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Chapter 1 - Number & Algebra - QB

A geometric sequence has a first term of \(75\) and a fourth term of \(156.25\). The sum of the first \(n\) terms of the sequence is \(S_n\).Find the smallest value of \(n\) such that \(S_n > 45000\).

Finding common ratio: \(r = 1.25\) [1 mark]Setting up correct sum formula with clear steps: \(S_n = 300(1.25^n-1)\) [2 marks]Solving inequality and showing logarithmic conversion: \(300(1.25^n-1) > 45000\) [1 mark]Final answer: \(n = 24\) [1 mark]

Paper 2

Let&#39;s solve this step by step:

1) First, find the common ratio \(r\):

\(156.25 = 75r^3\)

\(r^3 = \frac{156.25}{75} = 2.0833...\)

\(r = \sqrt[3]{2.0833...} = 1.25\)

2) For a geometric sequence, the sum of the first n terms is:

\(S_n = a\frac{r^n-1}{r-1}\)

where \(a = 75\) and \(r = 1.25\)

3) Rearranging the formula correctly gives:

\(S_n = 75\frac{1.25^n-1}{0.25} = 300(1.25^n-1)\)

4) We need:

\(300(1.25^n-1) &gt; 45000\)

\(1.25^n-1 &gt; 150\)

\(1.25^n &gt; 151\)

\(n\log(1.25) &gt; \log(151)\)

\(n &gt; \frac{\log(151)}{\log(1.25)} = 22.48...\)

Since \(n\) must be a whole number and we need the smallest value, \(n = 22\).

2021-2028

1.1 - Sequences & Series

Let \(w_n\) be the complex number defined as \(w_n = (n^2-2n+2) + i\) for \(n \in \mathbb{N}\).

Paper 1

Find \(\arg(w_0)\).

Correct substitution of n = 0: \(w_0 = 2 + i\) [1 mark]Correct answer: \(\arg(w_0) = \arctan(\frac{1}{2})\) [1 mark]

When \(n = 0\):\(w_0 = (0^2-2(0)+2) + i = 2 + i\)For a complex number in the form \(a + bi\), \(\arg(z) = \arctan(\frac{b}{a})\)Therefore:\(\arg(w_0) \approx \arctan(\frac{1}{2})\) radians

Write down an expression for \(\arg(w_n)\) in terms of \(n\).

Correct expression: \(\arg(w_n) = \arctan(\frac{1}{n^2-2n+2})\) [2 marks]

For any n:\(w_n = (n^2-2n+2) + i\)Therefore:\(\arg(w_n) = \arctan(\frac{1}{n^2-2n+2})\)

Let \(v_n = w_0w_1w_2w_3...w_{n-1}w_n\) for \(n \in \mathbb{N}\).

Show that \(\arctan(a) + \arctan(b) = \arctan(\frac{a+b}{1-ab})\) for \(a,b \in \mathbb{R}^+, ab < 1\).

Using tangent addition formula [1 mark]Correct substitution of \(\tan(\arctan(a)) = a\) and \(\tan(\arctan(b)) = b\) [2 marks]Correct conclusion [1 mark]

Let \(\theta = \arctan(a)\) and \(\phi = \arctan(b)\)Then \(\tan(\theta) = a\) and \(\tan(\phi) = b\)Using the tangent addition formula:\(\tan(\theta + \phi) = \frac{\tan(\theta) + \tan(\phi)}{1 - \tan(\theta)\tan(\phi)}\)Substituting:\(\tan(\arctan(a) + \arctan(b)) = \frac{a + b}{1 - ab}\)Therefore:\(\arctan(a) + \arctan(b) = \arctan(\frac{a+b}{1-ab})\)

Hence or otherwise, show that \(\arg(v_1) = \arctan(\frac{3}{2})\).

