IB Mathematics: Applications and Interpretation Quiz: Iterative And Numerical Methods

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Iterative And Numerical MethodsQuestion 1 of 16

A company's revenue $R$ (in millions of dollars) is modelled by $R(x) = x^2 e^{-0.1x}$ , where $x$ is the amount spent on marketing (in millions of dollars), for $x \ge 0$ . Find the marketing expenditure $x$ required to achieve a revenue of $10 million. Give your answer to 3 significant figures.

3.58

5.36

10.0

There is no such value.

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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Iterative And Numerical Methods

Practice Iterative And Numerical Methods in IB Mathematics: Applications and Interpretation with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Iterative And Numerical Methods, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

3.58 (correct answer)
5.36
10.0
There is no such value.

Explanation: We need to solve the equation

R(x) = 10

, which is

x^2 e^{-0.1x} = 10

. This is a transcendental equation that must be solved numerically. Using a GDC, we can graph

y_1 = x^2 e^{-0.1x}

and

y_2 = 10

and find their intersections. There are two intersection points. The first is at

x \approx 3.575

and the second is at

x \approx 10.97

. The question asks for 'the' marketing expenditure, which is slightly ambiguous. However, looking at the options, only 3.58 (which is 3.575 rounded to 3 s.f.) is present. This is a common way IB questions specify a unique answer from multiple solutions. The maximum revenue is at

x=20

, R(20)=54. So 10M is achievable.

Question 2

To find a root of $f(x) = x^4 - 8x^2 - x + 14 = 0$ , an iterative method is used. The equation has four real roots. To which root is the method most likely to converge if the starting value is $x_0 = 2.0$ ?

-2.685
-1.356
1.218 (correct answer)
2.823

Explanation: First, use a numerical solver to find all four real roots of the equation

x^4 - 8x^2 - x + 14 = 0

. The roots are approximately

x \approx -2.685

x \approx -1.356

x \approx 1.218

, and

x \approx 2.823

. Iterative methods, such as Newton-Raphson, generally converge to the root that is 'closest' to the starting value

x_0

. We need to find which of the four roots is closest to

x_0 = 2.0

. Let's calculate the distance from 2.0 to each root:

|2.0 - (-2.685)| = 4.685

|2.0 - (-1.356)| = 3.356

|2.0 - 1.218| = 0.782

|2.0 - 2.823| = 0.823

The smallest distance is 0.782, which corresponds to the root

x \approx 1.218

. Therefore, the iteration is most likely to converge to 1.218.

Question 3

The bisection method is used to find a root of an equation in the interval $[a, b]$ . If the length of the initial interval is $L = b-a$ , what is the length of the interval containing the root after $n$ iterations?

$\frac{L}{n}$
$\frac{L}{2^n}$ (correct answer)
$\frac{L}{n+1}$
$\frac{L}{2^{n+1}}$

Explanation: The bisection method is a numerical technique for finding roots where you repeatedly halve an interval until you narrow down to the root's location. Understanding how the interval shrinks with each iteration is crucial for analyzing the method's efficiency. In each iteration of the bisection method, you find the midpoint of your current interval and determine which half contains the root based on sign changes of the function. This means you're always keeping exactly half of the previous interval and discarding the other half. Starting with interval length

L = b-a

, after one iteration you have length

\frac{L}{2}

. After the second iteration, you halve again to get

\frac{L}{4} = \frac{L}{2^2}

. After the third iteration:

\frac{L}{8} = \frac{L}{2^3}

. The pattern shows that after

n

iterations, the interval length is

\frac{L}{2^n}

, making B correct. Option A,

\frac{L}{n}

, represents linear reduction rather than the exponential halving that actually occurs. This would be far too slow for the bisection method. Option C,

\frac{L}{n+1}

, also suggests linear reduction and incorrectly adds 1 to the denominator. Option D,

\frac{L}{2^{n+1}}

, represents the interval length after

n+1

iterations, not

n

iterations—this is an off-by-one error. Remember: the bisection method always halves intervals, so look for powers of 2 in the denominator. The exponent should match the number of iterations exactly, with no adjustments like

+1

Question 4

The value of a machine, $V$ , in thousands of dollars, depreciates according to the model $V(t) = 150e^{-0.15t}$ , where $t$ is the time in years. The cost of maintenance, $C$ , in thousands of dollars, is modelled by $C(t) = 2t^2 + 5$ . Using a numerical solver, find the time $t$ at which the value of the machine is equal to its maintenance cost.

