IB Mathematics Analysis & Approaches HL Practice Exams

Sit short skills checks and paper-style sets written for this course. Published sets include the questions and a mark scheme.

Practice Exams

Resources for IB Mathematics Analysis & Approaches HL

Topic skills check — calculus and algebra

Four original short questions (about 20 minutes) with a mark scheme. Non-calculator.

iStudyPlus original. This set was written for iStudyPlus. It is not an official IB examination, it does not use a real session code, and it is not copied from an IB past paper. Official IB examination papers and markschemes remain the property of the International Baccalaureate Organization.

Course: IB Mathematics Analysis and Approaches HL. Time: about 20 minutes. Calculator: not used. Total: 15 marks.

Attempt every part before you read the mark scheme. Keep exact form.

1. Differentiate $(2x-1)^{5}$ with respect to $x$. [3]

2. Solve the equations. [4]

  1. $2^{x} = 16$. [2]
  2. $log_{3}(x) = 2$. [2]

3. Find $int (6x^{2}-4),dx$. [3]

4. The curve $y = x^{2}$ has a tangent at $x = 1$. Find the equation of the normal to the curve at that point. [5]

Mark scheme

1. Chain rule: derivative of the inside is $2$ [1]. Factor $5(2x-1)^{4}$ [1]. Final answer $10(2x-1)^{4}$ [1].

2. (a) $16 = 2^{4}$, so $x = 4$ [2]. (b) $x = 3^{2} = 9$ [2].

3. $2x^{3} – 4x$ [2] plus a constant of integration [1].

4. Point $(1,1)$ [1]. $frac{dy}{dx} = 2x$, so the tangent gradient is $2$ [1]. Normal gradient $-frac{1}{2}$ [1]. Equation $y – 1 = -frac{1}{2}(x-1)$ [1], simplified to $y = -frac{1}{2}x + frac{3}{2}$ or $x + 2y – 3 = 0$ [1].

Paper 1 style set (non-calculator)

Original 46-mark non-calculator set: sequences, logarithms, trigonometry, calculus, and the binomial theorem.

iStudyPlus original. This set was written for iStudyPlus. It is not an official IB examination, it does not use a real session code, and it is not copied from an IB past paper. Official IB examination papers and markschemes remain the property of the International Baccalaureate Organization.

Sitting: shortened Paper 1 style. Time: 55 minutes. Calculator: not used. Total: 46 marks. Show the method a marker would need to follow.

1. An arithmetic sequence has first term $3$ and common difference $2$. [8]

  1. Find the 12th term. [2]
  2. Find the sum of the first 12 terms. [2]
  3. Find the least $n$ such that the sum of the first $n$ terms is greater than $200$. [4]

2. Let $f(x) = ln(2x-4)$, for $x gt 2$. [9]

  1. Find the exact value of $f(3)$. [2]
  2. Find an expression for $f^{-1}(x)$ and state its domain. [4]
  3. Solve $f(x) = ln 6$. [3]

3. Solve $sqrt{3}sin x – cos x = 1$ for $0 le x le 2pi$. Give exact answers. [9]

4. Let $f(x) = x^{3} – 6x^{2} + 9x + 1$. [12]

  1. Find $f'(x)$. [2]
  2. Find the coordinates of the stationary points and determine their nature. [5]
  3. Find the equation of the tangent to $y = f(x)$ at $x = 2$. [3]
  4. Find the exact value of $int_{0}^{1} f(x),dx$. [2]

5. [8]

  1. Find the coefficient of $x^{3}$ in the expansion of $(2-x)^{5}$. [3]
  2. Solve $log_{2}(x) + log_{2}(x-2) = 3$, where $x gt 2$. [5]

Mark scheme

1. (a) $u_{12} = 3 + 11 times 2 = 25$ [2]. (b) $S_{12} = frac{12}{2}(3+25) = 168$ [2]. (c) $S_{n} = n(n+2)$ [1]. $13 times 15 = 195$, which is not greater than $200$ [1]. $14 times 16 = 224$ [1]. Least $n$ is $14$ [1].

2. (a) $f(3) = ln 2$ [2]. (b) $y = ln(2x-4)$ leads to $x = frac{e^{y}+4}{2}$ [2], so $f^{-1}(x) = frac{e^{x}+4}{2}$ with domain $mathbb{R}$ [2]. (c) $2x-4 = 6$, so $x = 5$, and it satisfies $x gt 2$ [3].