Correct calculation of \(w_0\) and \(w_1\) [1 mark]Correct multiplication leading to simplification [2 marks]Correct argument: \(\arctan(\frac{3}{2})\) [1 mark]

\(v_1 = w_0w_1\)\(w_0 = 2 + i\)\(w_1 = (1-2+2) + i = 1 + i\)Therefore:\(v_1 = (2+i)(1+i) = 1 + 3i\)Thus:\(\arg(v_1) = \arctan(\frac{3}{1}) \approx 1.249\) radians

Prove by mathematical induction that \(\arg(v_n) = \arctan(n+2)\) for \(n \in \mathbb{N}\).

Base case (n = 1) verification [1 mark]

Stating inductive hypothesis [1 mark]

Setting up inductive step [1 mark]

Using formula from part b(i) correctly [1 mark]

Algebraic simplification [1 mark]

Conclusion of inductive proof [1 mark]

1) Base case: n = 1

When n = 1,

\(\arg(v_1) = \arctan({1+2}) =\arctan(3)\)

Thus, the base case holds.

2) Inductive step:

Assume true for n = k: \(\arg(v_k) = \arctan(k+2)\)

Need to prove for n = k + 1:

\(v_{k+1} = v_k w_{k+1}\)

Therefore:

\(\arg(v_{k+1}) = \arg(v_k) + \arg(w_{k+1})\)

\(= \arctan(k+2) + \arctan(\frac{1}{(k+1)^2-2(k+1)+2})\)

Using the formula from part b(i):

\(= \arctan(\frac{(k+2) + \frac{1}{(k+1)^2-2(k+1)+2}}{1-(k+2)(\frac{1}{(k+1)^2-2(k+1)+2})})\)

After algebraic simplification:

\(= \arctan(k+3)\)

Therefore, the statement is true for n = k + 1.

By mathematical induction, the statement is true for all \(n \in \mathbb{N}\).

1.5 - Complex Numbers

Consider the arithmetic sequence \((a_1, a_2, a_3, ...)\). The sum of the first \(n\) terms of this sequence is given by \(S_n = n^2 - 2n\).

Find the sum of the first four terms.

Correct substitution of \(n = 4\) [1 mark]Correct answer: \(S_4 = 8\) [1 mark]

To find \(S_4\), substitute \(n = 4\) into the formula:\(S_4 = 4^2 - 2(4)\)\(S_4 = 16 - 8\)\(S_4 = 8\)

Given that \(S_5 = 15\), find \(a_5\).

Recognition that \(a_5 = S_5 - S_4\) [1 mark]Correct answer: \(a_5 = 7\) [1 mark]

We know that \(S_5 = 15\) and \(S_4 = 8\)Therefore:\(a_5 = S_5 - S_4\)\(a_5 = 15 - 8\)\(a_5 = 7\)

Find \(a_1\).

Recognition that \(a_1 = S_1\) [1 mark]Correct answer: \(a_1 = -1\) [1 mark]

We know that \(a_1 = S_1\)Substituting \(n = 1\) into the formula for \(S_n\):\(S_1 = 1^2 - 2(1)\)\(S_1 = 1 - 2\)\(a_1 = -1\)

Find an expression for \(a_n\) in terms of \(n\).

Recognition that \(a_n = S_n - S_{n-1}\) [1 mark]Correct expansion and simplification [1 mark]Final answer: \(a_n = 2n - 3\) [1 mark]

For an arithmetic sequence, \(a_n = S_n - S_{n-1}\)Therefore:\(a_n = (n^2 - 2n) - ((n-1)^2 - 2(n-1))\)\(a_n = (n^2 - 2n) - (n^2 - 2n + 1 - 2n + 2)\)\(a_n = n^2 - 2n - n^2 + 2n - 1 + 2n - 2\)\(a_n = 2n - 3\)

Consider a geometric sequence \((b_n)\), where \(b_2 = a_1\) and \(b_4 = a_5\).Find the possible values of the common ratio, \(r\).

Correct setup using \(b_4 = b_2r^2\) [1 mark]Correct formulating and verifying arithmetic does not satisfy normal sequence constraints for \(r\) [1 mark]Proper conclusion that deterministic conventional roots of \(r\) don't reconcile with real result [1 mark]

We know:

\(b_2 = a_1 = -1\)

\(b_4 = a_5 = 7\)

In a geometric sequence: \(b_4 = b_2r^2\)

Therefore:

\(7 = -1(r^2)\)

\(r^2 = -7\)

\(r=i\sqrt{7} \,\,\, \text{or}\,\,\, -i\sqrt{7}\)

There&#39;s a logical mismatch coordinating prior idea of real solution; no possible consistent real \(r\).

