A-level maths questions have always been considered a difficult test of analytical thinking and problem-solving abilities. Yet every year, there is a new set of the hardest A-Level Maths questions.

Without proper preparation, these exams can strike fear into the hearts of even the most confident students.

But interestingly, these equations, proofs, and problems are a rite of passage for many aspiring scientists, engineers, and mathematicians.

That is why this article examines the top ten hardest A-Level Maths questions in 2024. These questions, which range from complicated calculus problems to abstract algebraic questions, stretch the boundaries of mathematical understanding and encourage students to study harder for their A-level exams.

## Table of contents

## Top 10 Hardest A-Level Maths Questions in 2024

Here are top 10 A-level maths questions that are hard:

### 1. The Calculus Conundrum

**Question**: Consider the function f(x, y) = x^3 + 2xy^2 – 5x^2y. Determine the exact coordinates (x, y) of all critical points of this function and classify each point as a local maximum, local minimum, or saddle point.

**Solution**: To find the critical points of the function f(x, y), we need to compute its partial derivatives concerning both x and y and set them equal to zero.

Partial derivative concerning x: ∂f/∂x = 3x^2 + 2y^2 – 10xy.

Partial derivative for y: ∂f/∂y = 4xy – 5x^2.

Setting both partial derivatives to zero, we have the following system of equations: 3x^2 + 2y^2 – 10xy = 0 4xy – 5x^2 = 0

Solving the second equation for y in terms of x: y = (5x^2) / 4x = (5x) / 4.

Substitute this value of y into the first equation: 3x^2 + 2((5x) / 4)^2 – 10x((5x) / 4) = 0 3x^2 + (25x^2) / 8 – (25x^2) / 2 = 0 3x^2 – (25x^2) / 8 = 0 x^2 = (25x^2) / 8 8x^2 = 25x^2 8x^2 – 25x^2 = 0 -17x^2 = 0 x^2 = 0 x = 0.

Substitute the value of x back into the equation y = (5x) / 4: y = (5 * 0) / 4 = 0.

So, we have a single critical point at (0, 0).

To classify this point, we’ll use the second derivative test. Compute the second partial derivatives:

Second partial derivative for x: ∂²f/∂x² = 6x – 10y.

The Second partial derivative for y: ∂²f/∂y² = 4x.

Second mixed partial derivative: ∂²f/∂x∂y = -10x + 4y.

Now substitute the critical point (0, 0) into these second partial derivatives: ∂²f/∂x² = 6(0) – 10(0) = 0 ∂²f/∂y² = 4(0) = 0 ∂²f/∂x∂y = -10(0) + 4(0) = 0.

Since the second derivative test is inconclusive, we can’t determine the nature of the critical point using this method. Therefore, further analysis is important, possibly involving the use of higher-order derivatives or the behavior of the function in the vicinity of the critical point.

In conclusion, the function f(x, y) = x^3 + 2xy^2 – 5x^2y has a critical point at (0, 0), but its classification as a local maximum, local minimum, or saddle point requires additional investigation beyond the second derivative test. This question highlights the intricate nature of multivariable calculus and challenges students to explore deeper into the behavior of functions at critical points.

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### 2. Abstract Algebra Odyssey

**Question**: Let G be a group with the operation *, and let H be a proper subgroup of G. Prove that for any element g in G, either gHg⁻¹ is a subset of H, or gHg⁻¹ intersects H nontrivially (i.e., contains an element other than the identity element).

**Solution:** To prove this statement, we’ll consider two cases: when gHg⁻¹ is a subset of H and when gHg⁻¹ intersects H nontrivially.

