Algebra 3 |
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- Inequalities use symbols such as \(>\), \(\ge\), \(<\), \(\le\) to describe ranges of values that satisfy a condition.
- Basic rules:
- Adding or subtracting the same number on both sides leaves the inequality unchanged.
- Multiplying or dividing both sides by a positive number preserves the direction.
- Multiplying or dividing by a negative number reverses the direction.
- Solutions can be expressed in interval notation or shown on a number line.
- Compound (double) inequalities, e.g. \(-2 < x \le 5\), describe continuous ranges.
- Graphical and algebraic methods are used to determine solution sets efficiently.
Solving Linear Inequalities
Example 2: Solve \(\tfrac16(x-1)\ge\tfrac13(x-4),\; x\in\mathbb R\), and graph your solution.
Example 3: Solve \(-9<3-4x\le1,\; x\in\mathbb R\), and graph your solution.
Example 4:
(i) Find \(A=\{x\mid 7\le 10-3x,\; x\in\mathbb R\}\).
(ii) Find \(B=\{x\mid 2>\tfrac43-2x,\; x\in\mathbb R\}\).
(iii) Find \(A\cap B\) and graph the solution on the number line.
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Inequalities — Interactive Questions
Solution.
\[ 3x + 7 > x + 2 \;\Rightarrow\; 2x > -5 \;\Rightarrow\; x > -\tfrac{5}{2}. \]
Hence, \(\boxed{x > -2.5}\).
Solution.
Multiply through by \(6\) (positive):
\[ x - 1 > 2(x - 4) \;\Rightarrow\; x - 1 > 2x - 8 \;\Rightarrow\; -1 + 8 > x \;\Rightarrow\; x < 7. \]
Thus, \(\boxed{x < 7}\).
Solution.
Subtract \(3\) from each part:
\[ -12 < -4x < -2. \]
Divide through by \(-4\) (reverse the inequality signs):
\[ 3 > x > \tfrac{1}{2} \;\;\Rightarrow\;\; \boxed{\tfrac{1}{2} < x < 3}. \]
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Solve the inequality \(\;3x + 7 \ge x + 2,\; x \in \mathbb{Z}\;\) and plot the solution on a number line.
Find the set of values for which \(\,3(x - 2) > x - 4\,\) and \(\,4x + 12 > 2x + 17,\; x \in \mathbb{R},\) and plot your answers on a number line.
- Write the inequality in the form \(ax^{2}+bx+c \ \,\square\,\ 0\) where \(\square \in \{<, \le, >, \ge\}\).
- Find the critical values by solving \(ax^{2}+bx+c=0\) (by factorising, completing the square, or the quadratic formula).
- Use a sign chart or a sketch of the parabola \(y=ax^{2}+bx+c\):
- If \(a>0\) (opens upward): the quadratic is positive outside the roots and negative between the roots.
- If \(a<0\) (opens downward): the quadratic is negative outside the roots and positive between the roots.
- Include endpoints when the inequality is non-strict (\(\le\) or \(\ge\)); exclude them for strict inequalities (\(<\) or \(>\)).
- If the discriminant is negative (\(\Delta<0\)):
- For \(a>0\): \(ax^{2}+bx+c>0\) for all \(x\); \(ax^{2}+bx+c<0\) has no real solution.
- For \(a<0\): \(ax^{2}+bx+c<0\) for all \(x\); \(ax^{2}+bx+c>0\) has no real solution.
- If there is a repeated root (\(\Delta=0\)), the quadratic touches the axis once; for strict inequalities the single point is excluded, for non-strict it is included.
- Avoid multiplying both sides by an expression in \(x\). Instead, keep one side \(0\) and reason from the sign of \(ax^{2}+bx+c\).
Solve the inequality \(\;12 - 5x - 2x^2 > 0\;\).
Solve \(\,\dfrac{2x-7}{x+3} < 1,\; x \ne -3\,\).
- \(|x|\) is the distance of \(x\) from \(0\): \[ |x|=\begin{cases} x,&x\ge0\\ -x,&x<0 \end{cases} \]
- To solve \(|ax+b|=k\) (\(k\ge0\)): solve \(ax+b=k\) or \(ax+b=-k\).