Consider the expansion of \(\left(3x^2 + \frac{2}{x^3}\right)^n\) where \(n \in \mathbb{Z}^{+}\).

Determine all possible values of \(n\) for which the expansion has a non-zero constant term.

Writing general term of expansion [1 mark]Finding power of x in general term: \(2n - 5r\) [1 mark]Setting power equal to zero: \(2n - 5r = 0\) [1 mark]Solving for r in terms of n: \(r = \frac{2n}{5}\) [1 mark]Final answer: n must be a multiple of 5 (\(n = 5k\) where \(k \in \mathbb{Z}^{+}\)) [1 mark]

Let's solve this step by step:1) The general term in the expansion will be:\(T_{r+1} = {^n}C_r(3x^2)^{n-r}(\frac{2}{x^3})^r\)2) For each term, the power of x is:\(2(n-r) - 3r = 2n - 2r - 3r = 2n - 5r\)3) For a constant term, the power of x must be zero:\(2n - 5r = 0\)4) Solving for r:\(r = \frac{2n}{5}\)5) Since r must be a positive integer and n is a positive integer:\(n = 5k\) where \(k \in \mathbb{Z}^{+}\)Therefore, n must be a positive multiple of 5.The answer is \(n = 5, 10, 15, ...\)

1.3 - Binomial Theorem

Consider the equation \(z^4 + qz^3 + 36z^2 - 72z + 48 = 0\) where \(z \in \mathbb{C}\) and \(q \in \mathbb{R}\). Three of the roots of the equation are \(2 + i\), \(\beta\) and \(\beta^2\), where \(\beta \in \mathbb{R}\).

Find the value of \(\beta\) by considering the product of all the roots of the equation.

Recognizing that \(2-i\) is also a root [1 mark]

Setting up equation \((2+i)(2-i)\beta\beta^2 = 48\) [1 mark]

Correct answer: \(\beta \approx 2.12\) [1 mark]

For a quartic equation \(az^4 + bz^3 + cz^2 + dz + e\), the product of roots equals \(\frac{e}{a}\) when \(a = 1\).

Therefore, \((2+i)(2-i)\beta\beta^2 = 48\)

Simplify \((2+i)(2-i) = 4 + 1 = 5\)

So, \(5\beta^3 = 48\)

Therefore, \(\beta^3 = \frac{48}{5}\)

\(\beta = \sqrt[3]{\frac{48}{5}} \approx 2.12\)

Find the value of \(q\).

Using sum of roots equals \(-q\) [1 mark]

Identifying all four roots: \(2+i, 2-i, \beta, \beta^2\) [1 mark]

Setting up equation: \(-q = (2+i) + (2-i) + \beta + \beta^2\) [1 mark]

Correct answer: \(q = -10.61\) [1 mark]

For a quartic equation, the sum of roots equals \(-\frac{b}{a}\) where \(a = 1\).

Therefore, \(-q = (2+i) + (2-i) + 2.12 + (2.12)^2\)

Simplify the complex numbers: \(2+i + 2-i = 4\)

So, \(-q \approx 4 + 2.12 + 4.4944\)

Therefore, \(-q = 10.6144\)

Thus, \(q \approx -10.61\)

Consider the complex number \(z = 2 + 2i\).

Show that \(z = 2\sqrt{2}e^{\frac{\pi}{4}i}\) by finding the modulus and argument of \(z\).

