Basic Mathematics (2009–2014 AD)

$\vec a-\vec b=(1+2,\,3-3)=(3,0)$, with magnitude $3$.

Unit vector $=\dfrac{(3,0)}{3}=(1,0)$.

For $z=-1$: modulus $r=1$, argument $\theta=\pi$.

$-1=\cos\pi+i\sin\pi$

Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0$ and both are non-zero, $\cos\theta=0$, so $\theta=90^\circ$. The vectors are perpendicular (orthogonal).

For $z=i=0+1i$: $r=1$, $\theta=\dfrac\pi2$.

$i=\cos\frac\pi2+i\sin\frac\pi2$

Since $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0$ with both vectors non-zero, $\cos\theta=0$, so $\theta=90^\circ$. The vectors are perpendicular (orthogonal).

$4\vec a-5\vec b=(4-15)\vec i+(8+5)\vec j+(-4-10)\vec k=-11\vec i+13\vec j-14\vec k$

$|4\vec a-5\vec b|=\sqrt{121+169+196}=\sqrt{486}=9\sqrt6$

$\text{Unit vector}=\frac{-11\vec i+13\vec j-14\vec k}{9\sqrt6}$

$A=\{\dots,-2,-1,0\}$ and $B=\{0,1,2,\dots\}$.

$A-B=\{\dots,-2,-1\}\ \text{(negative integers)},\qquad B-A=\{1,2,3,\dots\}\ \text{(positive integers)}$

$\{x\in R: -3<x\le 2\}=(-3,\,2]$

$\frac{1}{2+3i}=\frac{2-3i}{(2+3i)(2-3i)}=\frac{2-3i}{13}$

So the conjugate is $\dfrac{2+3i}{13}=\dfrac{2}{13}+\dfrac{3}{13}i$.

From $y^2-2x=-1$ we get $y=\pm\sqrt{2x-1}$. For $x=5$, $y=\pm3$, so one value of $x$ gives two values of $y$. Hence the relation is not a function of $x$.

As $n\to\infty$, $2^{n-1}\to\infty$, so $\dfrac{1}{2^{n-1}}\to0$.

$\lim_{n\to\infty}f(n)=2-0=2$

$C'(x)=4(x-3)$.

• At $x=2$: slope $=4(-1)=-4$
• At $x=3$: slope $=0$
• At $x=4$: slope $=4$

$\int_1^2\frac{2x}{1+x^2}dx=\Big[\ln(1+x^2)\Big]_1^2=\ln5-\ln2=\ln\frac52$

• Diagonal matrix: a square matrix whose off-diagonal elements are all zero (diagonal elements may be any numbers), e.g. $\begin{bmatrix}2&0\\0&5\end{bmatrix}$.
• Scalar matrix: a diagonal matrix whose diagonal elements are all equal, e.g. $\begin{bmatrix}3&0\\0&3\end{bmatrix}$.

Every scalar matrix is diagonal, but a diagonal matrix is scalar only if all its diagonal entries are the same.

$y=\tfrac52x^2+7x+C\ \Rightarrow\ \frac{dy}{dx}=\tfrac52(2x)+7=5x+7$

This equals the right-hand side of the equation, so $y$ is a solution.

The cardinal number of a set $A$, written $n(A)$, is the number of distinct elements in $A$.

Finding $n(A\cap B)$: $n(A\cup B)=n(A)+n(B)-n(A\cap B)\Rightarrow 80=75+40-n(A\cap B)$ $n(A\cap B)=35$

Finding $n(A-B)$: $n(A-B)=n(A)-n(A\cap B)=75-35=40$

If $B\subset A$: then $A\cap B=B$, so $n(A\cap B)=n(B)=40$.

(a) $4(\cos120^\circ+i\sin120^\circ)=4\left(-\tfrac12+i\tfrac{\sqrt3}{2}\right)=-2+2\sqrt3\,i$

(b) For $z=\sqrt3+i$: $r=\sqrt{3+1}=2$, $\theta=30^\circ$. So $z=2(\cos30^\circ+i\sin30^\circ)$.

By De Moivre's theorem: $z^6=2^6(\cos180^\circ+i\sin180^\circ)=64(-1+0)=-64$

Let $f(x)=x+\sqrt x$.