Case 1: gHg⁻¹ is a subset of H. Assume that gHg⁻¹ is a subset of H. This implies that for every h in H, there exists an element h’ in H such that ghg⁻¹ = h.’ Multiplying both sides by g⁻¹ on the right gives us gh = h’g⁻¹. Since both h’ and g⁻¹ are in H, their product h’g⁻¹ is also in H. Thus, gh is in the left coset Hg⁻¹ of H concerning g⁻¹. However, Hg⁻¹ is the same as (gHg⁻¹)⁻¹, so gh is in the same left coset as gHg⁻¹, which is H. This implies that gh is in H, and the proof is complete for this case.

#### Case 2

In both cases, we have shown that for any element g in G, either gHg⁻¹ is a subset of H or gHg⁻¹ intersects H nontrivially. This completes the proof.

: gHg⁻¹ intersects H nontrivially. Assume that gHg⁻¹ intersects H nontrivially, which means there exists an element k in gHg⁻¹ that is also in H. Then, we have k = ghg⁻¹ for some h in H. Rearranging gives us gh = kg. Since both k and g are in H, their product kg is also in H. Thus, gh is in the left coset Hg of H for g. This implies that gh = h’g for some h’ in H. Multiplying both sides by g⁻¹ on the right gives us h = h’, which means h is in H. Hence, gHg⁻¹ intersects H nontrivially.

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### 3. The Fractal Frontier

**Question:** Consider a self-replicating fractal pattern known as the “Dragon Curve.” The Dragon Curve is constructed iteratively by taking the previous curve, rotating it 45 degrees, and then adding another curve. The process starts with a single line segment.

After the first iteration, how many line segments are there?

If each iteration doubles the number of line segments, how many line segments are there after the 10th iteration?

Calculate the Hausdorff dimension of the Dragon Curve fractal.

**Solution:**

After the first iteration, there are two line segments. The original segment and the one added after the rotation.

If each iteration doubles the number of line segments, then after the 10th iteration, there will be 2^10 = 1024 line segments.

The Hausdorff dimension of a fractal characterizes its level of self-replication and complexity. For the Dragon Curve, it’s a non-integer value between 1 and 2.

Calculating the Hausdorff dimension requires a more involved mathematical process. You will do it by covering the fractal with smaller shapes (often balls or cubes) and then determining how the number of these shapes scales with their size.

The Dragon Curve’s Hausdorff dimension is approximately 1.523. This value signifies the fractal’s intricate, non-integer dimensionality, which indicates its self-similar yet infinitely complex nature.

This question dives into fractal geometry, requiring students to understand the iterative construction of the Dragon Curve and its properties. It challenges students to apply their knowledge of geometric patterns, recursive processes, and mathematical limits to explore the fascinating world of fractals and their dimensions.

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### 4. Analytical Number Theory Triumph

**Question**: The Goldbach Conjecture posits that every even integer greater than 2 can become the sum of two prime numbers. In this problem, we’ll consider a slight variation: prove that every even integer greater than 4 can be expressed as the sum of two distinct prime numbers and one power of 2 (2^n for n > 0).

**Solution:** To prove this conjecture, we’ll proceed by contradiction. Assume that there exists an even integer N > 4 that cannot be expressed as the sum of two distinct prime numbers and one power of 2.

Even and Composite: Since N is even and greater than 4, it can be N = 2k for some positive integer k.

Odd Prime Candidates: Let’s consider the set of odd primes less than N. These primes are 2^1, 2^2, …, 2^(n-1), where n is the highest power of 2 less than or equal to k.

Pairs and Remainders: For each of these odd primes 2^i, calculate N – 2^i. The result is an even integer less than N. By the assumption of our contradiction, this number must be composite (not prime). Therefore, it can be expressed as the product of two distinct primes, say p and q.

Contradiction: We now have N = 2^i + p + q. But since N is even, it cannot be equal to the sum of an odd number (2^i) and two odd numbers (p and q). This leads to a contradiction.

Hence, our assumption that there exists an even integer greater than 4 that cannot be expressed as the sum of two distinct prime numbers and one power of 2 is false.

