IB Math AI HL Paper 2 Exam Style Practice Questions

IB HL Paper 2 Sample Questions

Set 1: Number and Algebra • Set 2: Functions

⭐ About These Sample Sets

These sample Paper 2 questions are designed to show readers what calculator-active IB Mathematics HL questions look like. Each set contains three questions with full working, clear structure, and markscheme-style guidance.

Paper 2 often rewards method, interpretation, technology use, and accurate communication of results.

Set 1 — Number and Algebra

Question 1

An arithmetic sequence has first term $12$ and common difference $5$ .

(a) Find the 30th term.

(b) Find the sum of the first 30 terms.

(c) Determine the smallest value of $n$ such that the sum of the first $n$ terms exceeds $2000$ .

Solution

For an arithmetic sequence,

$u_n=u_1+(n-1)d$

and

$S_n=\frac{n}{2}\left(2u_1+(n-1)d\right)$

(a) 30th term

$u_{30}=12+(30-1)5$

$=12+145=157$

$\boxed{157}$

📊 Marks: M1 A1

(b) Sum of the first 30 terms

$S_{30}=\frac{30}{2}\left(2(12)+(30-1)5\right)$

$=15(24+145)$

$=15(169)=2535$

$\boxed{2535}$

📊 Marks: M1 A1

(c) Smallest $n$ with $S_n>2000$

Solve

$\frac{n}{2}\left(24+5(n-1)\right)>2000$

$\frac{n}{2}(19+5n)>2000$

$n(19+5n)>4000$

$5n^2+19n-4000>0$

Using the quadratic formula:

$n=\frac{-19\pm\sqrt{19^2-4(5)(-4000)}}{10}$

$=\frac{-19\pm\sqrt{361+80000}}{10}$

$=\frac{-19\pm\sqrt{80361}}{10}$

$\sqrt{80361}\approx283.48$

$n\approx\frac{-19+283.48}{10}=26.448$

Thus the smallest integer value is

$\boxed{27}$

📊 Marks: M1 M1 A1

Question 2

A population model is given by

$P(t)=8500(1.034)^t$

where $t$ is the number of years after 2020.

(a) Find the population in 2028.

(b) Find the annual percentage growth rate.

(c) Determine the year in which the population first exceeds $11000$ .

Solution

(a) Population in 2028

Since 2028 is 8 years after 2020,

$t=8$

$P(8)=8500(1.034)^8$

$\approx8500(1.30674)$

$\approx11107.3$

So the population is approximately

$\boxed{11107}$

📊 Marks: M1 A1

(b) Annual percentage growth rate

Since the multiplier is

$1.034$

the annual growth rate is

$0.034=3.4\%$

$\boxed{3.4\%}$

📊 Marks: M1 A1

(c) First year population exceeds 11000

Solve

$8500(1.034)^t>11000$

$(1.034)^t>\frac{11000}{8500}$

$(1.034)^t>1.29412$

Take logarithms:

$t>\frac{\ln(11000/8500)}{\ln(1.034)}$

$t>\frac{\ln(1.29412)}{\ln(1.034)}$

$t>7.67$

So the first whole year is

$t=8$

Therefore the year is

$2020+8=2028$

$\boxed{2028}$

📊 Marks: M1 M1 A1

Question 3

Let

$z_1=4+3i,\qquad z_2=2-5i$

(a) Find $z_1z_2$ .

(b) Express $\dfrac{z_1}{z_2}$ in the form $a+bi$ .

(c) Find the modulus and argument of $z_1$ .

Solution

(a)

$z_1z_2=(4+3i)(2-5i)$

$=8-20i+6i-15i^2$

$=8-14i+15$

$=23-14i$

$\boxed{23-14i}$

📊 Marks: M1 A1

(b)

$\frac{4+3i}{2-5i}$

Multiply numerator and denominator by the conjugate $2+5i$ :

$\frac{4+3i}{2-5i}\cdot\frac{2+5i}{2+5i} = \frac{(4+3i)(2+5i)}{2^2+5^2}$

$= \frac{8+20i+6i+15i^2}{29}$

$= \frac{8+26i-15}{29}$

$= \frac{-7+26i}{29}$

$\boxed{-\frac{7}{29}+\frac{26}{29}i}$

📊 Marks: M1 M1 A1

(c) Modulus and argument of $z_1$

$|z_1|=\sqrt{4^2+3^2}=\sqrt{25}=5$

Since $z_1$ lies in quadrant I,

$\arg(z_1)=\tan^{-1}\left(\frac{3}{4}\right)$

$\approx0.644\text{ rad}$

$\boxed{|z_1|=5,\qquad \arg(z_1)\approx0.644\text{ rad}}$

📊 Marks: M1 A1

Set 2 — Functions

Question 1

Consider the function

$f(x)=x^2-4x-5$

(a) Find the coordinates of the vertex.

(b) Find the zeros of $f$ .

(c) Hence sketch the graph, showing the intercepts and the axis of symmetry.

