5 Hardest IB Math HL Questions to Practice (With Solutions & Tips)

Topic: Number & Algebra + Calculus | Paper: Paper 1 (non-calculator) | Marks: ~16

The Question:

Consider the function $f(x) = (1 - ax)^{-1/2}$.

• (a) Prove by mathematical induction that $f^{(n)}(x) = \frac{a^n(2n-1)!(1-ax)^{-(2n+1)/2}}{2^{2n} \cdot n! \cdot (n-1)!}$, for $n \in \mathbb{Z}^+$.
• (b) Show that the Maclaurin series for $f(x)$ up to the $x^2$ term is $1 + \frac{1}{2}ax + \frac{3}{8}a^2x^2$.
• (c) Hence show that $(1-2x)^{-1/2}(1-4x)^{-1/2} \approx \frac{2 + 6x + 19x^2}{2}$.

Why It's Challenging:

The 6-mark induction proof requires manipulating nested factorials and fractional exponents during the inductive step. A single sign error cascades through the entire proof. Parts (c) onwards demand creative application of series to approximate $\sqrt{3}$, a conceptual leap many students struggle with under exam pressure.

Solution Approach:

For the induction proof, verify the base case $n=1$ by computing $f'(x)$ directly. Assume the formula holds for $n=k$, then differentiate to obtain $f^{(k+1)}(x)$ and show it matches the formula for $n=k+1$ through careful factorial manipulation.

For the Maclaurin series, evaluate $f(0)$, $f'(0)$, and $f''(0)$ using your proven formula, then construct the polynomial. Multiply the two series with $a=2$ and $a=4$, collecting terms up to $x^2$.

Common Mistakes:

Failing to write the concluding statement ("Therefore, by the principle of mathematical induction, the statement is true for all $n \in \mathbb{Z}^+$") costs at least 1 mark. Students also frequently work backward from the "show that" answer, which earns zero marks.

Topic: Complex Numbers + Functions | Paper: Paper 3 | Marks: 28 out of 55

The Question:

This question asks you to explore cubic polynomials of the form $(x-r)(x^2-2ax+a^2+b^2)$ and corresponding cubic equations with one real root and two complex roots.

In parts (a) to (c), let $r=1$, $a=4$, $b=1$.

• (a)(i) Given that 1 and $4+i$ are roots of $(z-1)(z^2-8z+17) = 0$, write down the third root.
• (a)(ii) Verify that the mean of the two complex roots is 4.
• (b) Show that the line $y = x-1$ is tangent to the curve $y = f(x)$ at the point $A(4,3)$.
• (c) Sketch the curve $y = f(x)$ and the tangent to the curve at point A.

For general $g(x) = (x-r)(x^2-2ax+a^2+b^2)$:

• (d)(i) Show that $g'(x) = 2(x-r)(x-a) + x^2-2ax+a^2+b^2$.
• (d)(ii) Hence prove that the tangent to $y = g(x)$ at $A(a, g(a))$ intersects the x-axis at $R(r, 0)$.
• (e) Deduce that the complex roots can be expressed as $a \pm \sqrt{g'(a)} \cdot i$.

Why It's Challenging:

Worth over half the marks on Paper 3, this investigation demands sustained mathematical reasoning across multiple domains. Part (d)(ii) requires proving a geometric relationship (that the tangent at a specific point passes through the real root on the x-axis) using algebraic differentiation. Part (e) then asks you to deduce that complex roots can be expressed as $a \pm \sqrt{g'(a)} \cdot i$, a creative leap connecting differentiation to imaginary components. Most students have never seen complex roots connected to tangent lines geometrically, making this genuinely novel under exam conditions.

Solution Approach:

For part (a), the third root is the conjugate $4-i$, and the mean of $4+i$ and $4-i$ equals 4. For part (b), calculate $f(4) = 3$ and $f'(4) = 1$, confirming the tangent equation $y = x-1$.

For part (d)(i), apply the product rule to $g(x) = (x-r)(x^2-2ax+a^2+b^2)$. For part (d)(ii), find the tangent line equation at $x=a$, substitute $x=r$, and show $y=0$. For part (e), recognise that $g'(a) = b^2$, so complex roots are $a \pm bi = a \pm \sqrt{g'(a)} \cdot i$.

Common Mistakes:

Students waste time on part (b) trying to verify tangency by solving the cubic rather than simply checking that $f(4)=3$ and $f'(4)=1$. In Paper 3, not reading instructions carefully and missing the scaffolded structure costs significant marks.

Topic: Geometry & Trigonometry | Paper: Paper 1 (non-calculator) | Marks: ~16

The Question:

Consider the three planes:

• $\Pi_1: 2x - y + z = 4$
• $\Pi_2: x - 2y + 3z = 5$
• $\Pi_3: -9x + 3y - 2z = 32$

• (a) Show that the three planes do not intersect.
• (b)(i) Verify that the point $P(1, -2, 0)$ lies on both $\Pi_1$ and $\Pi_2$.
• (b)(ii) Find a vector equation of $L$, the line of intersection of $\Pi_1$ and $\Pi_2$.
• (c) Find the distance between $L$ and $\Pi_3$.

Why It's Challenging:

This multi-step 3D geometry problem chains four distinct techniques together with no calculator. Part (c) is the killer. You must recognise that line $L$ is parallel to $\Pi_3$ by verifying the direction vector is perpendicular to $\Pi_3$'s normal vector, then apply the point-to-plane distance formula. The entire question requires manual arithmetic, making computational errors devastatingly common.