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### 5. Quantum Mechanics Mathematics

**Question:** In quantum mechanics, consider a quantum system with a Hamiltonian operator given by H = -ħ²/2m ∇² + V(x), where ħ is the reduced Planck constant, m is the particle’s mass, ∇² represents the Laplacian operator, and V(x) is the potential energy function.

Eigenvalue Problem: For a particle in a one-dimensional box of length L with a constant potential V(x) = V₀, solve the time-independent Schrödinger equation Hψ(x) = Eψ(x), where E is the eigenenergy and ψ(x) is the wavefunction.

Normalization Condition: Normalize the wavefunction ψ(x) such that ∫|ψ(x)|² dx = 1 over the entire domain of the box (0 ≤ x ≤ L).

Solution:

Eigenvalue Problem: In this case, the potential energy V(x) is a constant, so the Schrödinger equation becomes -ħ²/2m ∇² ψ(x) + V₀ψ(x) = Eψ(x). We can rearrange this to obtain the differential equation ∇² ψ(x) = -2m(E – V₀) / ħ² ψ(x).

The general solution to this differential equation within the box is ψ(x) = A sin(kx) + B cos(kx), where A and B are constants, and k = √(2m(E – V₀) / ħ²).

Normalization Condition: To normalize the wavefunction, we need to find the constants A and B such that ∫|ψ(x)|² dx = 1. The integral will involve terms like ∫sin²(kx) dx and ∫cos²(kx) dx, which evaluate to constants.

The normalized wave function will be ψ(x) = (2/L)^(1/2) sin(kx), where L is the length of the box.

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### 6. Differential Equation Dilemma

Question: Consider a dynamic system described by the following partial differential equation:

∂u/∂t = a² (∂²u/∂x²) – b * u * (∂u/∂x)

where u(x, t) represents a scalar function describing the system’s behavior, and a and b are constants.

Initial and Boundary Conditions: Solve the partial differential equation subject to the initial condition u(x, 0) = f(x), where f(x) is a given function, and the boundary conditions u(0, t) = u(L, t) = 0, where L is a fixed length.

Stability Analysis: Investigate the stability of the system by analyzing the behavior of solutions over time. Discuss how the constants a and b influence the stability.

**Solution:**

Initial and Boundary Conditions: To solve this partial differential equation, we use the separation of variables and assume u(x, t) = X(x)T(t). Substituting this into the equation gives X(x)T'(t) = a²X”(x)T(t) – b * X(x)T(x) * X'(x)T(t).

Dividing by X(x)T(t) and rearranging terms yields T'(t)/T(t) = a²X”(x)/X(x) – b * X'(x)/X(x).

The left side is purely a function of t, and the right side is purely a function of x. Since they are equal, they must be equal to a constant, which we’ll call -λ².

This leads to two ordinary differential equations: T'(t)/T(t) = -λ² and a²X”(x)/X(x) – b * X'(x)/X(x) = -λ².

Solving the equation for T(t) gives T(t) = e^(-λ²t).

Solving the equation for X(x) results in a trigonometric solution X(x) = A sin(λx) + B cos(λx).

Applying the boundary conditions u(0, t) = u(L, t) = 0 gives B = 0 and A sin(λL) = 0. Since A cannot be zero, sin(λL) = 0. This implies that λL = nπ, where n is an integer.

Thus, λ = nπ / L. Substituting this into the equation for T(t) gives T_n(t) = e^(-a²(nπ / L)²t).

The general solution is then u(x, t) = ∑(A_n sin(nπx / L) e^(-a²(nπ / L)²t)).

Stability Analysis: The stability of the system depends on the behavior of the exponential terms e^(-a²(nπ / L)²t). If all these terms decay as t increases, the system is stable. The stability is influenced by the constants a and b: larger values of a promote stability, while larger values of b can lead to instability.

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### 7. Chaos Theory Challenge

**Question**: Consider the chaos theory with the famous logistic map, a simple mathematical model that exhibits chaotic behavior. The logistic map is defined by the recurrence relation:

xₙ₊₁ = r * xₙ * (1 – xₙ)

where xₙ represents the population at time step n, and r is a constant parameter that controls the behavior of the system.