Solution

(a) Vertex

For

$f(x)=ax^2+bx+c$

the vertex occurs at

$x=-\frac{b}{2a}=-\frac{-4}{2(1)}=2$

Now

$f(2)=2^2-4(2)-5=4-8-5=-9$

So the vertex is

$\boxed{(2,-9)}$

📊 Marks: M1 A1

(b) Zeros

Solve

$x^2-4x-5=0$

Factorise:

$(x-5)(x+1)=0$

So the zeros are

$\boxed{x=5,\qquad x=-1}$

📊 Marks: M1 A1

(c) Sketch features

The $y$ -intercept is

$f(0)=-5$

so the graph passes through

$(0,-5)$

The axis of symmetry is

$x=2$

So the key features are:

$\boxed{\text{vertex }(2,-9),\quad x\text{-intercepts }(-1,0)\text{ and }(5,0),\quad y\text{-intercept }(0,-5),\quad \text{axis }x=2}$

📊 Marks: M1 A1

Question 2

Let

$f(x)=\frac{3x+1}{x-2}$

(a) Find $f^{-1}(x)$ .

(b) Verify that $f(f^{-1}(x))=x$ .

(c) State the restriction on the domain of $f^{-1}$ .

Solution

(a) Find the inverse

Let

$y=\frac{3x+1}{x-2}$

Multiply both sides by $(x-2)$ :

$y(x-2)=3x+1$

$yx-2y=3x+1$

$yx-3x=2y+1$

$x(y-3)=2y+1$

$x=\frac{2y+1}{y-3}$

Swap $x$ and $y$ :

$\boxed{f^{-1}(x)=\frac{2x+1}{x-3}}$

📊 Marks: M1 M1 A1

(b) Verify

$f\left(\frac{2x+1}{x-3}\right) = \frac{3\left(\frac{2x+1}{x-3}\right)+1}{\left(\frac{2x+1}{x-3}\right)-2}$

Simplify numerator:

$3\left(\frac{2x+1}{x-3}\right)+1 = \frac{6x+3}{x-3}+\frac{x-3}{x-3} = \frac{7x}{x-3}$

Simplify denominator:

$\left(\frac{2x+1}{x-3}\right)-2 = \frac{2x+1}{x-3}-\frac{2x-6}{x-3} = \frac{7}{x-3}$

$f\left(\frac{2x+1}{x-3}\right) = \frac{\frac{7x}{x-3}}{\frac{7}{x-3}}=x$

Hence

$\boxed{f(f^{-1}(x))=x}$

📊 Marks: M1 M1 A1

(c) Restriction

For

$f^{-1}(x)=\frac{2x+1}{x-3}$

the denominator cannot be zero, so

$\boxed{x\ne3}$

📊 Marks: M1 A1

Question 3

A Ferris wheel is modelled by

$h(t)=16\cos\bigl(45(t-4)\bigr)+19$

where $h$ is the height in metres and $t$ is the time in minutes.

(a) State the amplitude, midline, and period.

(b) Find the maximum and minimum heights.

(c) Find the time when the rider first reaches the top.

(d) State a reasonable domain and range for one complete revolution.

Solution

(a) Amplitude, midline, period

From

$h(t)=16\cos\bigl(45(t-4)\bigr)+19$

the amplitude is

$\boxed{16}$

The midline is

$\boxed{h=19}$

Since the model is in degrees, the period is

$\frac{360}{45}=8$

So the period is

$\boxed{8\text{ minutes}}$

📊 Marks: M1 A1

(b) Maximum and minimum heights

$\text{Maximum}=19+16=35$

$\text{Minimum}=19-16=3$

$\boxed{35\text{ m},\quad 3\text{ m}}$

📊 Marks: M1 A1

(c) First time rider reaches the top

The cosine function is at a maximum when its angle is $0^\circ$ .

$45(t-4)=0$

$t=4$

Thus the rider first reaches the top at

$\boxed{4\text{ minutes}}$

📊 Marks: M1 A1

(d) Reasonable domain and range for one revolution

One full revolution lasts 8 minutes, so a reasonable domain is

$\boxed{0\le t\le 8}$

The heights range from 3 m to 35 m, so

$\boxed{3\le h\le 35}$

📊 Marks: M1 A1

📘 Want More Full IB HL Practice?

These are sample Paper 2 sets for readers. For full unit-by-unit lessons, structured practice, markscheme-style solutions, and more complete exam preparation, please visit the course page.

IB Math AI HL Paper 2 Exam Style Practice Questions

IB HL Paper 2 Sample Questions

⭐ About These Sample Sets

Set 1 — Number and Algebra

Solution

Solution

Solution

Set 2 — Functions

Solution

Solution

Solution

📘 Want More Full IB HL Practice?

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IB HL Paper 2 Sample Questions

⭐ About These Sample Sets

Set 1 — Number and Algebra

Solution

Solution

Solution

Set 2 — Functions

Solution

Solution

Solution

📘 Want More Full IB HL Practice?

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