- \(|ax+b|
- \(|ax+b|>k \Rightarrow ax+b>k\) or \(ax+b<-k\)\, (two rays).
- When both sides contain moduli, square both sides (non-negative) or use identities — always verify solutions.
Solution. \(\lvert 2x-5\rvert=7 \iff 2x-5=7\) or \(2x-5=-7\).
\[ 2x=12 \Rightarrow x=6;\qquad 2x=-2 \Rightarrow x=-1. \]
Hence, \(\boxed{x\in\{-1,6\}}.\)
Solution. \(\lvert x+3\rvert<5 \iff -5 \[
-8 Hence, \(\boxed{x\in(-8,2)}.\)
Solution. Both sides are non-negative; square:
\[ (2x+1)^2 \ge (x-4)^2. \]
Difference of squares:
\[ \big[(2x+1)-(x-4)\big]\big[(2x+1)+(x-4)\big] \ge 0 \;\Rightarrow\; (x+5)(3x-3)\ge 0. \]
Critical points: \(-5\) and \(1\). Sign analysis \(\Rightarrow (-\infty,-5]\cup[1,\infty)\).
Therefore, \(\boxed{x\le -5\ \text{ or }\ x\ge 1}.\)
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- A mathematical proof is a logical argument demonstrating that a statement holds for all cases, using definitions, prior results, and valid reasoning.
- Common methods:
- Direct proof: deduce the conclusion from assumptions via algebra and definitions.
- Contradiction: assume the negation of the claim and derive an impossibility.
- Induction: prove a base case and an inductive step to establish truth for all \(n\in\mathbb{N}\).
- Good proofs state assumptions, proceed in clear logical steps, and end with a justified conclusion.
- Useful symbols: \(\Rightarrow\) (implies), \(\Leftrightarrow\) (iff), \(\therefore\) (therefore), Q.E.D.
- A mathematical proof is a logical argument showing that a statement is always true using accepted facts and reasoning.
- Common methods:
- Direct proof — apply definitions and algebra to show a result follows directly.
- Proof by contradiction — assume the opposite of what you want to prove and find a contradiction.
- Proof by induction — used to prove statements for all positive integers.
- Each proof should clearly show assumptions, logical steps, and a conclusion.
- Symbols such as “⇒”, “⇔”, “∴”, and “Q.E.D.” denote logical flow and completion.
Mathematical Proof — Interactive Questions
Solution — Direct Proof.
\[ n=2k\text{ for some integer }k \;\Rightarrow\; n^{2}=(2k)^{2}=4k^{2}=2(2k^{2}). \]
Since \(2k^{2}\) is an integer, \(n^{2}\) is even. ∴ proved.
Solution — Proof by Contradiction.
Assume \(\sqrt{2}\) is rational ⇒ \(\sqrt{2}=\tfrac{p}{q}\) in lowest terms (\(p,q\in\mathbb{Z}, q\ne0\)).
\[ 2=\frac{p^{2}}{q^{2}}\;\Rightarrow\;p^{2}=2q^{2}. \]
Then \(p^{2}\) even ⇒ \(p\) even ⇒ \(p=2k\). Substitute: \(4k^{2}=2q^{2}\Rightarrow q^{2}=2k^{2}\Rightarrow q\) even. Hence both \(p\) and \(q\) even ⇒ contradiction (not in lowest terms). ∴ \(\sqrt{2}\) is irrational.
Simple Example of Direct Proof
Simple proof by contradiction
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Proof root is irrational by contradiction (this is mentioned in the syllabus!)
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- Common proof strategies for inequalities include:
- Using non-negativity of squares, e.g. \((x-y)^{2}\ge 0\).
- Completing the square to compare expressions.
- Applying standard results such as the AM–GM or triangle inequalities.
- AM–GM (Arithmetic Mean–Geometric Mean Inequality):
For non-negative numbers \(x,y\), \[ \frac{x+y}{2} \ge \sqrt{xy}, \] with equality only when \(x=y\). It states that the arithmetic mean of two (or more) non-negative numbers is always at least as great as their geometric mean. - When proving inequalities, clearly identify when equality holds and specify any domain restrictions (e.g. \(x,y\ge0\)).