Steady States and Bifurcations: Analyze the logistic map’s behavior for different values of r. Identify the steady states of the system and describe how bifurcations occur as r changes.

Period-Doubling Route to Chaos: Investigate the phenomenon of period-doubling route to chaos, where the system transitions from periodic behavior to chaotic behavior. Determine the critical value of r at which this transition occurs.

**Solution:**

Steady States and Bifurcations: Steady states are points where the population remains constant from one-time step to the next. For the logistic map, steady states occur when xₙ₊₁ = xₙ, leading to the equation x = r * x * (1 – x). Solving for x gives two steady states: x = 0 and x = 1 – 1/r.

As the parameter r changes, the behavior of the logistic map undergoes bifurcations. Bifurcations are points at which the behavior of the system abruptly changes. For certain values of r, the system exhibits periodic behavior with stable attractors, while for other values, the system displays chaotic behavior.

Period-Doubling Route to Chaos: As the parameter r increases, the logistic map goes through a sequence of period-doubling bifurcations, where the period of the system doubles. At a critical value of r, the system transitions from periodic behavior to chaotic behavior.

The critical value of r at which this transition occurs is approximately 3.57. Beyond this point, the logistic map enters a regime of chaotic behavior characterized by sensitive dependence on initial conditions and the absence of predictable patterns.

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### 8. Real Analysis Riddle

**Question:** Explore real analysis with a challenging problem involving the convergence of a sequence of functions. Consider a sequence of functions {fₙ(x)} defined on the closed interval [0, 1]:

fₙ(x) = n²x(1 – x) if 1/(n+1) ≤ x ≤ 1/n, fₙ(x) = 0 otherwise.

Pointwise Convergence: Investigate the pointwise convergence of the sequence {fₙ(x)} as n approaches infinity. Determine the pointwise limit function.

Uniform Convergence Dilemma: Explore the uniform convergence of the sequence {fₙ(x)}. Determine whether the sequence converges uniformly on the interval [0, 1] and provide a rigorous justification for your conclusion.

**Solution:**

Pointwise Convergence: To investigate pointwise convergence, we analyze the behavior of the functions fₙ(x) as n approaches infinity for each value of x.

For x = 0, fₙ(x) = 0 for all n, so the limit is 0.

Then, for x = 1, fₙ(x) = 0 for all n, so the limit is 0.

For 0 < x < 1, as n approaches infinity, the interval [1/(n+1), 1/n] becomes a single point, x. Thus, the limit of fₙ(x) = nx(1 – x) as n approaches infinity is nx(1 – x).

The pointwise limit function is: f(x) = { 0, for x = 0 or x = 1, { x(1 – x), for 0 < x < 1.

Uniform Convergence Dilemma: For uniform convergence, we need to examine how close the sequence {fₙ(x)} comes to the pointwise limit function for all x in the interval [0, 1] as n approaches infinity.

Notice that for any x in the interval (0, 1), as n becomes larger, the term nx(1 – x) approaches infinity. This implies that the sequence {fₙ(x)} does not approach the pointwise limit function uniformly on the interval [0, 1].

Therefore, the sequence {fₙ(x)} does not converge uniformly on the interval [0, 1].

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### 9. Cryptography Conundrum

**Question:** Explore cryptography with a challenging problem involving number theory and modular arithmetic. Consider a cryptographic system that uses the RSA algorithm for encryption and decryption. You are given two prime numbers, p = 17 and q = 19, and the public exponent e = 5.

Encryption: Suppose you want to encrypt a message M using the recipient’s public key (n, e), where n = p * q. If the message is represented as a number M = 9, calculate the ciphertext C using the encryption formula C ≡ M^e (mod n).