3. Write $sqrt{3}sin x – cos x = 2sinleft(x-frac{pi}{6}right)$ [4]. Then $sinleft(x-frac{pi}{6}right) = frac{1}{2}$ [1]. Solutions in range: $x = frac{pi}{3}$ and $x = pi$ [4].

4. (a) $f'(x) = 3x^{2}-12x+9$ [2]. (b) $f'(x) = 3(x-1)(x-3)$, so stationary points at $x = 1$ and $x = 3$ [2]. Coordinates $(1,5)$ and $(3,1)$ [2]. $f”(x) = 6x-12$, so $(1,5)$ is a local maximum and $(3,1)$ is a local minimum [1]. (c) $f(2) = 3$, $f'(2) = -3$, tangent $y = -3x+9$ [3]. (d) $left[frac{x^{4}}{4}-2x^{3}+frac{9}{2}x^{2}+xright]_{0}^{1} = frac{15}{4}$ [2].

5. (a) Term in $x^{3}$ is $binom{5}{3} 2^{2}(-x)^{3} = -40x^{3}$, so the coefficient is $-40$ [3]. (b) $log_{2}left(x(x-2)right) = 3$ [1], so $x(x-2) = 8$ [1], $x^{2}-2x-8 = 0$ [1], $(x-4)(x+2) = 0$ [1]. Only $x = 4$ is allowed [1].

Paper 2 style set (calculator)

Original 40-mark calculator set: a cooling model, a normal distribution, and an exact-plus-numerical calculus question.

iStudyPlus original. This set was written for iStudyPlus. It is not an official IB examination, it does not use a real session code, and it is not copied from an IB past paper. Official IB examination papers and markschemes remain the property of the International Baccalaureate Organization.

Sitting: shortened Paper 2 style. Time: 50 minutes. Calculator: required for decimals; keep an exact line before you round. Total: 40 marks. Give numerical answers to 3 significant figures unless the question says otherwise.

1. The temperature of a drink, in $^{circ}mathrm{C}$, $t$ minutes after it is placed in a room is modelled by $T(t) = 22 + 58e^{-0.08t}$, for $t ge 0$. [14]

  1. Write down the initial temperature. [1]
  2. Find the temperature after 15 minutes. [3]
  3. Find the time at which the temperature is $30^{circ}mathrm{C}$. [4]
  4. State the temperature the model approaches as $t$ becomes large, and interpret it. [2]
  5. Find $T'(10)$ and interpret it in context, with units. [4]

2. The volume of juice dispensed by a machine, $X$ ml, is modelled by $X sim N(500, 4^{2})$. [12]

  1. Find $P(X lt 494)$. [3]
  2. Find $P(496 lt X lt 506)$. [4]
  3. Find the value $k$, to the nearest millilitre, such that $P(X gt k) = 0.05$. [5]

3. Let $y = xe^{-x/2}$ for $x ge 0$. [14]

  1. Find $frac{dy}{dx}$. [3]
  2. Find the exact coordinates of the maximum point on the curve. [4]
  3. Show that $int xe^{-x/2},dx = -2e^{-x/2}(x+2) + C$, and hence find the exact area bounded by the curve, the $x$-axis, and the lines $x = 0$ and $x = 4$. Give the area also to 3 significant figures. [5]
  4. Solve $xe^{-x/2} = 0.5$ for $x gt 0$. [2]

Mark scheme

1. (a) $T(0) = 80$, so $80^{circ}mathrm{C}$ [1]. (b) $T(15) = 39.5^{circ}mathrm{C}$ [3]. (c) $8 = 58e^{-0.08t}$ [1], $t = -frac{1}{0.08}lnleft(frac{8}{58}right)$ [2], $t = 24.8$ minutes [1]. (d) The model approaches $22^{circ}mathrm{C}$, the room temperature [2]. (e) $T'(t) = -4.64e^{-0.08t}$ [2], $T'(10) = -2.08$ [1]. At 10 minutes the drink is cooling at about $2.08^{circ}mathrm{C}$ per minute [1].

2. (a) $z = -1.5$, probability $0.0668$ [3]. (b) $0.775$ (calculator value about $0.7745$ before rounding to 3 s.f.) [4]. (c) $z approx 1.64$ for the upper 5% [2], $k = 500 + 1.64485 times 4 approx 506.6$ [2], nearest millilitre $507$ [1].

