor and cancel: Factor numerator and denominator, cancel common (x − a) factors.
Conjugate multiplication: For expressions with square roots, multiply by the conjugate: (√A − √B)(√A + √B) = A − B.
L'Hôpital's Rule: If lim f(x)/g(x) = 0/0 (or ∞/∞), then lim f(x)/g(x) = lim f'(x)/g'(x), provided the latter limit exists.
Common tricks: Multiply/divide by √(x), rewrite as a single fraction, or substitute u = x − a.

Example (MID 2025 Q1): lim_x→2 (x³ − 3x² + 4)/(x³ − 4x² + 4x). Factor: both have (x−2) factor. After cancellation or by L'Hôpital: limit = (3·4 − 6)/(3·4 − 8 + 4) = 6/8 = 3/4.

11. Asymptotes (Vertical, Horizontal, Oblique) High

MID 2020 Q5 • 2023 Q14 • 2024 Q12 • 2025 Q14 | RES 2024 S1 • 2025 S2

Find all asymptotes of a given function: vertical (where the function blows up), horizontal (behavior as x → ±∞), and oblique/slant (linear behavior at ±∞).

Methods to Solve

Vertical asymptotes: Find where the denominator is zero (or ln argument is zero, etc.). Check that the limit is ±∞ -- if the limit is finite, it's a removable singularity, not an asymptote.
Horizontal asymptotes: Compute lim_x→∞ f(x) and lim_x→−∞ f(x). If either is finite, y = L is a horizontal asymptote.
Oblique asymptotes: If no horizontal asymptote, compute a = lim_x→±∞ f(x)/x and b = lim_x→±∞ (f(x) − ax). If a ≠ 0 and b is finite, y = ax + b is an oblique asymptote.
Conjugate trick: For functions with √(x² + ...) − x, multiply by the conjugate to simplify the limit at infinity.
L'Hôpital for removable singularities: Use L'Hôpital to check if a candidate vertical asymptote is actually removable (finite limit).

Example (MID 2025 Q14): h(x) = (√(x⁶+3x⁵) − x³)/(x²−2x+1). Vertical asymptote at x = 1. Horizontal asymptote y = 3/2 as x → +∞. Oblique asymptote y = −2x − 11/2 as x → −∞.

12. Integration Techniques (Substitution, By Parts) High

MID 2020 Q9-10 • 2023 Q6 • 2024 Q7a, Q9, Q11 • 2025 Q7 | RES 2024 S4 • 2025 S4

Evaluate integrals using u-substitution, integration by parts, or a combination. Includes reduction formulas and LIATE ordering.

Methods to Solve

U-substitution: Identify an inner function u = g(x) such that g'(x) appears in the integrand. Replace dx = du/g'(x).
Integration by parts: ∫ u dv = uv − ∫ v du. Use LIATE to choose u: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential.
Reduction formulas: Express I_n in terms of I_n−1 by applying IBP once.
Combined techniques: Sometimes substitute first to simplify, then apply IBP. E.g., ∫ 2x³cos(x²)dx: let u = x², then IBP on ∫ u cos(u) du.
Definite integrals: Update limits when substituting. Check if symmetry (odd/even) can simplify.

Example (MID 2024 Q11): ∫ x² arctan(x) dx. LIATE: u = arctan(x), dv = x²dx. Then du = 1/(1+x²)dx, v = x³/3. Result: (x³/3)arctan(x) − ∫ x³/(3(1+x²))dx. Simplify the remaining integral by polynomial division.

13. Taylor Series for Anti-Derivative + Error Bound High

END 2023 O4 • 2024 O4 • 2025 O4 | RES 2023 O4 • 2024 S3 • 2025 O4

Given a function f(x) (often without a closed-form antiderivative), find the Taylor series of ∫ f(x) dx, then use it to approximate ∫_a^b f(x) dx with a given error bound.

Methods to Solve

Start from a known series: Use known expansions (e^x, ln(1+x), 1/(1−x), sin, cos) and substitute to get the series for f(x).
Integrate term-by-term: ∫ ∑ c_n xⁿ dx = C + ∑ c_n xⁿ⁺¹/(n+1). Set C using initial conditions if given.
Approximate the integral: ∫_a^b f(x) dx = ∑ c_n (bⁿ⁺¹ − aⁿ⁺¹)/(n+1). Truncate the series.
Error bound (alternating series): For alternating series, the error is bounded by the absolute value of the first omitted term: |error| ≤ |a_N+1|.
Error bound (positive series): Use comparison with a geometric series or integral remainder estimate.

