Mathematics important questions

(a) From $x-y=0$: $y=x$. From $-3x+z+2=0$: $z=3x-2$. Substitute in the second: $$ \begin{aligned} 2x-y+4z &= 18\Rightarrow2x-x+4(3x-2) \ &= 18\Rightarrow x+12x-8 \ &= 18\Rightarrow13x \ &= 26\Rightarrow x \ &= 2. \end{aligned} $$ Then $y=2$ and $z=3(2)-2=4$. $$\boxed{x=2,\ y=2,\ z=4}$$ Check: $2(2)-2+4(4)=4-2+16=18$. Correct.

(b) If $\alpha,\beta$ are the roots of $x^2-x+1=0$, then $\alpha+\beta=1,\ \alpha\beta=1$. The reciprocal roots $\tfrac1\alpha,\tfrac1\beta$ have sum $\dfrac{\alpha+\beta}{\alpha\beta}=1$ and product $\dfrac1{\alpha\beta}=1$. Required equation: $$x^2-x+1=0\quad(\text{same as the original reciprocal equation}).$$

(a) Subtract the first equation from the second and third: $$ \begin{aligned} (x+2y+3z)-(x+y+z) &= 4-1\Rightarrow y+2z \ &= 3. \ (x+3y+7z)-(x+2y+3z) &= 13-4\Rightarrow y+4z \ &= 9. \end{aligned} $$ Subtract: $2z=6\Rightarrow z=3$, then $y=3-2(3)=-3$, and $x=1-(-3)-3=1$. $$\boxed{x=1,\ y=-3,\ z=3}$$ Check: $1+3(-3)+7(3)=1-9+21=13$. Correct.

(b) Let the roots of $ax^2+bx+c=0$ be $\alpha$ and $\alpha^2$. Then $\alpha+\alpha^2=-\dfrac ba$ and $\alpha^3=\dfrac ca$. Cube the sum: $(\alpha+\alpha^2)^3=\alpha^3+\alpha^6+3\alpha^3(\alpha+\alpha^2)$: $$ \begin{aligned} \left(-\frac ba\right)^3 &= \frac ca+\left(\frac ca\right)^2+3\cdot\frac ca\left(-\frac ba\right)\Rightarrow-\frac{b^3}{a^3} \ &= \frac ca+\frac{c^2}{a^2}-\frac{3bc}{a^2}. \end{aligned} $$ Multiplying by $a^3$: $-b^3=a^2c+ac^2-3abc$, hence $\boxed{b^3+a^2c+ac^2=3abc}$.

(a) From $x+z=1$: $z=1-x$. Substitute in $2y+z=2$: $2y+1-x=2\Rightarrow x=2y-1$. Substitute in $5x-9y+3=0\ (\Rightarrow 5x-9y=-3)$: $5(2y-1)-9y=-3\Rightarrow 10y-5-9y=-3\Rightarrow y=2$. Then $x=2(2)-1=3$ and $z=1-3=-2$. $$\boxed{x=3,\ y...

(a) From the third equation $z=4y-7$. Substitute in the first: $$ \begin{aligned} x+4y+(4y-7) &= 18\Rightarrow x+8y \ &= 25\Rightarrow x \ &= 25-8y. \end{aligned} $$ Substitute both in the second: $3(25-8y)+3y-2(4y-7)=2$ $$ \begin{alig...

(a) Use $\sin^2\theta=1-\cos^2\theta$: $$ \begin{aligned} 1-\cos^2\theta-2\cos\theta+\tfrac14 &= 0\Rightarrow \cos^2\theta+2\cos\theta-\tfrac54 \ &= 0. \ \cos\theta &= \frac{-2\pm\sqrt{4+5}}{2} \ &= \frac{-2\pm3}{2} \ &= \tfrac12\ \t...

(a) Write the system as $-5x+9y=3,\ x+z=1,\ 2y+z=2$. From $x+z=1$: $z=1-x$. Substitute in $2y+z=2$: $2y+1-x=2\Rightarrow x=2y-1$. Substitute in $-5x+9y=3$: $-5(2y-1)+9y=3\Rightarrow -10y+5+9y=3\Rightarrow y=2$. Then $x=2(2)-1=3$ and $z=1-3=-2$. $$\boxed{x=3,\ y=2,\ z=-2}$$ Check: $z+2y=-2+4=2$. Correct.