Abstract Inequalities — Interactive Questions
Solution.
\[ (a-b)^{2}\ge 0 \;\Rightarrow\; a^{2}-2ab+b^{2}\ge 0 \;\Rightarrow\; a^{2}+b^{2}\ge 2ab. \]
Equality holds iff \((a-b)^{2}=0\), i.e. \(\boxed{a=b}\).
Solution.
Start with a non-negative square: \((\sqrt{x}-\sqrt{y})^{2}\ge 0\).
\[ x-2\sqrt{xy}+y\ge 0 \;\Rightarrow\; x+y\ge 2\sqrt{xy} \;\Rightarrow\; \frac{x+y}{2}\ge \sqrt{xy}. \]
Equality occurs when \(\sqrt{x}=\sqrt{y}\), i.e. \(\boxed{x=y}\).
Solution.
Square both sides (non-negative quantities):
\[ \lvert a+b\rvert^{2}=(a+b)^{2}=a^{2}+2ab+b^{2} \le a^{2}+2\lvert a\rvert\lvert b\rvert+b^{2} =(\lvert a\rvert+\lvert b\rvert)^{2}. \]
Taking square roots gives \(\lvert a+b\rvert\le \lvert a\rvert+\lvert b\rvert.\)
Equality holds when \(a,b\) have the same sign (or one is zero).
- Indices (or exponents/powers) indicate repeated multiplication: \[ a^{n} = \underbrace{a\times a\times\cdots\times a}_{n\text{ times}}. \]
- Index Laws (for \(a,b\ne0\), \(m,n\in\mathbb{R}\)):
- \(a^{m}\,a^{n}=a^{m+n}\)
- \(\dfrac{a^{m}}{a^{n}}=a^{m-n}\)
- \((a^{m})^{n}=a^{mn}\)
- \((ab)^{n}=a^{n}b^{n}\)
- \(\left(\dfrac{a}{b}\right)^{n}=\dfrac{a^{n}}{b^{n}}\)
- \(a^{0}=1\) (for \(a\ne0\))
- \(a^{-n}=\dfrac{1}{a^{n}}\)
- \(a^{\tfrac{1}{n}}=\sqrt[n]{a}\), so \(a^{\tfrac{m}{n}}=(\sqrt[n]{a})^{m}\).
- Indices follow the same algebraic rules for rational and real powers, provided \(a>0\) when roots are involved.
Indices — Interactive Questions
Solution.
Multiply coefficients and add indices of like bases:
\[ 3x^{2}y^{3}\times2x^{4}y = (3\times2)x^{2+4}y^{3+1}=6x^{6}y^{4}. \]
\(\boxed{6x^{6}y^{4}}\)
Solution.
Subtract exponents of same bases:
\[ a^{-3-2}b^{4-(-1)}=a^{-5}b^{5} =\frac{b^{5}}{a^{5}}. \]
\(\boxed{\dfrac{b^{5}}{a^{5}}}\)
Solution.
Apply \((a^{m})^{n}=a^{mn}\):
\[ (16x^{8})^{\tfrac{1}{4}} =16^{\tfrac{1}{4}}x^{8\tfrac{1}{4}} =2x^{2}. \]
\(\boxed{2x^{2}}\)
- Use laws of indices to rewrite with a common base where possible. If \(a>0\) and \(a\ne1\), then \(a^{x}=a^{y}\iff x=y\).
- When bases differ, take logarithms (any base): from \(a^{f(x)}=b\) we get \(f(x)=\dfrac{\ln b}{\ln a}\) or generally \(f(x)\ln a=\ln b\).
- Factor when an equation contains repeated terms (e.g. \(2^{x}\) appearing in several places).
- Always consider domains (exponentials are positive) and check for extraneous results introduced by algebraic manipulation.
Exponential Equations — Interactive Questions
Solution.
Since \(27=3^{3}\), we have \(3^{x+1}=3^{3}\Rightarrow x+1=3\Rightarrow \boxed{x=2}.\)
Solution.