Example (END 2025 O4): f(x) = e^x². Series: ∑ x²ⁿ/n!. Anti-derivative: F(x) = ∑ x²ⁿ⁺¹/((2n+1)n!). To approximate ∫₀¹ with error < 0.06, sum terms until |a_N+1| < 0.06.

14. Absolute Max/Min on Closed Domain (Multivariable) High

END 2023 O2b • 2024 O2b • 2025 O2b | RES 2023 O2b • 2024 O2b • 2025 O2b

Find the absolute maximum and minimum of f(x,y) on a closed and bounded domain D. By the Extreme Value Theorem, these must occur at critical points in the interior or on the boundary.

Methods to Solve

Step 1 -- Interior critical points: Find all points where ∇f = 0 inside D. Evaluate f at each.
Step 2 -- Boundary: Parametrize each boundary segment (edge, curve). Reduce to a single-variable optimization on each segment.
Step 3 -- Corners/vertices: Evaluate f at all corner points of D.
Step 4 -- Compare: The largest value among all candidates is the absolute max; the smallest is the absolute min.
Common domains: Rectangles (4 edges), triangles (3 edges), disks (parametrize boundary with x = r cosθ, y = r sinθ).

Example (RES 2023 O2b): f(x,y) = 2x² − 4xy + 4y on D = {0.5 ≤ x ≤ 3, 0 ≤ y ≤ 2}. Interior critical point: (1,1), f = 2. Check all 4 edges and corners. Compare to find absolute max and min.

15. Gradient Vector High

END 2019 S5b • 2020 S6a • 2023 S3a • 2024 S2a • 2025 S1a | RES 2024 S2a

Compute the gradient vector ∇f = (f_x, f_y) of a multivariable function and evaluate at a given point. Often a sub-part leading to directional derivative or tangent plane questions.

Methods to Solve

Compute partial derivatives: f_x = ∂f/∂x (treat y as constant), f_y = ∂f/∂y (treat x as constant).
Chain rule: For composite functions like e^xy, use ∂/∂x[e^xy] = y e^xy.
Evaluate: Substitute the given point (a,b) into the partial derivatives.
Geometric meaning: ∇f points in the direction of steepest ascent; |∇f| gives the maximum rate of change.

16. Series Convergence / Divergence Classification High

END 2020 S3, O8 • 2024 O6 • 2025 O7 | RES 2024 O4a • 2025 S6

Determine whether a given series is absolutely convergent, conditionally convergent, or divergent. Choose the appropriate convergence test.

Methods to Solve

Divergence Test: If lim a_n ≠ 0, the series diverges. Always check this first.
Geometric series: ∑ rⁿ converges iff |r| < 1. Sum = 1/(1−r).
p-series: ∑ 1/n^p converges iff p > 1.
Ratio Test: If lim |a_n+1/a_n| < 1, converges absolutely. Good for factorials and exponentials.
Root Test: If lim |a_n|^1/n < 1, converges absolutely. Good for n-th powers.
Comparison / Limit Comparison: Compare with a known convergent/divergent series.
Alternating Series Test: ∑ (−1)ⁿ b_n converges if b_n is decreasing and lim b_n = 0. This gives conditional convergence.
Integral Test: ∑ f(n) converges iff ∫₁^∞ f(x) dx converges, when f is positive, continuous, and decreasing.

Example (END 2024 O6): (a) ∑(−1)ⁿ 3ⁿ/((n+1)4ⁿ): ratio test gives 3/4 < 1, absolutely convergent. (b) ∑ n e^−n²: comparison with e⁻ⁿ, convergent. (c) ∑(−1)ⁿ n/(3n²−n−1): lim a_n = 0, alternating, but check |a_n| ~ 1/(3n) diverges ⇒ conditionally convergent.

17. Match Functions to Graphs / Contour Plots High

END 2019 S6 • 2020 S7 • 2023 S4 | RES 2023 S3 • 2025 S5

Match multivariable function formulas to their 3D surface plots or 2D contour plots (level curves).