(b) Let the roots be $\alpha$ and $\alpha^2$. Then $$ \begin{aligned} \alpha+\alpha^2 &= -\frac{b}{a}, \ \qquad \alpha\cdot\alpha^2 &= \alpha^3 \ &= \frac{c}{a}. \end{aligned} $$ Cube the sum: $(\alpha+\alpha^2)^3=\alpha^3+\alpha^6+3\alpha^3(\alpha+\alpha^2)$, i.e. $$ \begin{aligned} \left(-\frac{b}{a}\right)^3 &= \frac{c}{a}+\left(\frac{c}{a}\right)^2+3\cdot\frac{c}{a}\left(-\frac{b}{a}\right). \ -\frac{b^3}{a^3} &= \frac{c}{a}+\frac{c^2}{a^2}-\frac{3bc}{a^2}. \end{aligned} $$ Multiply by $a^3$: $-b^3=a^2c+ac^2-3abc$, hence $\boxed{b^3+a^2c+ac^2=3abc}$.

(a) Here $p$ (2 is even) is true and $q$ (4 is even) is true. The biconditional $p\Leftrightarrow q$ is true when both have the same truth value, so its truth value is T (true). (b) Let $y=2x-3\Rightarrow x=\dfrac{y+3}{2}$. Hence $f^{-1}...

(a) Let $\tan^{-1}x=\theta\Rightarrow\tan\theta=x$. Then $\sin\theta=\dfrac{x}{\sqrt{1+x^2}}$, so $$\tan^{-1}x=\sin^{-1}!\left(\frac{x}{\sqrt{1+x^2}}\right).$$ (b) $P(n):\ 1+3+5+\cdots+(2n-1)=n^2$. $n=1$: $1=1^2$. True. Assume $P(k)$: s...

(a) Write $3x+2y=-9,\ 2x-3y=-6$. $$ \begin{aligned} D &= \begin{vmatrix}3&2\2&-3\end{vmatrix} \ &= -9-4 \ &= -13, \ \ Dx &= \begin{vmatrix}-9&2\-6&-3\end{vmatrix} \ &= 27+12 \ &= 39, \ \ Dy &= \begin{vmatrix}3&-9\2&-6\end{vmatri...

Function. A relation $f$ from set $A$ to set $B$ is a function if every element of $A$ is associated with exactly one element of $B$.

Relation vs function. A relation is any set of ordered pairs; a function is a special relation in which no first component is repeated with different second components. Example: $R={(1,2),(1,3)}$ is a relation but not a function (input $1$ has two outputs); $f={(1,2),(2,3)}$ is a function.

Domain and range of $f(x)=\sqrt{x^2-2x-8}$. Require $x^2-2x-8\ge0\Rightarrow(x-4)(x+2)\ge0\Rightarrow x\le-2\text{ or }x\ge4$. $$\text{Domain}=(-\infty,-2]\cup[4,\infty).$$ Since a square root is non-negative and the radicand ranges over $[0,\infty)$ on the domain, $f(x)\ge0$ and takes all such values: $$\text{Range}=[0,\infty).$$

Let the two unequal positive numbers be $a$ and $b$. Then $$ \begin{aligned} A &= \frac{a+b}{2} \ \qquad G &= \sqrt{ab} \ \qquad H &= \frac{2ab}{a+b}. \end{aligned} $$

(ii) $G^2=A\cdot H$: $A\cdot H=\dfrac{a+b}{2}\cdot\dfrac{2ab}{a+b}=ab=(\sqrt{ab})^2=G^2$. Hence $G^2=AH$, i.e. $G$ is the geometric mean of $A$ and $H$.

(i) $A>G>H$: Since $a\neq b$, $(\sqrt a-\sqrt b)^2>0\Rightarrow a+b>2\sqrt{ab}\Rightarrow\dfrac{a+b}{2}>\sqrt{ab}$, so $A>G$. From $G^2=AH$ with $A>G>0$: $H=\dfrac{G^2}{A}<\dfrac{G^2}{G}=G$, so $G>H$. Therefore $A>G>H$.