3. (a) Product rule: $e^{-x/2}left(1-frac{x}{2}right)$ [3]. (b) Stationary when $x = 2$ [1], and it is a maximum because the factor $left(1-frac{x}{2}right)$ changes from positive to negative [1]. Coordinates $left(2, frac{2}{e}right)$ [2]. (c) Integration by parts as shown [3]. Area $= 4 – 12e^{-2}$ [1], which is $2.38$ to 3 s.f. [1]. (d) $x = 0.715$ and $x = 4.31$ [2].

Full mock run-sheet

A one-day plan that uses the iStudyPlus Paper 1 and Paper 2 sets on this page, plus an error log.

iStudyPlus original. This set was written for iStudyPlus. It is not an official IB examination, it does not use a real session code, and it is not copied from an IB past paper. Official IB examination papers and markschemes remain the property of the International Baccalaureate Organization.

Run this as a practice day for Analysis and Approaches HL. The papers are the iStudyPlus sets on this page. They are shorter than a real session, so keep the breaks, but do not stretch the sittings past the times printed on those sets.

  1. Before you start (10 min). Put the calculator away for the first sitting. Write your name and the start time. Skim the five Paper 1 questions and star the one you will leave until last.
  2. Paper 1 style set (55 min). Non-calculator. Stop when the time ends, even if question 5 is unfinished. Box every final answer.
  3. Break (20 min). Do not open the mark scheme yet.
  4. Paper 2 style set (50 min). Calculator on. For every decimal, write the equation or inverse-normal command you used, then the rounded value.
  5. Same evening (40 min). Mark both sets from the schemes on this page. Fill one row of an error log per mark you lost: question, topic, error type (setup, algebra, calculator, incomplete method, or misread command), and the next drill.
  6. Next study block. If calculus or the normal model cost marks, use the Topic skills check on this page, then the calculus rescue bootcamp. Do not sit another full paper the same night.

Optional HL extension on a second day: the Paper 3 style investigation on this page, 45 minutes, calculator allowed only to check a decimal you already have in exact form.

Paper 3 style investigation

An original short investigation of $f_a(x) = x^{3} - ax$: stationary points, an odd function, and one parameter to find.

iStudyPlus original. This set was written for iStudyPlus. It is not an official IB examination, it does not use a real session code, and it is not copied from an IB past paper. Official IB examination papers and markschemes remain the property of the International Baccalaureate Organization.

Sitting: shortened Paper 3 style investigation. Time: 45 minutes. Calculator: not required. Total: 22 marks. Later parts use earlier results; box anything you will reuse.

Consider the family of functions $f_{a}(x) = x^{3} – ax$, where $a$ is a real constant.

  1. Show that $f_{a}$ is an odd function. [2]
  2. Find $f_{a}'(x)$. [1]
  3. Find the $x$-coordinates of any stationary points. State clearly the values of $a$ that give none, one, or two stationary points. [5]
  4. When there are two stationary points, determine their nature. [3]
  5. Show that, when $a gt 0$, the local maximum value is $2left(frac{a}{3}right)^{3/2}$. [4]
  6. Find the value of $a$ for which that local maximum value equals $2$. [3]
  7. For the value of $a$ found in part (f), write down the intercepts on the axes and the coordinates of both stationary points. [4]

Mark scheme

(a) $f_{a}(-x) = -x^{3} + ax = -f_{a}(x)$ [2].

(b) $f_{a}'(x) = 3x^{2} – a$ [1].

(c) Stationary when $3x^{2} = a$ [1]. If $a lt 0$, no real stationary point [1]. If $a = 0$, one stationary point at $x = 0$ [1]. If $a gt 0$, $x = pmsqrt{a/3}$ [2].

(d) $f_{a}”(x) = 6x$ [1]. At $x = -sqrt{a/3}$, $f” lt 0$, so a local maximum [1]. At $x = sqrt{a/3}$, a local minimum [1].

(e) Let $t = sqrt{a/3} gt 0$, so the local maximum is at $x = -t$ and $a = 3t^{2}$ [1]. $f_{a}(-t) = -t^{3} + at = 2t^{3}$ [2]. $2t^{3} = 2(a/3)^{3/2}$ [1].

(f) $2t^{3} = 2$, so $t = 1$ [1] and $a = 3$ [2].

(g) $f_{3}(x) = x^{3}-3x = x(x^{2}-3)$ [1]. Intercepts $(0,0)$, $(sqrt{3},0)$, $(-sqrt{3},0)$ [1]. Stationary points $(-1, 2)$ and $(1, -2)$ [2].