Methods to Solve

Check symmetry: f(x,y) = f(y,x) means symmetric about y = x. f(−x,y) = −f(x,y) means odd in x (antisymmetric).
Set one variable to zero: Look at cross-sections f(x,0) and f(0,y) to identify shape along axes.
Level curves f = c: Set f(x,y) = c and identify the shape (lines, circles, hyperbolas, etc.).
Check special points: Evaluate f at origin, along axes, and along diagonal y = x.
Identify function type: Quadratic forms (paraboloids, saddles), products (xy gives saddle), trigonometric (periodic contours).

18. Complex Number Arithmetic (a + bi Form) High

END 2019 S2 • 2020 S1 | RES 2019 S7a • 2024 O5b • 2025 O5b

Perform arithmetic operations on complex numbers (division, powers, modulus, argument) and express results in a + bi form or polar form.

Methods to Solve

Division: Multiply numerator and denominator by the conjugate of the denominator: (a+bi)/(c+di) = (a+bi)(c−di)/(c²+d²).
Powers: Convert to polar form z = r e^iθ, then zⁿ = rⁿ e^inθ.
Modulus: |z| = √(a² + b²). |z₁z₂| = |z₁||z₂|, |z₁/z₂| = |z₁|/|z₂|.
Argument: arg(z₁z₂) = arg(z₁) + arg(z₂), arg(z̄) = −arg(z).
Conjugate properties: z̄ = a − bi. z · z̄ = |z|².

Example (END 2020 S1): z = (2−i)/(3+i). Multiply by conjugate: z = (2−i)(3−i)/((3+i)(3−i)) = (6−2i−3i+i²)/(9+1) = (5−5i)/10 = 1/2 − i/2.

19. Indeterminate Form Limits (1^∞, 0⁰, ∞⁰) High

MID 2020 Q4 • 2023 Q15 • 2024 Q6 • 2025 Q4

Compute limits of the form [f(x)]^g(x) where the base and exponent create an indeterminate form (1^∞, 0⁰, or ∞⁰).

Methods to Solve

Exponential rewrite: Write f(x)^g(x) = e^{g(x) ln(f(x))}. The limit becomes e^{lim g(x) ln(f(x))}.
Reduce to 0/0 or ∞/∞: Rewrite g(x) ln(f(x)) as ln(f(x))/(1/g(x)) to get an indeterminate form suitable for L'Hôpital.
Apply L'Hôpital's Rule: On the exponent g(x) ln(f(x)), not the original expression.
Special case (1 + u)^1/u → e: Recognize lim_u→0 (1+u)^1/u = e. Rewrite to match this pattern when possible.

Example (MID 2020 Q4): lim_x→0 (1−3x)^1/x = e^{lim (1/x)ln(1−3x)} = e^{lim ln(1−3x)/x}. L'Hôpital: e^{lim −3/(1−3x) / 1} = e⁻³.

20. Higher-Order Derivative from Power Series High

END 2023 O1b • 2024 O1b • 2025 S2 | RES 2023 O1b

Given a power series representation g(x) = ∑ c_n(x−a)ⁿ, compute a specific higher-order derivative g^(k)(a).

Methods to Solve

Key identity: If g(x) = ∑ c_n(x−a)ⁿ, then g^(k)(a) = k! · c_k.
Find c_k: Identify the coefficient of (x−a)^k in the series.
Multiply: g^(k)(a) = k! · c_k.
For products: If f(x) = x² arctan(x), find the Taylor series of arctan(x) first, multiply by x², then read off the coefficient.

Example (END 2025 S2): f(x) = x² arctan(x). arctan(x) = ∑ (−1)ⁿx²ⁿ⁺¹/(2n+1). So f(x) = ∑ (−1)ⁿx²ⁿ⁺³/(2n+1). The x⁹ term (n=3): c₉ = (−1)³/7 = −1/7. f⁽⁹⁾(0) = 9! · (−1/7) = −9!/7.

21. Geometric / Telescoping Series (Find Sum) High

END 2020 S3 • 2024 S4 | RES 2019 S2 • 2025 S6a

Identify a series as geometric (or related to geometric via differentiation/integration) and compute its exact sum.