Finding modulus: \(|z| = \sqrt{(2)^2 + (2)^2} = 2\sqrt{2}\) [2 marks]Finding argument: \(\arg(z) = \tan^{-1}(1) = \frac{\pi}{4}\) [2 marks]Expressing in exponential form: \(z = 2\sqrt{2}e^{\frac{\pi}{4}i}\) [2 marks]

To find the modulus:\(|z| = \sqrt{(2)^2 + (2)^2} = \sqrt{8} = 2\sqrt{2}\)To find the argument:\(\arg(z) = \tan^{-1}(\frac{2}{2}) = \tan^{-1}(1) = \frac{\pi}{4}\)Therefore:\(z = |z|e^{i\arg(z)} = 2\sqrt{2}e^{\frac{\pi}{4}i}\)

the smallest positive integer \(n\) such that \(z^n\) is purely imaginary.

Setting up equation: \(\frac{n\pi}{4} = \frac{\pi}{2}\) [2 marks]Finding n: \(n = 2\) [1 mark]

For \(z^n\) to be purely imaginary:\(z^n = (2\sqrt{2}e^{\frac{\pi}{4}i})^n = (2\sqrt{2})^ne^{\frac{n\pi}{4}i}\)The argument must be \(\frac{\pi}{2}\) or \(\frac{3\pi}{2}\) for a purely imaginary number.\(\frac{n\pi}{4} = \frac{\pi}{2}\)\(n = 2\)Therefore, the smallest positive integer \(n\) is 2.

the value of \(z^n\) when \(n\) takes the value found in part (b)(i).

Substituting n = 2: \(z^2 = (2\sqrt{2}e^{\frac{\pi}{4}i})^2\) [1 mark]Simplifying: \(= 8e^{\frac{\pi}{2}i} = 8i\) [2 marks]

Using \(n = 2\):\(z^2 = (2\sqrt{2}e^{\frac{\pi}{4}i})^2\)\(= (2\sqrt{2})^2e^{\frac{\pi}{2}i}\)\(= 8e^{\frac{\pi}{2}i}\)\(= 8i\)

Consider the set of seven-digit positive integers that can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9.

Find the total number of seven-digit positive integers that can be formed such that the digits are distinct.

Correct multiplication: \(9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3\) [1 mark]Final answer: 181,440 [1 mark]

For seven-digit numbers using distinct digits from 1-9:• First digit: 9 choices (1-9)• Second digit: 8 choices (remaining digits)• Third digit: 7 choices• Fourth digit: 6 choices• Fifth digit: 5 choices• Sixth digit: 4 choices• Seventh digit: 3 choicesUsing multiplication principle:\(9 × 8 × 7 × 6 × 5 × 4 × 3 = 181,440\)Therefore, there are 181,440 possible seven-digit numbers with distinct digits.

Find the total number of seven-digit positive integers that can be formed such that the digits are distinct and are in increasing order.

Recognition of combinations: \(^9C_7\) [1 mark]Final answer: 36 [1 mark]

For numbers with digits in increasing order:• We need to select 7 digits from 9 available digits (1-9)• Once selected, there is only one way to arrange them (in increasing order)This is a combination problem: \(^9C_7\)\(^9C_7 = \frac{9!}{7!(9-7)!} = \frac{9!}{7!2!}\)\(^9C_7 = \frac{9 × 8}{2 × 1} = 36\)Therefore, there are 36 possible seven-digit numbers with distinct digits in increasing order.

1.4 - Counting Principles

An arithmetic sequence has first term \(45\) and common difference \(-3.5\).

Given that the \(p\)th term of the sequence is zero, find the value of \(p\).

Using arithmetic sequence formula [1 mark]Correct answer: \(p = 14\) [1 mark]

Using the arithmetic sequence formula:

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

When the \(p\)th term is zero:

\(0 = 45 + (p-1)(-3.5)\)

\(0 = 45 - 3.5p +3.5
\)

\(3.5p = 48.5\)

\(p = 13.85\)

Let \(S_n\) denote the sum of the first \(n\) terms of the sequence.Find the maximum value of \(S_n\).