$f'(x)=\lim_{h\to0}\frac{(x+h)+\sqrt{x+h}-x-\sqrt x}{h}=\lim_{h\to0}\left[1+\frac{\sqrt{x+h}-\sqrt x}{h}\right]$

Rationalising, $\frac{\sqrt{x+h}-\sqrt x}{h}=\frac{1}{\sqrt{x+h}+\sqrt x}\to\frac{1}{2\sqrt x}$

$f'(x)=1+\frac{1}{2\sqrt x}$

(a) (i) $y=\log\left(x+\sqrt{x^2+a^2}\right)$

$\frac{dy}{dx}=\frac{1+\frac{x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}}=\frac{\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}}=\frac{1}{\sqrt{x^2+a^2}}$

(b) $y=x^2e^{3x}$

$y'=2xe^{3x}+3x^2e^{3x}$ $y''=2e^{3x}+6xe^{3x}+6xe^{3x}+9x^2e^{3x}=e^{3x}\left(9x^2+12x+2\right)$

(a) (i) Put $t=e^x$, $dt=e^xdx$: $\int\frac{e^{2x}}{1+e^x}dx=\int\frac{t}{1+t}dt=\int\left(1-\frac1{1+t}\right)dt=t-\ln(1+t)+C$ $=e^x-\ln(1+e^x)+C$

(b) $\text{Area}=\int_1^3 3^x\,dx=\left[\frac{3^x}{\ln3}\right]_1^3=\frac{27-3}{\ln3}=\frac{24}{\ln3}\approx21.85\text{ sq. units}$

$\vec{AB}=(2,2,2)$, $\vec{AC}=(0,2,4)$.

$\vec{AB}\times\vec{AC}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&2&2\\0&2&4\end{vmatrix}=(8-4)\hat i-(8-0)\hat j+(4-0)\hat k=(4,-8,4)$

$|\vec{AB}\times\vec{AC}|=\sqrt{16+64+16}=\sqrt{96}=4\sqrt6$

$\text{Area}=\tfrac12\cdot4\sqrt6=2\sqrt6\approx4.90\text{ sq. units}$

Expanding along the first row:

$\Delta=(1+a)\big[(1+b)(1+c)-1\big]-1\big[(1+c)-1\big]+1\big[1-(1+b)\big]$ $=(1+a)(b+c+bc)-c-b$ $=b+c+bc+ab+ac+abc-b-c=ab+bc+ca+abc$

Factoring out $abc$: $\Delta=abc\left(\frac1c+\frac1a+\frac1b+1\right)=abc\left(\frac1a+\frac1b+\frac1c+1\right)$

Differentiate $x^2+2xy+3y^2=1$ with respect to $x$: $2x+2y+2xy'+6yy'=0\Rightarrow y'=-\frac{x+y}{x+3y}$

Differentiate again: $y''=-\frac{(1+y')(x+3y)-(x+y)(1+3y')}{(x+3y)^2}$

Numerator: $(x+3y)-(x+y)+y'\big[(x+3y)-3(x+y)\big]=2y-2xy'$.

Substituting $y'$: $2y+\frac{2x(x+y)}{x+3y}=\frac{2y(x+3y)+2x(x+y)}{x+3y}=\frac{2(x^2+2xy+3y^2)}{x+3y}=\frac{2}{x+3y}$

$y''=-\frac{2}{(x+3y)^3}$

Equilibrium: $20-5q=4+3q\Rightarrow8q=16\Rightarrow q_0=2$, and $p_0=20-10=10$.

Consumer's surplus: $CS=\int_0^2(20-5q)\,dq-p_0q_0=\Big[20q-\tfrac52q^2\Big]_0^2-20=(40-10)-20=\text{Rs }10$

Producer's surplus: $PS=p_0q_0-\int_0^2(4+3q)\,dq=20-\Big[4q+\tfrac32q^2\Big]_0^2=20-(8+6)=\text{Rs }6$

$(6y-12)dx=dy\Rightarrow\dfrac{dy}{dx}-6y=-12$, which is linear.

Integrating factor: $e^{\int-6dx}=e^{-6x}$.

