Algebra: logic and set; real numbers; functions (inverse, composite, algebraic/transcendental); curve sketching; sequences and series (AP/GP/HP, means, infinite GP); matrices and determinants; quadratic equations; complex numbers

$p$: "2 is an even number" is true, and $q$: "4 is an even number" is true.

The biconditional $p\leftrightarrow q$ ("2 is even if and only if 4 is even") is true when $p$ and $q$ have the same truth value. Here both are true, so $p\leftrightarrow q$ is true.

Let $y=f(x)=2x-3$. Solving for $x$:

$y+3=2x\;\Rightarrow\; x=\frac{y+3}{2}.$

Interchanging the roles of $x$ and $y$,

$f^{-1}(x)=\frac{x+3}{2}.$

Check: $f\!\left(\dfrac{x+3}{2}\right)=2\cdot\dfrac{x+3}{2}-3=x$.

Let $f(x)=8x^2$. Then

$f(-x)=8(-x)^2=8x^2=f(x)\quad\text{for all }x\in\mathbb{R}.$

So the function is even. The graph of an even function is symmetric about the $y$-axis: the point $(x,y)$ lies on the curve if and only if $(-x,y)$ does.

It is not odd, since $f(-x)\neq -f(x)$ unless $x=0$, so it is not symmetric about the origin.

For $A=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, the adjoint is obtained by swapping the diagonal entries and changing the signs of the off-diagonal entries:

$\operatorname{adj}A=\begin{pmatrix}d&-b\\-c&a\end{pmatrix}.$

For $A=\begin{pmatrix}2&5\\3&-7\end{pmatrix}$:

$\operatorname{adj}A=\begin{pmatrix}-7&-5\\-3&2\end{pmatrix}.$

Check: $A\cdot\operatorname{adj}A=\begin{pmatrix}-14-15&-10+10\\-21+21&-15-14\end{pmatrix}=-29I=|A|\,I$, and $|A|=-14-15=-29$.

Proof of $A-(B\cup C)=(A-B)\cap(A-C)$.

Let $x\in A-(B\cup C)$. Then $x\in A$ and $x\notin B\cup C$, so $x\notin B$ and $x\notin C$. Thus $x\in A-B$ and $x\in A-C$, i.e. $x\in(A-B)\cap(A-C)$. Hence

$A-(B\cup C)\subseteq(A-B)\cap(A-C).$

Conversely, let $x\in(A-B)\cap(A-C)$. Then $x\in A$, $x\notin B$ and $x\notin C$. So $x\notin B\cup C$, and therefore $x\in A-(B\cup C)$. Hence

$(A-B)\cap(A-C)\subseteq A-(B\cup C).$

Both inclusions hold, so $A-(B\cup C)=(A-B)\cap(A-C)$. $\blacksquare$

OR

The absolute value of a real number $x$ is

$|x|=\begin{cases}x,&x\ge0\\-x,&x<0.\end{cases}$

Proof of $|x|-|y|\le|x+y|$. By the triangle inequality $|a+b|\le|a|+|b|$ (which follows from $\pm a\le|a|$, $\pm b\le|b|$, giving $\pm(a+b)\le|a|+|b|$), take $a=x+y$ and $b=-y$:

$|x|=|(x+y)+(-y)|\le|x+y|+|-y|=|x+y|+|y|.$

So $|x|-|y|\le|x+y|$ for all $x,y\in\mathbb{R}$. $\blacksquare$

$f(x)=(x-1)^3$ is the graph of $y=x^3$ shifted 1 unit to the right.

Characteristics

• Domain $=\mathbb{R}$, range $=\mathbb{R}$.
• $x$-intercept: $f(x)=0\Rightarrow x=1$, i.e. $(1,0)$. $y$-intercept: $f(0)=-1$, i.e. $(0,-1)$.
• $f'(x)=3(x-1)^2\ge0$, so the function is increasing everywhere and has no maximum or minimum. At $x=1$, $f'(1)=0$, so the tangent there is horizontal.
• $f''(x)=6(x-1)$ changes sign at $x=1$, so $(1,0)$ is the point of inflection (concave down for $x<1$, concave up for $x>1$).
• Neither even nor odd about the $y$-axis or origin; it is symmetric about the point $(1,0)$.
• One-one and onto, so it has an inverse $f^{-1}(x)=1+\sqrt[3]{x}$.