Methods to Solve

Geometric series: ∑_n=0^∞ rⁿ = 1/(1−r) for |r| < 1. Adjust for starting index and constant factors.
Recognize known series: ∑ xⁿ/n! = e^x, ∑ (−1)ⁿx²ⁿ⁺¹/(2n+1) = arctan(x), etc.
Differentiation trick: ∑ n xⁿ⁻¹ = d/dx[1/(1−x)] = 1/(1−x)². Useful for ∑ n rⁿ.
Partial fractions / telescoping: Decompose a_n into partial fractions; sum telescopes when consecutive terms cancel.

Example (END 2024 S4): ∑_n=1^∞ n/2ⁿ⁻¹. Use d/dx[∑ xⁿ] = ∑ nxⁿ⁻¹ = 1/(1−x)². Evaluate at x = 1/2: ∑ n/2ⁿ⁻¹ = 1/(1/2)² = 4.

22. Inverse Function (Formula + Domain) Low

MID 2023 Q1 • 2024 Q1 • 2025 Q12

Given a function f with a restricted domain that makes it one-to-one, find the formula for f⁻¹(x) and specify its domain.

Methods to Solve

Swap and solve: Write y = f(x), swap x and y, solve for y to get f⁻¹(x).
Quadratic formula: For quadratic f, use the quadratic formula after swapping. Choose the correct ± sign based on the restricted domain.
Domain of f⁻¹: equals the range of f. Compute f on its restricted domain to find the range.

Example (MID 2024 Q1): f(x) = −x² + x + 2 on [1,∞). Swap: x = −y² + y + 2. Solve: y = (1 − √(9−4x))/2 (choose minus sign since y ≥ 1 requires checking). Domain of f⁻¹ = range of f = (−∞, 2].

23. Inverse Trig Simplification Low

MID 2020 Q2 • 2025 Q2 | RES 2019 S1

Simplify expressions involving inverse trigonometric functions (arcsin, arccos, arctan) to algebraic expressions or exact values without inverse trig.

Methods to Solve

Triangle method: For tan(arcsin(x)), draw a right triangle where sin(θ) = x. Read off tan(θ) from the triangle.
Double angle formulas: cos(2θ) = 1 − 2sin²(θ). So cos(2 arcsin(x)) = 1 − 2x².
Range restrictions: arccos(cos(θ)) = θ only if θ ∈ [0, π]. Otherwise, use periodicity and symmetry to reduce θ to the correct range.
arctan(tan(θ)): = θ only if θ ∈ (−π/2, π/2). Reduce using θ − nπ.

Example (MID 2020 Q2a): tan(arcsin(1/√7)). Draw triangle: opposite = 1, hypotenuse = √7, adjacent = √6. So tan = 1/√6.

24. Extreme Values on Closed Interval (Single Variable) Low

MID 2023 Q4 • 2025 Q3 | END 2024 S1

Find the absolute maximum and minimum of a single-variable function on a closed interval [a,b].

Methods to Solve

Find critical points: Solve f'(x) = 0 and find where f'(x) is undefined, within (a,b).
Evaluate f at all candidates: Compute f at each critical point and at the endpoints x = a and x = b.
Compare: The largest value is the absolute maximum; the smallest is the absolute minimum.
Give values, not points: The question usually asks for the function values, not the x-coordinates.

25. Symmetry in Definite Integrals (Odd/Even Functions) Low

MID 2023 Q16 • 2024 Q10 • 2025 Q11

Recognize that an integrand is an odd function on a symmetric interval [−a, a], so the integral is zero.

Methods to Solve

Odd function test: If f(−x) = −f(x) for all x, then ∫_−a^a f(x) dx = 0.
Even function test: If f(−x) = f(x), then ∫_−a^a f(x) dx = 2 ∫₀^a f(x) dx.
Check products: odd × odd = even, odd × even = odd, even × even = even. Useful for products like x³e^x⁴ (odd times even = odd).
Split if needed: If the integrand is a sum, check each term separately.

Example (MID 2025 Q11): ∫_−π^π (x³e^x⁴ + x²sin(x)cos(x)) dx. Both terms are odd functions on [−π, π], so the integral = 0.

26. Sketch Maximal Domain (Multivariable) Low

END 2019 S5a • 2020 S5 | RES 2019 O3a

Given a multivariable function f(x,y), determine and sketch the maximal domain where f is defined.