$ye^{-6x}=\int-12e^{-6x}dx=2e^{-6x}+C$

$y=2+Ce^{6x}$

(a) Table (units per barrel)

(b) Model: Let $x,y,z$ be the barrels of P, Q, R. $x+2y+z=8,\qquad 2x+3y+2z=14,\qquad 3x+2y+2z=13$

(c) Gaussian elimination

Augmented matrix: $\left[\begin{array}{ccc|c}1&2&1&8\\2&3&2&14\\3&2&2&13\end{array}\right]$

$R_2\to R_2-2R_1$, $R_3\to R_3-3R_1$: $\left[\begin{array}{ccc|c}1&2&1&8\\0&-1&0&-2\\0&-4&-1&-11\end{array}\right]$

From $R_2$: $y=2$. From $R_3$: $-4(2)-z=-11\Rightarrow z=3$. From $R_1$: $x+4+3=8\Rightarrow x=1$.

$x=1,\ y=2,\ z=3$

(d) Check:

• A: $1+4+3=8$ correct
• B: $2+6+6=14$ correct
• C: $3+4+6=13$ correct

All values are non-negative whole numbers, so they make sense: 1 barrel of P, 2 barrels of Q and 3 barrels of R exactly meet the requirement.

Since $A,B$ are disjoint, $n(A\cup B)=n(A)+n(B)$. Given $n(A\cup B)=n(A)$, we get $n(B)=0$.

Hence $B$ is the empty set, $B=\varnothing$.

There is no possible value. The absolute value $|x|$ is the distance from zero, so $|x|\ge0$ for every real $x$ and can never equal $-3$. The solution set is empty.

$z=-3+i$, so $\bar z=-3-i$.

Real part $=-3$, imaginary part $=-1$.

We need $625-x^2\ge0\Rightarrow-25\le x\le25$. Since quantity cannot be negative, $x\ge0$.

Domain: $0\le x\le25$, i.e. $[0,25]$.

Divide numerator and denominator by $x$: $\lim_{x\to\infty}\frac{\sqrt x}{x+3}=\lim_{x\to\infty}\frac{x^{-1/2}}{1+3/x}=\frac{0}{1}=0$

$C'(x)=6-6x+3x^2$, $C''(x)=6x-6$.

$C''(x)=0\Rightarrow x=1$. Since $C''$ changes sign from negative to positive at $x=1$ (and $C'''=6\neq0$), the point of inflection occurs at $x=1$, where $C(1)=6-3+1=4$.

Since $|\vec a\times\vec b|=|\vec a||\vec b|\sin\theta=0$ and both vectors are non-zero, $\sin\theta=0$, so $\theta=0^\circ$ or $180^\circ$. Hence $\vec a$ and $\vec b$ are parallel (collinear).

$|A|=2(0)-6(-4)=24\neq0$.

$A^{-1}=\frac1{24}\begin{bmatrix}0&-6\\4&2\end{bmatrix}=\begin{bmatrix}0&-\tfrac14\\ \tfrac16&\tfrac1{12}\end{bmatrix}$

$\int_3^6(-3)\,dx=-3(6-3)=-9$

The line lies below the x-axis, so the area is the absolute value: 9 square units.

$y=ce^x\Rightarrow\dfrac{dy}{dx}=ce^x=y$, so it satisfies the equation. It contains one arbitrary constant $c$, the same as the order (1) of the equation, so it is the general solution.

(a) Let $E$ = failed in English, $S$ = failed in Science: $n(E)=56$, $n(S)=48$, $n(E\cap S)=18$.

(i) Failed in the examination (at least one subject): $n(E\cup S)=56+48-18=86\%$

(ii) Passed in both subjects: $100-86=14\%$.

(b) $|2x-7|>3$ means $2x-7>3\Rightarrow x>5\qquad\text{or}\qquad2x-7<-3\Rightarrow x<2$

Solution set: $(-\infty,2)\cup(5,\infty)$.

On the real line, shade to the left of $2$ and to the right of $5$ with open circles at $2$ and $5$ (the segment from $2$ to $5$ is not shaded).

(a) Let $\sqrt{3-4i}=a+bi$. Then $a^2-b^2=3$ and $2ab=-4$.

Also $a^2+b^2=|3-4i|=5$. So $a^2=4$, $b^2=1$. Since $ab<0$: $a=2,b=-1$ or $a=-2,b=1$.

Square roots: $\pm(2-i)$.

(b) $2+2\sqrt3i$: $r=\sqrt{4+12}=4$, $\theta=\tan^{-1}\sqrt3=60^\circ$.