4.08 years
4.21 years
4.52 years
4.33 years (correct answer)

Explanation: When you encounter problems involving two functions that need to intersect, you're looking for the point where they're equal. This requires setting up an equation and solving it, often numerically when analytical methods become complex. To find when the machine's value equals its maintenance cost, you need to solve

V(t) = C(t)

, which gives you

150e^{-0.15t} = 2t^2 + 5

. This equation combines exponential decay with quadratic growth, making it impossible to solve algebraically. This is exactly when you'd use a numerical solver or graphing technology. Using a numerical solver (like your calculator's equation solver or graphing the intersection), you'll find that these functions intersect at approximately

t = 4.33

years. You can verify this by substituting:

V(4.33) = 150e^{-0.15(4.33)} \approx 76.5

and

C(4.33) = 2(4.33)^2 + 5 \approx 76.5

. Looking at the wrong answers: Choice A (4.08 years) would give you a machine value of about 82.3 and maintenance cost of about 38.3 – these don't match. Choice B (4.21 years) yields approximately 79.1 for value and 40.4 for cost. Choice C (4.52 years) gives about 73.5 for value and 45.8 for cost. Remember that numerical solver problems on the IB exam often have answers that are close together, so precision matters. Always verify your answer by substituting back into both original functions to ensure they produce approximately the same result.

Question 5

The bisection method is used to find a root of $f(x)=0$ on the interval $[a,b]$ . Let $c = (a+b)/2$ be the midpoint. The search requires that $f(a)$ and $f(b)$ have opposite signs. If $f(a) < 0$ and $f(b) > 0$ , and it is found that $f(c) > 0$ , what is the next interval for the search?

[a, c] (correct answer)
[c, b]
[a, b]
The method requires recalculation with a different c.

Explanation: The core principle of the bisection method is to maintain an interval where the function values at the endpoints have opposite signs, guaranteeing a root lies within. The initial state is

f(a) < 0

and

f(b) > 0

. The midpoint is

c

, and we find that

f(c) > 0

. To maintain the sign change, we need to choose a new interval with one endpoint having a negative function value and the other having a positive one. Since

f(a)

is negative and

f(c)

is positive, the root must lie between

a

and

c

. Therefore, the new interval is

[a, c]

. The old endpoint

b

is replaced by

c

Question 6

4.08 years
4.21 years
4.52 years
4.33 years (correct answer)

V(t) = C(t)

, which gives you

150e^{-0.15t} = 2t^2 + 5

t = 4.33

years. You can verify this by substituting:

V(4.33) = 150e^{-0.15(4.33)} \approx 76.5

and

C(4.33) = 2(4.33)^2 + 5 \approx 76.5

Question 7

-2.685
-1.356
1.218 (correct answer)
2.823

Explanation: First, use a numerical solver to find all four real roots of the equation

x^4 - 8x^2 - x + 14 = 0

. The roots are approximately

x \approx -2.685

x \approx -1.356

x \approx 1.218

, and

x \approx 2.823

. Iterative methods, such as Newton-Raphson, generally converge to the root that is 'closest' to the starting value

x_0

. We need to find which of the four roots is closest to

x_0 = 2.0

. Let's calculate the distance from 2.0 to each root:

|2.0 - (-2.685)| = 4.685

|2.0 - (-1.356)| = 3.356

|2.0 - 1.218| = 0.782

|2.0 - 2.823| = 0.823

The smallest distance is 0.782, which corresponds to the root

x \approx 1.218

. Therefore, the iteration is most likely to converge to 1.218.