Conjugate. For $z=x+iy$, the conjugate is $\bar z=x-iy$ (the sign of the imaginary part is reversed).

Square roots of $-2+2\sqrt3,i$. Modulus $r=\sqrt{(-2)^2+(2\sqrt3)^2}=\sqrt{4+12}=4$. The point is in the second quadrant; reference angle $\tan^{-1}\dfrac{2\sqrt3}{2}=60^\circ$, so $\arg z=120^\circ=\dfrac{2\pi}{3}$. Thus $z=4\big(\cos120^\circ+i\sin120^\circ\big)$. By De Moivre's theorem the square roots ($k=0,1$) are $$z^{1/2}=2\left(\cos\frac{120^\circ+360^\circ k}{2}+i\sin\frac{120^\circ+360^\circ k}{2}\right).$$ $k=0$:

$$ \begin{aligned} 2(\cos60^\circ+i\sin60^\circ) &= 2\left(\tfrac12+\tfrac{\sqrt3}{2}i\right) \ &= 1+\sqrt3,i. \end{aligned} $$ $k=1$: $-(1+\sqrt3,i)$. $$\boxed{\pm(1+\sqrt3,i)}.$$

(a) By definition $A-\bar B=A\cap(\bar B)^c=A\cap B$ (since the complement of the complement of $B$ is $B$). Element proof: $x\in A-\bar B\iff x\in A\text{ and }x\notin\bar B\iff x\in A\text{ and }x\in B\iff x\in A\cap B$.

(b) $B\cup C={b,c,d}$, $B\cap C={c}$. $$ \begin{aligned} A\times(B\cup C) &= {(a,b),(a,c),(a,d),(b,b),(b,c),(b,d)}. \ A\times(B\cap C) &= {(a,c),(b,c)}. \end{aligned} $$

(c) $f(x)=\cos\pi x$. Since $\cos\theta$ has period $2\pi$, $\cos\pi(x+T)=\cos\pi x$ requires $\pi T=2\pi\Rightarrow T=2$. So $f$ is periodic with period $2$.

(a) Let $\sin^{-1}x=\theta\Rightarrow\sin\theta=x,\ \cos\theta=\sqrt{1-x^2}$. Then $\sin(2\sin^{-1}x)=\sin2\theta=2\sin\theta\cos\theta=2x\sqrt{1-x^2}$.

(b) $P(n):\ 2+2^2+\cdots+2^n=2(2^n-1)$. $n=1$: LHS $=2$, RHS $=2(2-1)=2$. True. Assume $P(k)$: $\sum_{r=1}^{k}2^r=2(2^k-1)$. Then $$ \begin{aligned} \sum_{r=1}^{k+1}2^r &= 2(2^k-1)+2^{k+1} \ &= 2^{k+1}-2+2^{k+1} \ &= 2\cdot2^{k+1}-2 \ &= 2(2^{k+1}-1), \end{aligned} $$ which is $P(k+1)$. Hence by induction the result holds for all $n\in\mathbb{N}$.

(c) $A=\begin{pmatrix}2&1\1&-2\end{pmatrix}$ is symmetric, so $A^{T}=\begin{pmatrix}2&1\1&-2\end{pmatrix}$. $$ \begin{aligned} AA^{T} &= \begin{pmatrix}2&1\1&-2\end{pmatrix}\begin{pmatrix}2&1\1&-2\end{pmatrix} \ &= \begin{pmatrix}4+1&2-2\2-2&1+4\end{pmatrix} \ &= \begin{pmatrix}5&0\0&5\end{pmatrix}. \end{aligned} $$

(a) $D=\begin{vmatrix}2&5\5&-2\end{vmatrix}=-4-25=-29,\ Dx=\begin{vmatrix}17&5\-1&-2\end{vmatrix}=-34+5=-29,\ Dy=\begin{vmatrix}2&17\5&-1\end{vmatrix}=-2-85=-87.$ $$ \begin{aligned} x &= \frac{-29}{-29} \ &= 1, \ \qquad y &= \frac{-...

