Mathematics High Level (HL) – International Baccalaureate (IB)

November 25, 2021October 24, 2022

202111251031 Solution to 1995-IB-MATH-HL-I-3

Find the value of $x$ such that

$\begin{array}{|cc|} x & 6 \\ x & x \end{array} = \begin{array}{|ccc|} 2 & 1 & 3 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{array}$ .

Working.

$\begin{aligned} \textrm{LHS} & = x^2 - 6x \\ \textrm{RHS} & = 2 \cdot \begin{array}{|cc|} 2 & 3 \\ 1 & 1 \end{array} - 1 \cdot \begin{array}{|cc|} 1 & 3 \\ 3 & 1 \end{array} + 3 \cdot \begin{array}{|cc|} 1 & 2 \\ 3 & 1 \end{array} \\ & = 2\big( (2)(1)-(1)(3) \big) - 1\big( (1)(1)-(3)(3) \big) + 3\big( (1)(1)-(3)(2) \big) \\ & = 2(-1) - 1(-8) + 3(-5) \\ & = -9 \\ \end{aligned}$

$\begin{aligned} \textrm{LHS} & = \textrm{RHS} \\ x^2-6x & = -9 \\ x^2-6x+9 & = 0 \\ (x-3)^2 & = 0 \\ x & = 3 \qquad\textrm{(double roots)} \\ \end{aligned}$

Answers.

$x=3$ .

November 25, 2021October 24, 2022

202111250959 Solution to 1995-IB-MATH-HL-I-2

Calculate $x$ and $y$ exactly, as rational numbers, if

$\begin{aligned} 5^{3x}\times 25^{y}& =\frac{1}{5} \\ \textrm{and} \qquad\qquad 7^{x}\times 49^{2y} & = 1 \\ \end{aligned}$ .

Working.

From the first equation,

$\begin{aligned} 5^{3x}\times 25^{y} & = \frac{1}{5} \\ 5^{3x} \times 5^{2y} & = 5^{-1} \\ 5^{3x+2y} & = 5^{-1} \\ 3x+2y & = -1 \\ \end{aligned}$ .

From the second equation,

$\begin{aligned} 7^x\times 49^{2y} & = 1 \\ 7^x\times 7^{4y} & = 7^{0} \\ 7^{x+4y} & = 7^0 \\ x+4y & = 0 \\ \end{aligned}$ .

We have two linear equations with two unknowns:

$\begin{cases} 3x+2y = -1 \\ x+4y = 0 \\ \end{cases}$

Substituting $-4y$ for $x$ ,

$\begin{aligned} 3(-4y)+2y & =-1 \\ -12y+2y & = -1 \\ -10 y & = -1 \\ y & = \frac{1}{10} \\ \end{aligned}$

Then,

$x=-4y=\displaystyle{-4\bigg(\frac{1}{10}\bigg)=-\frac{2}{5}}$ .

Answers.

$\begin{cases} \enspace x = - \displaystyle{\frac{2}{5}} \\ \enspace y = \displaystyle{\frac{1}{10}} \\ \end{cases}$

November 25, 2021October 24, 2022

202111250947 Solution to 1995-IB-MATH-HL-I-1

Given the complex numbers $z_1=3+\mathrm{i}$ and $z_2=2-\mathrm{i}$ , find

(a) $z_1z_2$ :

(b) $\displaystyle{\frac{z_1}{z_2}}$ .

Working.

(a)

$\begin{aligned} z_1z_2 & = (3+\mathrm{i})(2-\mathrm{i}) \\ & = 6-3\mathrm{i}+2\mathrm{i}-\mathrm{i}^2 \\ & = 6-\mathrm{i}-(-1) \\ & = 7-\mathrm{i} \\ \end{aligned}$

(b)

$\begin{aligned} \frac{z_1}{z_2} & = \frac{3+\mathrm{i}}{2-\mathrm{i}} \\ & = \frac{(3+\mathrm{i})(2+\mathrm{i})}{(2-\mathrm{i})(2+\mathrm{i})} \\ & = \frac{6+3\mathrm{i}+2\mathrm{i}+\mathrm{i}^2}{(2)^2-(\mathrm{i})^2} \\ & = \frac{5+5\mathrm{i}}{5} \\ & = 1 + \mathrm{i} \\ \end{aligned}$

Answers.

(a) $7-\mathrm{i}$ ; (b) $1 + \mathrm{i}$ .

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