Take natural logs (any log base works):
\[ x\ln 7=(x+1)\ln 3\ \Rightarrow\ x\ln 7=x\ln 3+\ln 3 \ \Rightarrow\ x(\ln 7-\ln 3)=\ln 3. \]
\[ \boxed{\,x=\dfrac{\ln 3}{\ln 7-\ln 3}\,}. \]
Solution.
Factor \(2^{x}\): \(2^{x}(1+2)=12\Rightarrow 3\cdot2^{x}=12\Rightarrow 2^{x}=4.\)
Hence \(x=\boxed{2}\).
- Exponential functions have the form \(f(x)=a^{x}\) (or more generally \(f(x)=A\,a^{x}\)) with \(a>0,\ a\neq1\).
- Key points: \(f(0)=1\Rightarrow(0,1)\), \(f(1)=a\Rightarrow(1,a)\), \(f(-1)=\tfrac{1}{a}\Rightarrow(-1,\tfrac{1}{a})\).
- Domain: \(\mathbb{R}\); Range: \(f(x)>0\); horizontal asymptote \(y=0\).
- Behaviour: growth if \(a>1\); decay if \(0<a<1\).
- Applications include growth/decay and other real-world models.
Exponential Functions — Interactive Questions
Solution.
(i) \((0,1),\ (1,a),\ \big(-1,\tfrac{1}{a}\big)\).
(ii) Horizontal asymptote: \(y=0\); range: \(f(x)>0\).
(iii) Growth if \(a>1\); decay if \(0<a<1\).
Solution.
(i) Since \(0.72<1\), this is decay as \(T\) increases.
(ii) \[ D(5)=18(0.72)^{5},\qquad D(2)=18(0.72)^{2},\qquad D(0)=18. \]
(iii) Solve \(18(0.72)^{T}\ge 5\Rightarrow (0.72)^{T}\ge \tfrac{5}{18}\). Taking logs: \(T\le\dfrac{\ln(5/18)}{\ln(0.72)}\) (note \(\ln(0.72)<0\), so the inequality reverses when dividing). Give \(T\) to the nearest whole degree as required.
Solution.
(i) \(\displaystyle P(200)=100(0.99988)^{200},\qquad P(500)=100(0.99988)^{500}.\)
(ii) Set \(100(0.99988)^{n}=50\Rightarrow (0.99988)^{n}=0.5\Rightarrow n=\dfrac{\ln 0.5}{\ln 0.99988}\) years (round appropriately).
(iii) Set \(100(0.99988)^{n}=79\Rightarrow n=\dfrac{\ln(0.79)}{\ln(0.99988)}\) years.
- Definition: for \(a>0,\ a\neq1,\ x>0\), \(y=\log_{a}x\iff a^{y}=x\).
- Laws of logs (same base \(a\)):
- \(\log_{a}(XY)=\log_{a}X+\log_{a}Y\)
- \(\log_{a}\!\left(\dfrac{X}{Y}\right)=\log_{a}X-\log_{a}Y\)
- \(\log_{a}(X^{k})=k\,\log_{a}X\)
- Change of base: \(\displaystyle \log_{a}x=\frac{\log_{b}x}{\log_{b}a}\).
- \(\ln x=\log_{e}x\). Logs are defined only for \(x>0\).
- Graph \(y=\log_{a}x\) (\(a>1\)): domain \(x>0\); range \(\mathbb{R}\); vertical asymptote \(x=0\); increasing if \(a>1\).
- Exponential and logarithmic functions are inverses: \(a^{\log_{a}x}=x\) and \(\log_{a}(a^{x})=x\).
Logarithms & Log Functions — Interactive Questions
(a) \(\log_{2}32-\log_{2}4+\log_{2}\!\left(\tfrac{1}{8}\right)\) (b) \(\log_{3}27^{\,2/3}\) (c) \(\dfrac{\log 50}{\log 2}-\dfrac{\log 25}{\log 2}\)
Solution.