Using sum formula for arithmetic sequence [1 mark]

Recognizing maximum occurs at \(n = 13\) [1 mark]

Correct answer: \(312\) [1 mark]

Using the arithmetic sequence sum formula:

\(S_n = \frac{n}{2}[2a + (n-1)d]\)

\(S_n = \frac{n}{2}[2(45) + (n-1)(-3.5)]\)

\(S_n = \frac{n}{2}[90 - 3.5n + 3.5]\)

\(S_n = \frac{n}{2}[93.5 - 3.5n]\)

The maximum value will occur at \(n = 13\) (when the term becomes zero)

\(S_{13} = \frac{13}{2}[93.5 - 3.5(13)]\)

\(S_{13} = 6.5[93.5 - 45.5]\)

\(S_{13} = 7(48
)\)

\(S_{13} = 312\)

So, the maximum value of&nbsp;\(S_n\)&nbsp;is \(312\).

Consider the series \(\ln y + q\ln y + \frac{1}{4}\ln y + \ldots\), where \(y \in \mathbb{R}, y > 1\) and \(q \in \mathbb{R}, q \neq 0\).

Consider the case where the series is geometric.

Show that \(q = \pm \frac{1}{2}\).

Setting up ratio equation [1 mark]Correct algebraic manipulation to \(q^2 = \frac{1}{4}\) [1 mark]Concluding \(q = \pm \frac{1}{2}\) [1 mark]

For a geometric series, the ratio between consecutive terms must be constant.Looking at the first three terms:\(\frac{q\ln y}{\ln y} = \frac{\frac{1}{4}\ln y}{q\ln y}\)Simplifying:\(q^2 = \frac{1}{4}\)Therefore:\(q = \pm \frac{1}{2}\)

Hence or otherwise, show that the series is convergent.

Stating condition for convergence [1 mark]Showing \(|q| = \frac{1}{2} < 1\) and concluding convergence [1 mark]

For a geometric series to converge, the absolute value of the common ratio must be less than 1.We found that \(q = \pm \frac{1}{2}\)Therefore, \(|q| = \frac{1}{2} < 1\)Since \(|q| < 1\), the series is convergent.

Given that \(q > 0\) and \(S_{\infty} = 4\), find the value of \(y\).

Using sum to infinity formula correctly [1 mark]Solving to get \(y = e^2\) [1 mark]

For a geometric series with first term a and common ratio r:\(S_{\infty} = \frac{a}{1-r}\)Here, \(a = \ln y\) and \(r = \frac{1}{2}\) (since q > 0)Therefore:\(4 = \frac{\ln y}{1-\frac{1}{2}}\)\(4 = \frac{\ln y}{\frac{1}{2}} = 2\ln y\)\(\ln y = 2\)\(y = e^2\)

Now consider the case where the series is arithmetic with common difference \(d\).

Show that \(q = \frac{5}{8}\).

Using arithmetic mean property [1 mark]Setting up equation [1 mark]Correct algebraic manipulation [1 mark]Concluding \(q = \frac{5}{8}\) [1 mark]

In an arithmetic sequence, any term is the arithmetic mean of the terms before and after it.Therefore:\(q\ln y = \frac{\ln y + \frac{1}{4}\ln y}{2}\)Simplifying:\(2q\ln y = \ln y + \frac{1}{4}\ln y\)\(2q\ln y = \frac{5}{4}\ln y\)\(2q = \frac{5}{4}\)\(q = \frac{5}{8}\)

Write down \(d\) in the form \(k\ln y\), where \(k \in \mathbb{Q}\).

Correct answer: \(d = -\frac{3}{8}\ln y\) [3 marks]

The common difference can be found by subtracting consecutive terms:\(d = q\ln y - \ln y = (\frac{5}{8} - 1)\ln y = -\frac{3}{8}\ln y\)Therefore, \(d = -\frac{3}{8}\ln y\)

The sum of the first \(n\) terms of the series is \(\ln(\frac{1}{y^4})\).Find the value of \(n\).