Solution Approach:

For part (a), eliminate variables between $\Pi_1$ and $\Pi_2$ to get $-3x + z = -3$, then between $\Pi_2$ and $\Pi_3$ to get $-3x + z = 44$. Since $-3 \neq 44$, no common solution exists.

For part (b)(i), substitute $P(1,-2,0)$ into both equations to verify. For part (b)(ii), find the direction vector using $\mathbf{d} = \mathbf{n}_1 \times \mathbf{n}_2 = (2,-1,1) \times (1,-2,3) = (-1,-5,-3)$, simplify to $(1,5,3)$. Line equation: $\mathbf{r} = (1,-2,0) + \lambda(1,5,3)$.

For part (c), verify $\mathbf{d} \cdot \mathbf{n}_3 = (1)(-9) + (5)(3) + (3)(-2) = 0$, confirming $L \parallel \Pi_3$. Calculate distance $= \frac{|-9(1) + 3(-2) - 2(0) - 32|}{\sqrt{81+9+4}} = \frac{47}{\sqrt{94}}$.

Common Mistakes:

Students often confuse the distance between a line and a parallel plane with the distance between two planes, or forget to verify that $L$ is actually parallel to $\Pi_3$ before applying the formula. Cross product computation errors (especially sign errors) are extremely frequent without a calculator.

Topic: Statistics & Probability | Paper: Paper 2 | Marks: ~15

The Question:

A medical screening test for a rare disease has the following properties:

• The disease affects 0.5% of the population
• If a person has the disease, the test is positive 98% of the time (sensitivity)
• If a person does not have the disease, the test is negative 95% of the time (specificity)

• (a) Construct a tree diagram showing all possibilities.
• (b) Find the probability that a randomly selected person tests positive.
• (c) Given that a person tests positive, find the probability that they actually have the disease.
• (d) Two people are selected at random. Given that both test positive, find the probability that exactly one of them has the disease.

Why It's Challenging:

Bayes' theorem questions consistently produce some of the lowest success rates on IB Math HL exams. The core difficulty is conceptual. Even with a "98% accurate" test, the probability of actually having a rare disease given a positive result is shockingly low (approximately 9%), a counterintuitive result that confuses most students. Part (d) compounds this by requiring conditional probability calculations for two independent events. Students must condition on both testing positive, then consider the combinations of exactly one having the disease.

Solution Approach:

Define events: $D$ = has disease, $T^+$ = tests positive. Given: $P(D) = 0.005$, $P(T^+|D) = 0.98$, $P(T^-|D') = 0.95$, so $P(T^+|D') = 0.05$.

Calculate $P(T^+) = P(T^+|D) \cdot P(D) + P(T^+|D') \cdot P(D') = 0.98(0.005) + 0.05(0.995) = 0.05465$.

Apply Bayes' theorem: $P(D|T^+) = \frac{P(T^+|D) \cdot P(D)}{P(T^+)} = \frac{0.0049}{0.05465} \approx 0.0897$ (only 9%).

For part (d), let $p = P(D|T^+) \approx 0.0897$. Using the binomial distribution: $P(\text{exactly one has disease | both positive}) = 2pq$ where $q = 1-p \approx 0.9103$, giving approximately 0.163.

Common Mistakes:

Drawing the tree diagram incorrectly (putting test result first instead of disease status first), confusing conditional probabilities, and not recognising that $P(D|T^+) \neq P(T^+|D)$. Part (d) trips students who fail to recognise they need to work within the already-conditioned probability space.

Topic: Number & Algebra | Paper: Paper 1 (non-calculator) | Marks: 12

The Question:

• (a) Use the binomial theorem to expand $(\cos \theta + i \sin \theta)^5$.
• (b) Hence use De Moivre's theorem to prove that $\sin 5\theta = 5\cos^4\theta \sin \theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta$.

Why It's Challenging:

This question requires bridging two major HL topics (the binomial theorem and De Moivre's theorem) in a single proof. The expansion of $(\cos \theta + i \sin \theta)^5$ produces six terms, each requiring careful tracking of powers of $i$ (where $i^0=1$, $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, $i^5=i$). Students must then separate real and imaginary parts and equate imaginary components using De Moivre's identity. Without a calculator, the arithmetic is unforgiving. A single sign error in any of the six binomial terms cascades through the entire proof.

Solution Approach:

Apply binomial theorem: $(\cos \theta + i \sin \theta)^5 = \sum_{k=0}^{5} \binom{5}{k}(\cos \theta)^{5-k}(i \sin \theta)^k$.

Expand all six terms:

• $k=0$: $\cos^5\theta$
• $k=1$: $5i \cos^4\theta \sin \theta$
• $k=2$: $-10\cos^3\theta \sin^2\theta$
• $k=3$: $-10i \cos^2\theta \sin^3\theta$
• $k=4$: $5\cos \theta \sin^4\theta$
• $k=5$: $i \sin^5\theta$

By De Moivre's theorem: $(\cos \theta + i \sin \theta)^5 = \cos 5\theta + i \sin 5\theta$.

Equate imaginary parts: $\sin 5\theta = 5\cos^4\theta \sin \theta - 10\cos^2\theta \sin^3\theta + \sin^5\theta$.

Common Mistakes:

Forgetting that $i^2 = -1$ (not $+1$), miscalculating binomial coefficients for the 5th power, and failing to correctly identify which terms are imaginary. Some students try to prove this using addition formulas for $\sin(A+B)$ applied repeatedly, a valid but far more laborious method that usually runs out of time.