One-to-one: Suppose $f(x1)=f(x2)$. Then $cx1+d=cx2+d\Rightarrow cx1=cx2\Rightarrow x1=x2$ (since $c\neq0$). So $f$ is injective. Onto: Let $y\in\mathbb{R}$. Solve $y=cx+d\Rightarrow x=\dfrac{y-d}{c}\in\mathbb{R}$, and $f!\left(\dfrac{y-...

Let the A.P. be $12-d,\ 12,\ 12+d$ (sum $=36\Rightarrow$ middle term $12$). After adding $1,4,43$: $$ \begin{aligned} 13-d,\quad16,\quad55+d\quad\text{form a G.P.}\Rightarrow16^2 &= (13-d)(55+d). \ 256 &= 715-42d-d^2\Rightarrow d^2+42d-459 \ &= 0\Rightarrow d \ &= \frac{-42\pm60}{2} \ &= 9\ \text{or}\ -51. \end{aligned} $$

• $d=9$: numbers $\boxed{3,\ 12,\ 21}$ (increased: $4,16,64$, G.P. with ratio $4$).
• $d=-51$: numbers $63,\ 12,\ -39$ (increased: $64,16,4$, G.P. with ratio $\tfrac14$).

De Moivre's theorem. For any integer $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. Cube roots of unity. Solve $z^3=1=\cos0+i\sin0$. The roots are $$zk=\cos\frac{2k\pi}{3}+i\sin\frac{2k\pi}{3},\quad k=0,1,2.$$ $k=0$: $1$. ...

(a) $x+2<4\iff -4<x+2<4\iff -6<x<2.$ (b) $R={(1,1),(2,2),(4,4)}$. Domain $={1,2,4}$ (first components); Range $={1,2,4}$ (second components). Each element maps to itself, so $R$ is the identity relation (which is reflexive, symmetr...

(a) Use $\csc^2x=1+\cot^2x$: $$ \begin{aligned} \cot^2x+1+\cot^2x &= 3\Rightarrow 2\cot^2x \ &= 2\Rightarrow\cot^2x \ &= 1\Rightarrow\cot x \ &= \pm1. \end{aligned} $$ Thus $\tan x=\pm1$. In $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right...

(a) $D=\begin{vmatrix}3&2\4&1\end{vmatrix}=3-8=-5,\ Dx=\begin{vmatrix}8&2\9&1\end{vmatrix}=8-18=-10,\ Dy=\begin{vmatrix}3&8\4&9\end{vmatrix}=27-32=-5.$ $$ \begin{aligned} x &= \frac{-10}{-5} \ &= 2, \ \qquad y &= \frac{-5}{-5} \ &=...

$$ \begin{aligned} (f\circ g)(x) &= f\big(\sin x\big) \ &= (\sin x)^3 \ &= \sin^3x. \ (g\circ f)(x) &= g\big(x^3\big) \ &= \sin\big(x^3\big). \end{aligned} $$ In general $\sin^3x\neq\sin(x^3)$ (e.g. at $x=1$: $\sin^3 1\approx0.596$ b...

The $r$-th term is $tr=r\big(n-(r-1)\big)=r(n-r+1)=r(n+1)-r^2$. $$ \begin{aligned} Sn &= \sum{r=1}^{n}\big[r(n+1)-r^2\big] \ &= (n+1)\sum{r=1}^{n}r-\sum{r=1}^{n}r^2. \ &= (n+1)\cdot\frac{n(n+1)}{2}-\frac{n(n+1)(2n+1)}{6} \ &= \frac{n(...

De Moivre's theorem. For any positive integer $n$, $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta$. Write $z=4+4\sqrt3,i$ in polar form: $r=\sqrt{4^2+(4\sqrt3)^2}=\sqrt{16+48}=8$, and $\arg z=\tan^{-1}\dfrac{4\sqrt3}{4}=\tan^{-1...

(a) Truth table of $\sim(p\lor q)\wedge q$: $p$ $q$ $p\lor q$ $\sim(p\lor q)$ $\sim(p\lor q)\wedge q$ --------------- T T T F F T F T F F F T T F F F F F T F The final column is always F, so the statement is a contradiction. (b) $(g\circ...