Table of points

Plot these points and join them with a smooth S-shaped curve that rises from bottom-left to top-right, flattening momentarily at $(1,0)$.

Let $\Delta$ be the given determinant. Multiply $R_1,R_2,R_3$ by $a,b,c$ respectively and divide the whole by $abc$:

$\Delta=\frac1{abc}\begin{vmatrix}a^3+a&a^2b&a^2c\\ab^2&b^3+b&b^2c\\ac^2&bc^2&c^3+c\end{vmatrix}.$

Take $a$ out of $C_1$, $b$ out of $C_2$, $c$ out of $C_3$; the factor $abc$ cancels:

$\Delta=\begin{vmatrix}a^2+1&a^2&a^2\\b^2&b^2+1&b^2\\c^2&c^2&c^2+1\end{vmatrix}.$

Apply $R_1\to R_1+R_2+R_3$. Every entry of the new first row becomes $1+a^2+b^2+c^2$:

$\Delta=(1+a^2+b^2+c^2)\begin{vmatrix}1&1&1\\b^2&b^2+1&b^2\\c^2&c^2&c^2+1\end{vmatrix}.$

Now $C_2\to C_2-C_1$ and $C_3\to C_3-C_1$:

$\Delta=(1+a^2+b^2+c^2)\begin{vmatrix}1&0&0\\b^2&1&0\\c^2&0&1\end{vmatrix}=(1+a^2+b^2+c^2)\cdot1.$

Hence $\Delta=1+a^2+b^2+c^2$. $\blacksquare$

Let $\alpha,\beta$ be the roots of $x^2-x+1=0$. The required equation has roots $\dfrac1\alpha,\dfrac1\beta$.

Here $\alpha+\beta=1$ and $\alpha\beta=1$. Then

$\text{sum}=\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}=1,\qquad \text{product}=\frac1{\alpha\beta}=1.$

The equation is $x^2-(\text{sum})x+\text{product}=0$:

$x^2-x+1=0.$

(Equivalently, put $x=\dfrac1y$ in $x^2-x+1=0$ to get $1-y+y^2=0$.) The equation is unchanged: the roots are complex and the reciprocal of each is the other root.

Function. A function $f$ from a set $A$ to a set $B$ is a rule that assigns to each element $x\in A$ exactly one element $f(x)\in B$.

Relation vs function. A relation from $A$ to $B$ is any subset of $A\times B$; an element of $A$ may be related to none, one or several elements of $B$. A function is a special relation in which every element of $A$ has exactly one image.

Example. Let $A=\{1,2,3\}$, $B=\{a,b,c\}$.

• $R=\{(1,a),(1,b),(2,c)\}$ is a relation but not a function: $1$ has two images and $3$ has none.
• $f=\{(1,a),(2,a),(3,c)\}$ is a function: each element of $A$ has exactly one image.

Domain and range of $f(x)=\sqrt{x^2-2x-8}$.

Domain. We need $x^2-2x-8\ge0$, i.e. $(x-4)(x+2)\ge0$, so $x\le-2$ or $x\ge4$.

$\text{Domain}=(-\infty,-2]\cup[4,\infty).$

Range. Let $y=\sqrt{x^2-2x-8}\ge0$. Then $y^2=x^2-2x-8$, i.e. $x^2-2x-(8+y^2)=0$. For real $x$, the discriminant must be non-negative:

$4+4(8+y^2)\ge0,$

which holds for every real $y$. Also $x=1\pm\sqrt{9+y^2}$, and for $y\ge0$ these satisfy $x\le-2$ or $x\ge4$ (since $\sqrt{9+y^2}\ge3$). So every $y\ge0$ is attained (e.g. $y=0$ at $x=-2,4$).

$\text{Range}=[0,\infty).$

Let the two unequal positive numbers be $a$ and $b$. Then

$A=\frac{a+b}{2},\qquad G=\sqrt{ab},\qquad H=\frac{2ab}{a+b}.$

(ii) $G^2=A\times H$.

$A\times H=\frac{a+b}{2}\cdot\frac{2ab}{a+b}=ab=G^2.$

(i) $A>G>H$.

