Basic Mathematics (2015 AD onward)

For $Z = x + iy$ with $x = 1$, $y = \sqrt3$:

$r = \sqrt{1 + 3} = 2,\qquad \theta = \tan^{-1}\!\left(\frac{\sqrt3}{1}\right) = 60^\circ = \frac{\pi}{3}$

(the point lies in the first quadrant). Hence

$Z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)$

The equation is linear: $\dfrac{dy}{dx}+ay=e^{mx}$.

Integrating factor: $e^{\int a\,dx}=e^{ax}$. Then

$\frac{d}{dx}\left(ye^{ax}\right)=e^{(a+m)x}\;\Rightarrow\; ye^{ax}=\frac{e^{(a+m)x}}{a+m}+C\quad(a+m\ne0).$

Using $x=0,\ y=0$: $0=\dfrac{1}{a+m}+C$, so $C=-\dfrac1{a+m}$.

$y=\frac{e^{mx}-e^{-ax}}{a+m}$

(If $a+m=0$, then $ye^{ax}=x+C$ and $C=0$, giving $y=xe^{mx}$.)

The left side is an exact derivative:

$(x^2-1)\frac{dy}{dx}+2xy=\frac{d}{dx}\big[(x^2-1)y\big]=1$

Integrating both sides:

$(x^2-1)y=x+c$

$y=\frac{x+c}{x^2-1}$

(Equivalently, the integrating factor of $\frac{dy}{dx}+\frac{2x}{x^2-1}y=\frac1{x^2-1}$ is $e^{\int \frac{2x}{x^2-1}dx}=x^2-1$.)

$z=(8+5i)(i-1)=8i-8+5i^2-5i=-8-5+3i=-13+3i$

So $A=-13,\ B=3$.

$A - B$ contains the elements of $A$ that are not in $B$.

$A - B = \{1, 2, 3\}$

$B - A$ contains the elements of $B$ that are not in $A$.

$B - A = \{8, 9, 10\}$

The interval $-3 \le x \le 9$ has centre $\dfrac{-3+9}{2} = 3$ and half-width $\dfrac{9-(-3)}{2} = 6$.

So $-6 \le x - 3 \le 6$, that is

$|x - 3| \le 6$

Divide the numerator and denominator by $x^2$:

$\lim_{x\to\infty}\frac{3 + \frac{3}{x} - \frac{1}{x^2}}{2 + \frac{5}{x} - \frac{6}{x^2}} = \frac{3 + 0 - 0}{2 + 0 - 0} = \frac{3}{2}$

Two vectors are orthogonal if their dot product is zero.

$(-3\vec i + 3\vec j - 2\vec k)\cdot(5\vec i + 7\vec j + 3\vec k) = (-3)(5) + (3)(7) + (-2)(3) = -15 + 21 - 6 = 0$

Since the dot product is $0$, the vectors are orthogonal.

Differentiate both sides with respect to $x$:

$2x - 2y\frac{dy}{dx} = 0 \quad\Rightarrow\quad \frac{dy}{dx} = \frac{x}{y}$

$A^T = \begin{pmatrix}3&4\\2&5\\6&7\end{pmatrix}$

$5A^T = \begin{pmatrix}15&20\\10&25\\30&35\end{pmatrix}$

$\text{Area} = \int_1^2 2x^2\,dx = \left[\frac{2x^3}{3}\right]_1^2 = \frac{2}{3}(8 - 1) = \frac{14}{3}$

The area is $\dfrac{14}{3}$ square units ($\approx 4.67$).

Separate variables and integrate:

$dy = (2x + 1)\,dx \quad\Rightarrow\quad y = \int (2x+1)\,dx = x^2 + x + C$

where $C$ is the arbitrary constant.

Expanding along the first row:

$\begin{vmatrix}4&7&2\\4&5&2\\6&4&3\end{vmatrix} = 4(15 - 8) - 7(12 - 12) + 2(16 - 30) = 28 - 0 - 28 = 0$

The value of the determinant is $0$.