Question 8

[a, c] (correct answer)
[c, b]
[a, b]
The method requires recalculation with a different c.

f(a) < 0

and

f(b) > 0

. The midpoint is

c

, and we find that

f(c) > 0

. To maintain the sign change, we need to choose a new interval with one endpoint having a negative function value and the other having a positive one. Since

f(a)

is negative and

f(c)

is positive, the root must lie between

a

and

c

. Therefore, the new interval is

[a, c]

. The old endpoint

b

is replaced by

c

Question 9

$\frac{L}{n}$
$\frac{L}{2^n}$ (correct answer)
$\frac{L}{n+1}$
$\frac{L}{2^{n+1}}$

L = b-a

, after one iteration you have length

\frac{L}{2}

. After the second iteration, you halve again to get

\frac{L}{4} = \frac{L}{2^2}

. After the third iteration:

\frac{L}{8} = \frac{L}{2^3}

. The pattern shows that after

n

iterations, the interval length is

\frac{L}{2^n}

, making B correct. Option A,

\frac{L}{n}

, represents linear reduction rather than the exponential halving that actually occurs. This would be far too slow for the bisection method. Option C,

\frac{L}{n+1}

, also suggests linear reduction and incorrectly adds 1 to the denominator. Option D,

\frac{L}{2^{n+1}}

, represents the interval length after

n+1

iterations, not

n

+1

Question 10

A loan of $50,000 is taken out with an annual interest rate $r$ , compounded monthly. The loan is repaid over 10 years with monthly payments of $550. The relationship between the loan amount $P$ , monthly payment $M$ , monthly interest rate $i = r/12$ , and number of payments $n$ is given by the amortization formula $P = M \left[ \frac{1 - (1+i)^{-n}}{i} \right]$ . Find the annual interest rate $r$ for this loan.

4.50%
4.67%
4.89% (correct answer)
5.12%

Explanation: We are given

P = 50000

M = 550

, and

n = 10 \times 12 = 120

. We need to find the monthly interest rate

i

by solving the equation:

50000 = 550 \left[ \frac{1 - (1+i)^{-120}}{i} \right]

This equation cannot be solved for

i

algebraically and requires a numerical solver. We can rewrite the equation as

550 \left[ \frac{1 - (1+i)^{-120}}{i} \right] - 50000 = 0

and use a GDC's equation solver function. Solving for

i

gives

i \approx 0.004074

. This is the monthly interest rate. To find the annual interest rate

r

, we multiply by 12:

r = i \times 12 \approx 0.004074 \times 12 \approx 0.048888

. As a percentage, this is approximately 4.89%.

Question 11

4.50%
4.67%
4.89% (correct answer)
5.12%

Explanation: We are given

P = 50000

M = 550

, and

n = 10 \times 12 = 120

. We need to find the monthly interest rate

i

by solving the equation:

50000 = 550 \left[ \frac{1 - (1+i)^{-120}}{i} \right]

This equation cannot be solved for

i

algebraically and requires a numerical solver. We can rewrite the equation as

550 \left[ \frac{1 - (1+i)^{-120}}{i} \right] - 50000 = 0

and use a GDC's equation solver function. Solving for

i

gives

i \approx 0.004074

. This is the monthly interest rate. To find the annual interest rate

r

, we multiply by 12:

r = i \times 12 \approx 0.004074 \times 12 \approx 0.048888

. As a percentage, this is approximately 4.89%.

Question 12

3.58 (correct answer)
5.36
10.0
There is no such value.

Explanation: We need to solve the equation

R(x) = 10

, which is

x^2 e^{-0.1x} = 10

. This is a transcendental equation that must be solved numerically. Using a GDC, we can graph

y_1 = x^2 e^{-0.1x}

and

y_2 = 10

and find their intersections. There are two intersection points. The first is at

x \approx 3.575

and the second is at

x \approx 10.97

x=20

, R(20)=54. So 10M is achievable.

Question 13

The Newton-Raphson method is applied to find a root of $f(x) = x^3 - 6x + 2 = 0$ . Using a starting value of $x_0 = \sqrt{2}$ , what is the outcome?

It converges to the root near $x = 2.33$ .
It converges to the root near $x = 0.34$ .
It enters a cycle, alternating between $\sqrt{2}$ and $-\sqrt{2}$ .
It fails because the derivative $f'(x_0)$ is zero. (correct answer)

Explanation: The Newton-Raphson formula is

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

. First, we need the derivative of

f(x)

f'(x) = 3x^2 - 6

. The method fails if the derivative is zero at any step. Let's evaluate the derivative at the starting value

x_0 = \sqrt{2}

f'(\sqrt{2}) = 3(\sqrt{2})^2 - 6 = 3(2) - 6 = 6 - 6 = 0

. Since the derivative is zero, the denominator in the Newton-Raphson formula is zero, and the method fails at the very first step. Division by zero is undefined.