Methods to Solve

Identify restrictions: Square roots require argument ≥ 0. Logarithms require argument > 0. Denominators ≠ 0.
Solve inequalities: Translate each restriction into an inequality in x and y.
Intersect: The domain is the intersection of all individual restrictions.
Sketch: Draw boundary curves and shade the valid region. Use dashed lines for strict inequalities, solid for non-strict.

27. Limits at Infinity / Squeeze Theorem Low

MID 2023 Q7 • 2025 Q5 | RES 2019 S3b

Compute limits as x → ∞ where L'Hôpital fails or is inapplicable (e.g., oscillating terms). Use algebraic manipulation or the squeeze theorem.

Methods to Solve

Squeeze Theorem: If g(x) ≤ f(x) ≤ h(x) and lim g = lim h = L, then lim f = L. Useful when f contains bounded oscillating terms like sin(x) or e^sin(x).
Bound oscillating parts: −1 ≤ sin(x) ≤ 1, e⁻¹ ≤ e^sin(x) ≤ e.
Divide by dominant term: For rational-like functions, divide numerator and denominator by x (or xⁿ).
Conjugate trick: For √(x²+ax) − x, multiply by conjugate to get ax/(√(x²+ax)+x) → a/2.
L'Hôpital pitfall: lim x/(x+sin(x)): applying L'Hôpital gives 1/(1+cos(x)) which does not exist. Use squeeze instead: divide by x to get 1/(1+sin(x)/x) → 1.

28. Limits via Taylor Series Expansion Low

END 2019 O1c • 2020 O9b

Compute a limit by substituting Taylor/Maclaurin expansions into the expression, then simplifying.

Methods to Solve

Expand to sufficient order: Replace functions with their Taylor series up to enough terms to cancel indeterminate forms.
Key expansions: e^x = 1 + x + x²/2 + ..., sin(x) = x − x³/6 + ..., cos(x) = 1 − x²/2 + ..., ln(1+x) = x − x²/2 + ...
Substitute and simplify: After expanding, cancel common factors in numerator and denominator, then evaluate the limit.

Example (END 2020 O9b): lim_x→0 f(x)/(cos(x²)−1). Expand cos(x²) = 1 − x⁴/2 + ..., so denominator ~ −x⁴/2. Expand f(x) from its power series, divide, and take the limit.

29. FTC Application / Critical Points via FTC Low

MID 2024 Q3 | END 2023 O3b

Use the Fundamental Theorem of Calculus to differentiate functions defined as integrals, g(x) = ∫_a^x f(t) dt, and find critical points or evaluate related integrals.

Methods to Solve

FTC Part 1: If g(x) = ∫_a^x f(t) dt, then g'(x) = f(x).
Chain rule variant: If g(x) = ∫_a^h(x) f(t) dt, then g'(x) = f(h(x)) · h'(x).
Critical points: Set g'(x) = f(x) = 0 and solve within the given domain.
Integration by parts on integral of integral: ∫_a^b G(x) dx where G(x) = ∫_c^x f(t) dt can be evaluated by interchanging order of integration.

30. Euler's Formula / Complex Trig Identities Low

END 2025 O6 | RES 2025 O5c

Prove Euler's formula e^iθ = cos(θ) + i sin(θ) using power series, or derive trigonometric identities from complex exponentials.

Methods to Solve

Power series proof: Expand e^iθ = ∑ (iθ)ⁿ/n!, separate real and imaginary parts to get cos(θ) and sin(θ) series.
Derive identities: From e^i(a+b) = e^ia · e^ib, expand both sides and equate real/imaginary parts to get angle addition formulas.
Double angle: Set a = b = x in the addition formula, or use (e^ix)² = e^2ix. Expand left side: (cos x + i sin x)² = cos²x − sin²x + 2i sin x cos x. Real part gives cos(2x) = cos²x − sin²x.

31. Area Between Curves Low

MID 2025 Q6

Find the area enclosed between two curves by integrating the difference of the upper and lower functions.

Methods to Solve

Find intersection points: Set f(x) = g(x) and solve for x. These are the limits of integration.
Determine which is on top: Check which function is larger on each sub-interval.
Integrate: A = ∫_a^b |f(x) − g(x)| dx. If f ≥ g on [a,b], this is ∫_a^b (f − g) dx.
Multiple regions: If the curves cross, split into sub-intervals and sum the areas.