$A>G$: $A-G=\frac{a+b-2\sqrt{ab}}{2}=\frac{(\sqrt a-\sqrt b)^2}{2}>0,$ because $a\ne b$. So $A>G$.

$G>H$: From (ii), $H=\dfrac{G^2}{A}$. Since $A>G>0$, we have $\dfrac G A<1$, so $H=G\cdot\frac GA<G.$ So $G>H$.

Therefore $A>G>H$ and $G^2=A\times H$. $\blacksquare$

By definition $X-Y=X\cap\overline{Y}$. Therefore

$A-\overline{B}=A\cap\overline{\overline{B}}=A\cap B,$

since the complement of the complement of $B$ is $B$. $\blacksquare$

(Elementwise: $x\in A-\overline B\iff x\in A$ and $x\notin\overline B\iff x\in A$ and $x\in B\iff x\in A\cap B$.)

$B\cup C=\{b,c,d\}$ and $B\cap C=\{c\}$.

$A\times(B\cup C)=\{a,b\}\times\{b,c,d\}=\{(a,b),(a,c),(a,d),(b,b),(b,c),(b,d)\}.$

$A\times(B\cap C)=\{a,b\}\times\{c\}=\{(a,c),(b,c)\}.$

A function is periodic if there is a positive $T$ with $f(x+T)=f(x)$ for all $x$.

$f(x+T)=\cos\pi(x+T)=\cos(\pi x+\pi T).$

This equals $\cos\pi x$ for all $x$ when $\pi T=2\pi$, the smallest positive such value, i.e. $T=2$. Indeed $\cos(\pi x+2\pi)=\cos\pi x$.

So $f(x)=\cos\pi x$ is periodic, with period $2$.

$A=\begin{pmatrix}2&1\\1&-2\end{pmatrix}$, so $A^T=\begin{pmatrix}2&1\\1&-2\end{pmatrix}$ (A is symmetric).

$AA^T=\begin{pmatrix}2&1\\1&-2\end{pmatrix}\begin{pmatrix}2&1\\1&-2\end{pmatrix}=\begin{pmatrix}4+1&2-2\\2-2&1+4\end{pmatrix}=\begin{pmatrix}5&0\\0&5\end{pmatrix}.$

Right side: $(1+i)(4-3i)=4-3i+4i-3i^2=4+i+3=7+i$.

Left side: $(y+1)+(x-1)i$.

Equating real and imaginary parts:

$y+1=7\Rightarrow y=6,\qquad x-1=1\Rightarrow x=2.$

So $x=2$ and $y=6$.

If $x=0$ is a root of $3x^2+7x+6-K=0$, substituting gives $6-K=0$, so

$K=6.$

Check: the equation becomes $3x^2+7x=0$, i.e. $x(3x+7)=0$, with roots $0$ and $-\tfrac73$.

Conjunction. The conjunction of two statements $p$ and $q$ is the compound statement "$p$ and $q$", written $p\wedge q$. It is true only when both $p$ and $q$ are true, and false in every other case.

Truth table of $(p\wedge q)\wedge\sim(p\vee q)$:

Conclusion. The last column is false for every combination of truth values of $p$ and $q$, so the statement is a contradiction (an absurdity). This is expected: $p\wedge q$ needs both true, while $\sim(p\vee q)$ needs both false.

OR: $|2x-1|\ge3$.

$|2x-1|\ge3$ means $2x-1\ge3$ or $2x-1\le-3$.

$2x-1\ge3\Rightarrow x\ge2$, and $2x-1\le-3\Rightarrow x\le-1$.

Solution set: $x\le-1$ or $x\ge2$, i.e. $(-\infty,-1]\cup[2,\infty)$.

Graph: on the number line draw closed (filled) dots at $-1$ and $2$. Shade the line to the left of $-1$ and to the right of $2$, leaving the part between $-1$ and $2$ unshaded.

$f(x)=x-x^2=-\left(x-\tfrac12\right)^2+\tfrac14$ is a quadratic whose graph is a parabola.

Characteristics

• The coefficient of $x^2$ is $-1<0$, so the parabola opens downward.
• Vertex: $\left(\tfrac12,\tfrac14\right)$, which is the maximum point. The axis of symmetry is $x=\tfrac12$.
• $x$-intercepts: $x-x^2=0\Rightarrow x(1-x)=0$, so $x=0$ and $x=1$.
• $y$-intercept: $f(0)=0$, so the curve passes through the origin.
• Domain $=\mathbb{R}$; range $=\left(-\infty,\tfrac14\right]$.
• $f$ is increasing for $x<\tfrac12$ and decreasing for $x>\tfrac12$.