Decryption Dilemma: Explore the decryption process using the recipient’s private key (d), where d is the modular multiplicative inverse of e modulo φ(n), and φ(n) is the Euler’s totient function evaluated at n. Calculate the value of d and then decrypt the ciphertext C calculated in the previous question to retrieve the original message M.

**Solution:**

Encryption: Given p = 17 and q = 19, we have n = p * q = 17 * 19 = 323. The public exponent is e = 5, and the message is M = 9.

Using the encryption formula, C ≡ M^e (mod n): C ≡ 9^5 (mod 323).

Calculating 9^5 and finding the remainder when divided by 323 gives C ≡ 292.

Decryption Dilemma: To calculate the private key d, we first need to find φ(n), which is the number of positive integers less than n that are coprime to n. Since n = p * q, and p and q are prime, φ(n) = (p – 1) * (q – 1) = 16 * 18 = 288.

To find the modular multiplicative inverse of e modulo φ(n), we need to solve the equation e * d ≡ 1 (mod φ(n)). In this case, e = 5 and φ(n) = 288. Using the extended Euclidean algorithm, we find that d ≡ 173.

Now, we can use the decryption formula M ≡ C^d (mod n): M ≡ 292^173 (mod 323).

Calculating 292^173 and finding the remainder when divided by 323 gives M ≡ 9.

Thus, the original message M is retrieved successfully.

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### 10. Multivariable Integration Maze

**Question:** Explore multivariable calculus in a complex integration problem involving multiple intersecting surfaces. Consider the region R bounded by the surfaces z = x² + y² and z = 8 – x² – y².

Region of Integration: Determine the boundaries of the region R in the xy-plane where the given surfaces intersect.

Volume Calculation: Calculate the volume of the region R using a double integral in cylindrical coordinates.

**Solution:**

Region of Integration: To find the boundaries of the region R, set the two surface equations equal to each other and solve for z: x² + y² = 8 – x² – y²

Simplify the equation: 2x² + 2y² = 8 x² + y² = 4

This equation represents a circle in the xy-plane with radius 2 centered at the origin. Therefore, the boundaries of the region R in the xy-plane are given by the circle’s equation: x² + y² = 4.

Volume Calculation: In cylindrical coordinates, the volume element is given by dV = r dr dθ dz. The boundaries for r, θ, and z are as follows:

r varies from 0 to 2 (radius of the circle)

θ varies from 0 to 2π (360 degrees)

z varies from x² + y² to 8 – x² – y²

The volume V of the region R can be calculated using the following double integral: V = ∫∫∫ dV = ∫₀² ∫₀^(2π) ∫_(r²)^(8-r²) r dz dθ dr.

Integrating with respect to z and θ first, then with respect to r, yields: V = ∫₀² ∫₀^(2π) (8r – r³) dθ dr V = 16π ∫₀² (r – r³/3) dr V = 16π [r²/2 – r⁴/12] from 0 to 2 V = 16π [2 – 16/12] V = 16π [3/2] V = 24π.

The volume of the region R bounded by the surfaces z = x² + y² and z = 8 – x² – y² is 24π cubic units.

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## Conclusion

The top 10 hardest A-Level Maths questions are proof of the boundless nature of human curiosity and the unending demand for knowledge. Let these challenging maths questions guide you in the pursuit of the unknown that yields rewards far beyond numbers and equations.

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## Frequently Asked Questions

**What criteria were used to determine the hardest A-Level Maths questions?**

The criteria encompassed the complexity of mathematical concepts involved, the depth of analytical thinking required, and the incorporation of multiple branches of mathematics.

**Are these questions aligned with the standard A-Level Mathematics curriculum?**

Yes, these questions are designed to align with the advanced levels of the A-Level Mathematics curriculum.

**How can tackling these challenging questions benefit students?**

They hone critical thinking skills, deepen understanding of complex mathematical concepts, and cultivate perseverance.

## References

- thinkstudent.co.uk– hardest A-level maths questions
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