(a) Since $\sqrt{-25} = 5i$ and $\sqrt{-36} = 6i$,

$z = \frac{2 - 5i}{1 - 6i}\cdot\frac{1 + 6i}{1 + 6i} = \frac{2 + 12i - 5i - 30i^2}{1 + 36} = \frac{32 + 7i}{37}$

So $z = \dfrac{32}{37} + \dfrac{7}{37}i$.

$|z| = \frac{\sqrt{32^2 + 7^2}}{37} = \frac{\sqrt{1073}}{37} \approx 0.885$

(Check: $|z| = \dfrac{\sqrt{4+25}}{\sqrt{1+36}} = \sqrt{\dfrac{29}{37}}$.)

(b) Let $\sqrt{2 + 2\sqrt3\,i} = a + ib$. Then $a^2 - b^2 = 2$ and $2ab = 2\sqrt3$. Also $|z| = \sqrt{4 + 12} = 4$, so $a^2 + b^2 = 4$.

Adding and subtracting: $a^2 = 3$, $b^2 = 1$. Since $ab = \sqrt3 > 0$, $a$ and $b$ have the same sign.

$\sqrt{z} = \pm(\sqrt3 + i)$

Check: $(\sqrt3 + i)^2 = 3 - 1 + 2\sqrt3\,i = 2 + 2\sqrt3\,i$.

A function is continuous at $x = 2$ if $\lim_{x\to2^-}f(x) = \lim_{x\to2^+}f(x) = f(2)$.

Left-hand limit: $\lim_{x\to2^-} 2x^2 = 2(4) = 8$.

Right-hand limit: $\lim_{x\to2^+} 4x = 8$.

So $\lim_{x\to2} f(x) = 8$, but $f(2) = 5 \ne 8$.

Hence $f(x)$ is not continuous at $x = 2$ (a removable discontinuity).

To make it continuous, redefine $f(2) = 8$, i.e.

$f(x)=\begin{cases}2x^2 & x<2\\ 8 & x=2\\ 4x & x>2\end{cases}$

Given $f(x) = \dfrac{ax - 2}{x + b}$.

Using $\lim_{x\to0} f(x) = 1$:

$\frac{-2}{b} = 1 \quad\Rightarrow\quad b = -2$

Using $\lim_{x\to\infty} f(x) = 3$: dividing by $x$,

$\lim_{x\to\infty}\frac{a - \frac{2}{x}}{1 + \frac{b}{x}} = a = 3$

So $f(x) = \dfrac{3x - 2}{x - 2}$, and

$f(6) = \frac{3(6) - 2}{6 - 2} = \frac{16}{4} = 4$

Hence $f(6) = 4$, as required.

(a) Rationalize the denominator:

$y = \frac{\sqrt{x+2} + \sqrt{x-2}}{(x+2) - (x-2)} = \frac{\sqrt{x+2} + \sqrt{x-2}}{4}$

$\frac{dy}{dx} = \frac14\left[\frac{1}{2\sqrt{x+2}} + \frac{1}{2\sqrt{x-2}}\right] = \frac18\left[\frac{1}{\sqrt{x+2}} + \frac{1}{\sqrt{x-2}}\right]$

(b) By the quotient rule:

$\frac{dy}{dx} = \frac{(1+x)e^x - e^x\cdot 1}{(1+x)^2} = \frac{x\,e^x}{(1+x)^2}$

(a) Put $u = 1 + \log x$, so $du = \dfrac{1}{x}dx$:

$\int \frac{1}{x(1+\log x)}\,dx = \int \frac{du}{u} = \log|1 + \log x| + C$

(b) Use $\int e^x f(x)\,dx$ by parts. First,

$\int x^2 e^x\,dx = x^2e^x - 2\int x e^x\,dx = x^2e^x - 2(xe^x - e^x) = e^x(x^2 - 2x + 2)$

and $\int e^x\,dx = e^x$. Therefore

$\int (x^2+1)e^x\,dx = e^x(x^2 - 2x + 2) + e^x + C = e^x(x^2 - 2x + 3) + C$

Three vectors are coplanar if the determinant of their coefficients is zero.