Table of values

Graph. Plot these points on the axes and join them with a smooth curve. It rises from the lower left through $(0,0)$ to the peak $\left(\tfrac12,\tfrac14\right)$, then falls through $(1,0)$ to the lower right. It is symmetric about the line $x=\tfrac12$.

Let $\Delta$ be the given determinant. Apply $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$:

$\Delta=\begin{vmatrix}0&x-y&x^2-y^2\\0&y-z&y^2-z^2\\1&z&z^2\end{vmatrix}.$

Take out the common factor $(x-y)$ from $R_1$ and $(y-z)$ from $R_2$:

$\Delta=(x-y)(y-z)\begin{vmatrix}0&1&x+y\\0&1&y+z\\1&z&z^2\end{vmatrix}.$

Expand along the first column; only the entry $1$ in the third row contributes, with sign $+$:

$\Delta=(x-y)(y-z)\begin{vmatrix}1&x+y\\1&y+z\end{vmatrix}=(x-y)(y-z)\big[(y+z)-(x+y)\big]=(x-y)(y-z)(z-x).$

Hence $\Delta=(y-z)(z-x)(x-y)$.

Let the roots of $ax^2+bx+c=0$ be $\alpha$ and $\alpha^2$. Then

$\alpha+\alpha^2=-\frac{b}{a},\qquad \alpha\cdot\alpha^2=\alpha^3=\frac{c}{a}.$

Cube the first relation, using $(m+n)^3=m^3+n^3+3mn(m+n)$ with $m=\alpha,\ n=\alpha^2$:

$(\alpha+\alpha^2)^3=\alpha^3+\alpha^6+3\alpha^3(\alpha+\alpha^2).$

Substitute $\alpha^3=\dfrac ca$, $\alpha^6=\dfrac{c^2}{a^2}$ and $\alpha+\alpha^2=-\dfrac ba$:

$-\frac{b^3}{a^3}=\frac ca+\frac{c^2}{a^2}-\frac{3bc}{a^2}.$

Multiply by $a^3$:

$-b^3=a^2c+ac^2-3abc\quad\Longrightarrow\quad b^3+a^2c+ac^2=3abc.$

$f:\mathbb{R}\to\mathbb{R}$, $f(x)=cx+d$, with $c\ne0$.

One-one. Let $x_1,x_2\in\mathbb{R}$ with $f(x_1)=f(x_2)$. Then

$cx_1+d=cx_2+d\ \Rightarrow\ cx_1=cx_2\ \Rightarrow\ x_1=x_2\quad(\text{since }c\ne0).$

Different inputs therefore give different outputs, so $f$ is one-one.

Onto. Let $y\in\mathbb{R}$ be any element of the codomain. Solve $y=cx+d$ for $x$:

$x=\frac{y-d}{c},$

which is a real number because $c\ne0$. Also $f\!\left(\dfrac{y-d}{c}\right)=c\cdot\dfrac{y-d}{c}+d=y$. So every $y$ has a pre-image in $\mathbb{R}$, and $f$ is onto.

Inverse. Since $f$ is one-one and onto, $f^{-1}$ exists. Put $y=cx+d$ and solve for $x$: $x=\dfrac{y-d}{c}$. Interchanging the names of the variables,

$f^{-1}(x)=\frac{x-d}{c}.$

Check: $f\big(f^{-1}(x)\big)=c\cdot\dfrac{x-d}{c}+d=x$.

Let the three numbers in A.P. be $a-d,\ a,\ a+d$.

Sum $=3a=36\Rightarrow a=12$. The numbers are $12-d,\ 12,\ 12+d$.

After increasing them by $1,4,43$ respectively, we get $13-d,\ 16,\ 55+d$, which are in G.P. So the middle term squared equals the product of the others:

$16^2=(13-d)(55+d)$

$256=715+13d-55d-d^2=715-42d-d^2$

$d^2+42d-459=0\ \Rightarrow\ (d-9)(d+51)=0\ \Rightarrow\ d=9\ \text{or}\ d=-51.$

Case $d=9$: the numbers are $3,\ 12,\ 21$. Check: $4,\ 16,\ 64$ is a G.P. with ratio $4$.