(a) \(\log_{2}32=5,\ \log_{2}4=2,\ \log_{2}\!\big(\tfrac{1}{8}\big)=-3\Rightarrow5-2-3=0.\ \boxed{0}\)
(b) \(\log_{3}27^{\,2/3}=\tfrac{2}{3}\log_{3}27=\tfrac{2}{3}\times3=\boxed{2}\)
(c) \(\dfrac{\log50}{\log2}-\dfrac{\log25}{\log2} =\dfrac{\log(50/25)}{\log2}=\log_{2}2=\boxed{1}\)
(i) \(5^{x}=140\) (ii) \(3^{2x-1}=10\)
Solution.
(i) Take logs (base 10 or \(e\)): \(x\log5=\log140\Rightarrow \boxed{x=\frac{\log140}{\log5}\approx3.12}\)
(ii) \((2x-1)\log3=\log10\Rightarrow2x-1=\frac{\log10}{\log3}\Rightarrow \boxed{x=\tfrac12\!\left(1+\frac{\log10}{\log3}\right)\approx1.55}\)
(i) Find the pH when \([H^{+}]=4.0\times10^{-5}\).
(ii) If the pH is 3.4, find \([H^{+}]\).
(iii) Explain how an increase of 1 pH unit affects \([H^{+}]\).
Solution.
(i) \(pH=-\log_{10}(4.0\times10^{-5}) =-(\log4.0+\log10^{-5})=-(0.6021-5)=4.398\Rightarrow\boxed{pH\approx4.40}\)
(ii) \([H^{+}]=10^{-pH}=10^{-3.4}=\boxed{4.0\times10^{-4}}\text{ mol L}^{-1}\)
(iii) Each increase of 1 in pH corresponds to \([H^{+}]\) dividing by 10 (ten-fold decrease in concentration).
Solution.
Domain: \(x>-1\) and \(2x+3>0\Rightarrow x>-1.5\); hence \(x>-1\).
\(\log_{5}\!\frac{2x+3}{x+1}=2\Rightarrow\frac{2x+3}{x+1}=25 \Rightarrow2x+3=25x+25\Rightarrow23x=-22\Rightarrow \boxed{x=-\tfrac{22}{23}}\).
Check domain: \(-\tfrac{22}{23}\approx-0.956>-1\) ✅ valid solution.
- Exponential and logarithmic functions are inverses: \[ y=a^{x}\;\Longleftrightarrow\;x=\log_{a}y. \]
- The graph of \(y=\log_{a}x\) is the reflection of \(y=a^{x}\) in the line \(y=x\).
- Exponential \(y=a^{x}\): domain \(\mathbb{R}\); range \((0,\infty)\); horizontal asymptote \(y=0\).
- Logarithmic \(y=\log_{a}x\): domain \((0,\infty)\); range \(\mathbb{R}\); vertical asymptote \(x=0\).
- Typical inverse pairs: \(y=10^{x}\) ↔ \(y=\log_{10}x\); \(y=2^{x}\) ↔ \(y=\log_{2}x\); \(y=e^{x}\) ↔ \(y=\ln x\).
- At \(x=1,\ y=0\) for all logs; at \(x=0,\ y=1\) for all exponentials.
Comparing Logarithmic & Exponential Functions — Interactive Questions
Solution.
- The graphs are inverses: each point \((x,y)\) on \(y=10^{x}\) corresponds to \((y,x)\) on \(y=\log_{10}x\).
- They are reflections in the line \(y=x\).
- They intersect where \(x=y\Rightarrow y=10^{y}\), giving approximately \((1.26,1.26)\).
- \(y=10^{x}\) has horizontal asymptote \(y=0\); \(y=\log_{10}x\) has vertical asymptote \(x=0\).
Solution.
| x | \(y=10^{x}\) | \(y=\log_{10}x\) |
|---|---|---|
| -2 | 0.01 | – undefined |
| -1 | 0.1 | – undefined |
| 0 | 1 | 0 |
| 1 | 10 | 1 |
| 2 | 100 | 2 |
The ordered pairs of one graph become reversed on the other: e.g. \(10^{1}=10\) ↔ \(\log_{10}10=1\). Thus, the two graphs are mirror images in \(y=x\).