32. Mean Value Theorem Application Low

MID 2023 Q8

Apply the Mean Value Theorem (MVT) to a word problem or function to conclude the existence of a point where f'(c) equals a specific value.

Methods to Solve

MVT statement: If f is continuous on [a,b] and differentiable on (a,b), then there exists c ∈ (a,b) such that f'(c) = (f(b)−f(a))/(b−a).
Word problems: Translate physical quantities (distance, time, velocity) into MVT: average velocity = instantaneous velocity at some point.
Verify hypotheses: Check continuity and differentiability before applying.

Example (MID 2023 Q8): Car drives 100 km in 2 hours. By MVT, there exists t₀ where v(t₀) = 100/2 = 50 km/h.

33. Newton-Raphson Method Low

END 2025 O5

Sketch or apply Newton's method to find roots of an equation, showing how the iterative sequence converges graphically.

Methods to Solve

Formula: x_n+1 = x_n − f(x_n)/f'(x_n).
Graphical interpretation: Draw the tangent line at (x_n, f(x_n)), find where it crosses the x-axis -- that's x_n+1.
Sketch: Draw f(x), pick x₀, draw tangent, show x₁, x₂, ... converging to the root.
Choose f(x): To find √[3]{6}, use f(x) = x³ − 6.

Generated from 13 exams: Midterms 2020, 2023, 2024, 2025 | Endterms 2019, 2020, 2023, 2024, 2025 | Resits 2019, 2023, 2024, 2025
CSE1200 Calculus, TU Delft

CSE1200 -- Exam Pattern Analysis

Table of Contents

1. Power Series: Interval / Radius of Convergence Very High

Methods to Solve

2. Critical Points + Second Derivative Test (Multivariable) Very High

Methods to Solve

3. Double Integrals (Setup, Evaluate, Interchange Order) Very High

Methods to Solve

4. Complex Roots (zn = c) Very High

Methods to Solve

5. Linearization / Linear Approximation Very High

Methods to Solve

6. Directional Derivative Very High

Methods to Solve

7. Improper Integrals Very High

Methods to Solve

8. Recursive Sequences (Monotonicity, Boundedness, Limit) Very High

Methods to Solve

9. Implicit Differentiation Very High

Methods to Solve

10. Limits (0/0 Indeterminate Form) High

Methods to Solve

11. Asymptotes (Vertical, Horizontal, Oblique) High

Methods to Solve

12. Integration Techniques (Substitution, By Parts) High

Methods to Solve

13. Taylor Series for Anti-Derivative + Error Bound High

Methods to Solve

14. Absolute Max/Min on Closed Domain (Multivariable) High

Methods to Solve

15. Gradient Vector High

Methods to Solve

16. Series Convergence / Divergence Classification High

Methods to Solve

17. Match Functions to Graphs / Contour Plots High

Methods to Solve

18. Complex Number Arithmetic (a + bi Form) High

Methods to Solve

19. Indeterminate Form Limits (1∞, 00, ∞0) High

Methods to Solve

20. Higher-Order Derivative from Power Series High

Methods to Solve

21. Geometric / Telescoping Series (Find Sum) High

Methods to Solve

22. Inverse Function (Formula + Domain) Low

Methods to Solve

23. Inverse Trig Simplification Low

Methods to Solve

24. Extreme Values on Closed Interval (Single Variable) Low

Methods to Solve

25. Symmetry in Definite Integrals (Odd/Even Functions) Low

Methods to Solve

26. Sketch Maximal Domain (Multivariable) Low

Methods to Solve

27. Limits at Infinity / Squeeze Theorem Low

Methods to Solve

28. Limits via Taylor Series Expansion Low

Methods to Solve

29. FTC Application / Critical Points via FTC Low

Methods to Solve

30. Euler's Formula / Complex Trig Identities Low

Methods to Solve

31. Area Between Curves Low

Methods to Solve

32. Mean Value Theorem Application Low

Methods to Solve

33. Newton-Raphson Method Low

Methods to Solve

4. Complex Roots (zⁿ = c) Very High

19. Indeterminate Form Limits (1^∞, 0⁰, ∞⁰) High