Case $d=-51$: the numbers are $63,\ 12,\ -39$. Check: $64,\ 16,\ 4$ is a G.P. with ratio $\tfrac14$.

Hence the numbers are $3,\ 12,\ 21$ (or, taking the other value of $d$, $63,\ 12,\ -39$).

(i) $f(x)=x^3$ is a polynomial function (a cubic), so it is algebraic in nature.

(ii) The graph $y=x^3$ is a cubic curve (cubic parabola).

(iii) The curve has no vertex, because it has no turning point. It is strictly increasing, and the origin $(0,0)$ is its point of inflection.

(iv) One property for sketching: $f(-x)=-f(x)$, so the function is odd and the curve is symmetric about the origin. It passes through $(0,0)$ and lies in the first and third quadrants.

(v) Domain $=\mathbb{R}$, the set of all real numbers.

Let

$S_1 = 1 + 2a + 3a^2 + \ldots + na^{n-1},\qquad S_2 = a + 2a^2 + 3a^3 + \ldots + na^{n}.$

Each term of $S_2$ is $a$ times the corresponding term of $S_1$, so $S_2 = aS_1$. Hence

$S_1 - S_2 = (1-a)S_1.$

To find $S_1$ explicitly, subtract $S_2$ from $S_1$ term by term:

$S_1 - S_2 = 1 + a + a^2 + \ldots + a^{n-1} - na^n = \frac{1-a^n}{1-a} - na^n.$

Therefore, for $a \ne 1$,

$S_1 = \frac{1-(n+1)a^n + na^{n+1}}{(1-a)^2},\qquad S_2 = aS_1 = \frac{a-(n+1)a^{n+1}+na^{n+2}}{(1-a)^2}.$

Comparison: $S_2 = aS_1$.

• If $a>1$ (with $a>0$), then $S_2 > S_1$.
• If $0<a<1$, then $S_2 < S_1$.
• If $a=1$, both sums equal $\frac{n(n+1)}{2}$.

a) As typed, the right-hand side $a\sin\left(\frac B2+C\right)$ does not give an identity. For the right isosceles triangle with $A=90^\circ$, $B=C=45^\circ$, $a=\sqrt2$, $b=c=1$, the left side is $2\sin45^\circ=\sqrt2$ but the right side is $\sqrt2\sin67.5^\circ\ne\sqrt2$. The intended result is

$(b+c)\sin\frac A2 = a\sin\left(\frac A2 + C\right).$

Proof. By the sine rule, $b=2R\sin B$, $c=2R\sin C$, $a=2R\sin A$. Then

$(b+c)\sin\frac A2 = 2R(\sin B+\sin C)\sin\frac A2.$

Since $\sin B+\sin C = 2\sin\frac{B+C}{2}\cos\frac{B-C}{2}$ and $\frac{B+C}{2}=\frac{\pi}{2}-\frac A2$, we have $\sin\frac{B+C}{2}=\cos\frac A2$. So

$(b+c)\sin\frac A2 = 2R\cdot 2\sin\frac A2\cos\frac A2\cos\frac{B-C}{2} = 2R\sin A\cos\frac{B-C}{2} = a\cos\frac{B-C}{2}.$

Now $\frac A2 + C = \frac{\pi}{2}-\frac{B+C}{2}+C = \frac{\pi}{2}-\frac{B-C}{2}$, so $\sin\left(\frac A2+C\right)=\cos\frac{B-C}{2}$. Hence

$(b+c)\sin\frac A2 = a\sin\left(\frac A2+C\right).$

b) Let $\vec r = x\vec a + y\vec b$. Then $(4,7)=x(5,-4)+y(-2,5)$, giving

$5x-2y=4,\qquad -4x+5y=7.$

The determinant is $25-8=17$.

$x=\frac{4\cdot5-(-2)\cdot7}{17}=\frac{34}{17}=2,\qquad y=\frac{5\cdot7-(-4)\cdot4}{17}=\frac{51}{17}=3.$

So $\vec r = 2\vec a + 3\vec b$.

Check: $2(5,-4)+3(-2,5)=(10-6,\,-8+15)=(4,7)$.