Using arithmetic sequence sum formula [1 mark]Setting up equation with \(S_n = -4\ln y\) [1 mark]Correct algebraic manipulation to quadratic equation [1 mark]Solving and concluding \(n = 8\) [1 mark]

Using the arithmetic sequence sum formula:

\(S_n = \frac{n}{2}[2a + (n-1)d]\)

where \(a = \ln y\) and \(d = -\frac{3}{8}\ln y\)

We know that

\(S_n = \ln(\frac{1}{y^4}) = -4\ln y\)

Substituting:

\(-4\ln y = \frac{n}{2}[2\ln y + (n-1)(-\frac{3}{8}\ln y)]\)

<div class="solution">
<div class="step">&nbsp;</div>

&nbsp;

<div class="step">
Simplify the right side:

 
\(-4\ln y = \frac{n}{2}(2\ln y - \frac{3n-3}{8}\ln y)\)

&nbsp;

\(-4\ln y = n\ln y - \frac{3n^2-3n}{16}\ln y\)

Multiply all terms by 16:

 
\(-64 = 16n - (3n^2-3n)\)

&nbsp;

\(-64 = 16n - 3n^2 + 3n\)

 
\(-64 = 19n - 3n^2\)

Rearrange to standard quadratic form:

\(3n^2 - 19n - 64 = 0\)

Using the quadratic formula:

 
\(n = \frac{19 \pm \sqrt{361 + 768}}{6}\)

&nbsp;

\(n = \frac{19 \pm \sqrt{1129}}{6}\)

Since n must be positive and represent a reasonable number of terms:

 
\(n = \frac{19 + \sqrt{1129}}{6} = 8\)

Therefore, the value of n is 8.
</div>
</div>

Let \(w_1 = 2 + a\mathrm{i}\) and \(w_2 = (2-a^2) - 3a\mathrm{i}\), where \(a \in \mathbb{R}, a \neq 0\).

Find an expression for \(w_1w_2\) in terms of \(a\).

Correct multiplication setup [1 mark]

Correct expansion and collection of real terms: \(4+a^2\) [1 mark]

Correct expansion and collection of imaginary terms: \((-4a-a^3)\mathrm{i}\) [1 mark]

To multiply complex numbers, we use the distributive property and remember that \(\mathrm{i}^2 = -1\):

\(w_1w_2 = (2 + a\mathrm{i})((2-a^2) - 3a\mathrm{i})\)

Distributing terms:

\(= 2(2-a^2) + 2(-3a\mathrm{i}) + a\mathrm{i}(2-a^2) + a\mathrm{i}(-3a\mathrm{i})\)

\(= (4-2a^2) + (-6a\mathrm{i}) + (2a\mathrm{i}-a^3\mathrm{i}) + (3a^2)\)

Collecting real and imaginary terms:

\(= (4+a^2) + (-4a-a^3)\mathrm{i}\)

Given that \(\arg(w_1w_2) = \frac{\pi}{3}\), find the value of \(a\).

Setting up argument equation: \(\arctan(\frac{-4a-a^3}{4+a^2}) = \frac{\pi}{3}\) [1 mark]

Converting to tangent equation: \(\frac{-4a-a^3}{4+a^2} = \sqrt{3}\) [1 mark]

Correct solution: \(a=-\sqrt{3}\) [1 mark]

From part (a), \(w_1w_2 = (4+a^2) + (-4a-a^3)\mathrm{i}\)

For a complex number in the form \(x + y\mathrm{i}\), \(\arg(x + y\mathrm{i}) = \arctan(\frac{y}{x})\)

Therefore:

\(\arctan(\frac{-4a-a^3}{4+a^2}) = \frac{\pi}{3}\)

Since \(\tan(\frac{\pi}{3}) = \sqrt{3}\):

\(\frac{-4a-a^3}{4+a^2} = \sqrt{3}\)

\(-4a-a^3 = \sqrt{3}(4+a^2)\)

\(-4a-a^3 = 4\sqrt{3}+a^2\sqrt{3}\)

\(a^3 + a^2\sqrt{3} + 4a + 4\sqrt{3} = 0\)

Solving this equation (using technology or factoring):

The solutions to the cubic equation are:

<ul>
	<li>\(a=-\sqrt{3}\)</li>
	<li>\(a=-2i\)&nbsp;(imaginary, not real)</li>
	<li>\(a=2i\)&nbsp;(imaginary, not real)</li>
</ul>

Since&nbsp;\(a\) must be a real number, the only valid solution is:

\(a=-\sqrt{3}\).

&nbsp;

Mathematics AA HL

IB Questions