$(2+2\sqrt3i)^3=4^3(\cos180^\circ+i\sin180^\circ)=-64$

(a) $f(2)=\dfrac{0}{0}$ is undefined. Although $\lim_{x\to2}\frac{x^2-4}{x-2}=\lim_{x\to2}(x+2)=4$ exists, $f$ is not defined at $x=2$, so $f$ is discontinuous at $x=2$ (a removable discontinuity).

(b) $f'(x)=\lim_{h\to0}\frac{[2(x+h)+3]-[2x+3]}{h}=\lim_{h\to0}\frac{2h}{h}=2$

(a) $y=\log\left(x+\sqrt{x^2-a^2}\right)$

$\frac{dy}{dx}=\frac{1+\frac{x}{\sqrt{x^2-a^2}}}{x+\sqrt{x^2-a^2}}=\frac{1}{\sqrt{x^2-a^2}}$

(b) $y=e^{2\log x}=e^{\log x^2}=x^2$.

$y'=2x,\quad y''=2,\quad y'''=0$

So the third derivative is zero.

(a) Put $u=3x-1$, so $x=\frac{u+1}{3}$, $2x+1=\frac{2u+5}{3}$, $dx=\frac{du}{3}$.

$\int\frac{(2u+5)}{9}u^{1/2}du=\frac19\left[\frac45u^{5/2}+\frac{10}{3}u^{3/2}\right]+C=\frac{4}{45}(3x-1)^{5/2}+\frac{10}{27}(3x-1)^{3/2}+C$

(b) Put $t=x^2+3x$, $dt=(2x+3)dx$: $\int e^t\,dt=e^{x^2+3x}+C$

(c) Integrating by parts: $\int_1^e\log x\,dx=\Big[x\log x-x\Big]_1^e=(e-e)-(0-1)=1$

(a) Multiply $A$ by its transpose:

$A^T=\begin{bmatrix}-1&3&2\\2&0&1\\1&-1&0\end{bmatrix},\quad AA^T=\begin{bmatrix}1+4+1&-3+0-1&-2+2+0\\-4&9+0+1&6+0+0\\0&6&4+1\end{bmatrix}=\begin{bmatrix}6&-4&0\\-4&10&6\\0&6&5\end{bmatrix}$

Here $a_{12}=a_{21}=-4$, $a_{13}=a_{31}=0$ and $a_{23}=a_{32}=6$, so $(AA^T)^T=AA^T$. Hence $AA^T$ is symmetric.

(b) Let $C_1,C_3$ be the first and third columns. $C_1=(42,28,14)^T=14(3,2,1)^T$ and $C_3=(6,4,2)^T=2(3,2,1)^T$, so $C_1=7C_3$.

Take $C_1\to C_1-7C_3$ (this does not change the determinant). The first column becomes $(0,0,0)^T$, so the determinant is $0$.

Scalar (dot) product: for vectors $\vec a,\vec b$ with angle $\theta$ between them, $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$. In components, if $\vec a=a_1\vec i+a_2\vec j+a_3\vec k$ and $\vec b=b_1\vec i+b_2\vec j+b_3\vec k$, then $\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3$. It is a scalar.

$\vec{AB}=B-A=(5-2,\,4-3,\,-1-4)=3\vec i+\vec j-5\vec k$ $\vec{CD}=D-C=(1-3,\,2-6,\,0-2)=-2\vec i-4\vec j-2\vec k$

$\vec{AB}\cdot\vec{CD}=3(-2)+1(-4)+(-5)(-2)=-6-4+10=0$

Since the scalar product is zero and both vectors are non-zero, $AB\perp CD$.

Revenue is the integral of marginal revenue, and $R(0)=0$.

$R(x)=\int\left[\frac{6}{(x+2)^2}+5\right]dx=-\frac{6}{x+2}+5x+c$

$R(0)=0\Rightarrow -3+c=0\Rightarrow c=3$, so

$R=5x+3-\frac{6}{x+2}=\frac{5x^2+13x}{x+2}$

Demand law (price $p=R/x$):

$p=\frac{5x+13}{x+2}=5+\frac{3}{x+2}$

$f'(x)=4x^3-24x^2+36x-7,\qquad f''(x)=12x^2-48x+36=12(x-1)(x-3)$