- Recognise exponential growth or decay models: \(A(t)=A_{0}(1+r)^{t}\) or \(A(t)=A_{0}a^{t}\) where \(a>0,\ a\neq1\).
- Use logarithms to solve for unknown time or exponent: from \(a^{x}=k\) we get \(x=\dfrac{\ln k}{\ln a}\).
- Always check that exponential quantities are positive and that logarithm arguments are valid.
- Include correct units and interpret results in context.
- For crossover problems, equate models and solve for the time when they are equal.
Problem-solving — Interactive Questions
Solution.
(i) The growth model is \[ A(t)=5000(1.032)^{t}. \] For \(t=6\): \[ A(6)=5000(1.032)^{6}\approx6040.3. \] Hence the balance after six years is €6,040 (approx.).
(ii) To find when the amount doubles: \[ 5000(1.032)^{t}=10000 \;\Longrightarrow\; (1.032)^{t}=2. \] Taking logs: \[ t=\frac{\ln2}{\ln(1.032)}\approx21.96. \] Therefore, the money doubles in about 22 years.
Solution.
Set \(\dfrac{m(t)}{m_{0}}=0.20\): \[ (0.87)^{t}=0.20 \;\Longrightarrow\; t=\frac{\ln(0.20)}{\ln(0.87)}. \] \[ t\approx11.5. \] Therefore, it takes about 12 days for the mass to fall to \(20\%\) of its original value.
Solution.
Set \(200(1.08)^{t}=500(0.92)^{t}\). Divide both sides by \(200\): \[ \left(\frac{1.08}{0.92}\right)^{t}=\frac{500}{200}=2.5. \] Taking logs: \[ t=\frac{\ln(2.5)}{\ln(1.08/0.92)}\approx5.7. \] Hence, after about 6 years, the growing quantity \(A\) overtakes \(B\).
- Induction structure: (i) Base case; (ii) Inductive hypothesis (assume true for \(n=k\)); (iii) Inductive step (prove for \(n=k+1\)); (iv) Conclude for all stated \(n\).
- Included exemplars: one identity, one inequality, one factorisation (divisibility) result.
- Clearly state domains (e.g. \(n\in\mathbb{N}\), or \(n\ge 4\)), and show every logical step.
Induction — Interactive Questions (3 Exemplars)
Proof (Induction).
Base case: \(n=1\): LHS \(=1\), RHS \(=\frac{1\cdot2}{2}=1\) ✓.
Inductive step: Assume for \(n=k\) that \(1+\cdots+k=\frac{k(k+1)}{2}\). Then \[ 1+\cdots+k+(k+1)=\frac{k(k+1)}{2}+(k+1) =(k+1)\!\left(\frac{k}{2}+1\right) =\frac{(k+1)(k+2)}{2}. \] Hence the result holds for \(n=k+1\). By induction, it holds for all \(n\in\mathbb{N}\).
Proof (Induction).
Base case: \(n=4\): \(2^{4}=16=4^{2}\) ✓.
Inductive step: Assume \(2^{k}\ge k^{2}\) for some \(k\ge4\). Then \[ 2^{k+1}=2\cdot2^{k}\ \ge\ 2k^{2}. \] It suffices to show \(2k^{2}\ge (k+1)^{2}\), i.e. \[ 2k^{2}-(k+1)^{2}=k^{2}-2k-1\ \ge\ 0\quad(\text{true for }k\ge4). \] Hence \(2^{k+1}\ge (k+1)^{2}\). By induction, \(2^{n}\ge n^{2}\) for all \(n\ge4\).
Proof (Induction).
Base case: \(n=1\): \(4^{1}-1=3\) is divisible by \(3\) ✓.
Inductive step: Assume \(4^{k}-1=3m\) for some \(m\in\mathbb{Z}\). Then \[ 4^{k+1}-1=4\cdot4^{k}-1 =4(4^{k}-1)+3 =4\cdot(3m)+3 =3(4m+1), \] which is divisible by \(3\). Therefore \(3\mid(4^{k+1}-1)\). By induction, \(3\) divides \(4^{n}-1\) for all \(n\in\mathbb{N}\).