Question 14

To find a root of $f(x) = x^3 - 2x - 5 = 0$ , the rearrangement $x_{n+1} = x_n^3 - x_n - 5$ is proposed. Starting with $x_0 = 2$ , what is the behavior of this iteration?

It converges to the root $x \approx 2.09$ .
It diverges to $-\infty$ . (correct answer)
It converges to a different root of the equation.
It diverges to $+\infty$ .

Explanation: When you encounter iterative methods for finding roots, you need to analyze whether the sequence converges or diverges by examining the behavior of successive terms. Let's trace this iteration starting with

x_0 = 2

using the formula

x_{n+1} = x_n^3 - x_n - 5

x_1 = 2^3 - 2 - 5 = 8 - 2 - 5 = 1

x_2 = 1^3 - 1 - 5 = 1 - 1 - 5 = -5

x_3 = (-5)^3 - (-5) - 5 = -125 + 5 - 5 = -125

x_4 = (-125)^3 - (-125) - 5 = -1,953,125 + 125 - 5 = -1,953,005

The values are becoming increasingly negative, indicating divergence to

-\infty

. Answer A is incorrect because while

x \approx 2.09

is indeed a root of the original equation, this particular iteration doesn't converge to it. Answer C is wrong because the sequence doesn't converge to any value at all—it diverges. Answer D incorrectly suggests divergence to positive infinity, but as we calculated, the terms become increasingly negative. The key insight is that not every rearrangement of an equation produces a convergent iteration. For convergence, you typically need

|g'(x)| < 1

near the root, where

g(x) = x^3 - x - 5

. Here,

g'(x) = 3x^2 - 1

, and near

x = 2

, this derivative has a large magnitude, causing divergence. Study tip: Always test the first few iterations when analyzing convergence behavior—the pattern usually becomes clear quickly, and you can avoid complex theoretical analysis during exams.

Question 15

To find a root of $f(x) = x^3 - 2x - 5 = 0$ , the rearrangement $x_{n+1} = x_n^3 - x_n - 5$ is proposed. Starting with $x_0 = 2$ , what is the behavior of this iteration?

It converges to the root $x \approx 2.09$ .
It diverges to $-\infty$ . (correct answer)
It converges to a different root of the equation.
It diverges to $+\infty$ .

x_0 = 2

using the formula

x_{n+1} = x_n^3 - x_n - 5

x_1 = 2^3 - 2 - 5 = 8 - 2 - 5 = 1

x_2 = 1^3 - 1 - 5 = 1 - 1 - 5 = -5

x_3 = (-5)^3 - (-5) - 5 = -125 + 5 - 5 = -125

x_4 = (-125)^3 - (-125) - 5 = -1,953,125 + 125 - 5 = -1,953,005

The values are becoming increasingly negative, indicating divergence to

-\infty

. Answer A is incorrect because while

x \approx 2.09

|g'(x)| < 1

near the root, where

g(x) = x^3 - x - 5

. Here,

g'(x) = 3x^2 - 1

, and near

x = 2

Question 16

The Newton-Raphson method is applied to find a root of $f(x) = x^3 - 6x + 2 = 0$ . Using a starting value of $x_0 = \sqrt{2}$ , what is the outcome?

It converges to the root near $x = 2.33$ .
It converges to the root near $x = 0.34$ .
It enters a cycle, alternating between $\sqrt{2}$ and $-\sqrt{2}$ .
It fails because the derivative $f'(x_0)$ is zero. (correct answer)

Explanation: The Newton-Raphson formula is

x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}

. First, we need the derivative of

f(x)

f'(x) = 3x^2 - 6

. The method fails if the derivative is zero at any step. Let's evaluate the derivative at the starting value

x_0 = \sqrt{2}

f'(\sqrt{2}) = 3(\sqrt{2})^2 - 6 = 3(2) - 6 = 6 - 6 = 0

. Since the derivative is zero, the denominator in the Newton-Raphson formula is zero, and the method fails at the very first step. Division by zero is undefined.