$\begin{vmatrix}-1&4&3\\2&-3&-5\\2&7&-3\end{vmatrix} = -1(9 + 35) - 4(-6 + 10) + 3(14 + 6)$

$= -44 - 16 + 60 = 0$

Since the determinant is $0$, the given vectors are coplanar.

$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$

This is homogeneous. Put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$v + x\frac{dv}{dx} = \frac{1 + v^2}{v} \quad\Rightarrow\quad x\frac{dv}{dx} = \frac{1}{v}$

Separating variables:

$v\,dv = \frac{dx}{x} \quad\Rightarrow\quad \frac{v^2}{2} = \log|x| + C$

Substituting $v = y/x$:

$\frac{y^2}{2x^2} = \log|x| + C \quad\Longrightarrow\quad y^2 = 2x^2\left(\log|x| + C\right)$

Let $M$, $S$, $E$ be the sets of students reading mathematics, statistics and economics, and let $x$ = number reading all three, $y$ = number reading mathematics and economics only.

Given: $n(M) = 24$, $n(S) = 28$, $n(E) = 22$, $M$ only $= 12$, $S$ only $= 14$, $M\cap S$ only $= 11$, $S\cap E$ only $= 11$.

(i) All three subjects

$n(S) = 14 + 11 + 11 + x = 28 \;\Rightarrow\; x = 28 - 36 = -8$

(The figures as printed are inconsistent, since the statistics regions already total 36 > 28; working formally with the given numbers:)

(ii) Mathematics and economics only

$n(M) = 12 + 11 + x + y = 24 \;\Rightarrow\; y = 24 - 23 + 8 = 9$

(iii) Economics only

$n(E) = 11 + x + y + e = 22 \;\Rightarrow\; e = 22 - 11 + 8 - 9 = 10$

(iv) All together

$n(M\cup S\cup E) = 12 + 14 + 10 + 11 + 11 + 9 + (-8) = 59$

Venn diagram regions: $M$ only $= 12$, $S$ only $= 14$, $E$ only $= 10$, $M\cap S$ only $= 11$, $S\cap E$ only $= 11$, $M\cap E$ only $= 9$, $M\cap S\cap E = x$, drawn as three overlapping circles inside the universal set with these numbers in the respective regions.

Let $Y$ = expenditure and $X = \text{year} - 2018$ (so $\sum X = 0$). Fit $Y = a + bX$.

Normal equations (with $\sum X = 0$):

$a = \frac{\sum Y}{n} = \frac{132}{7} = 18.857,\qquad b = \frac{\sum XY}{\sum X^2} = \frac{30}{28} = 1.071$

Equation of the line:

$Y = 18.857 + 1.071X \qquad (X = \text{year} - 2018)$

Estimates

2022: $X = 4$, $Y = 18.857 + 1.071(4) = 23.14$ (Rs millions)

2023: $X = 5$, $Y = 18.857 + 1.071(5) = 24.21$ (Rs millions)

Let $x$, $y$, $z$ be the units of fertilizers A, B, C.

Nitrate: $x + 3y + 2z = 11$

Potash: $2x + y + 3z = 10$

Phosphate: $3x + 2y + z = 9$

Cramer's rule

$D = \begin{vmatrix}1&3&2\\2&1&3\\3&2&1\end{vmatrix} = 1(1-6) - 3(2-9) + 2(4-3) = -5 + 21 + 2 = 18$

$D_x = \begin{vmatrix}11&3&2\\10&1&3\\9&2&1\end{vmatrix} = 11(-5) - 3(10-27) + 2(20-9) = -55 + 51 + 22 = 18$

$D_y = \begin{vmatrix}1&11&2\\2&10&3\\3&9&1\end{vmatrix} = 1(10-27) - 11(2-9) + 2(18-30) = -17 + 77 - 24 = 36$

$D_z = \begin{vmatrix}1&3&11\\2&1&10\\3&2&9\end{vmatrix} = 1(9-20) - 3(18-30) + 11(4-3) = -11 + 36 + 11 = 36$

$x = \frac{18}{18} = 1,\qquad y = \frac{36}{18} = 2,\qquad z = \frac{36}{18} = 2$

Check: $1 + 6 + 4 = 11$, $2 + 2 + 6 = 10$, $3 + 4 + 2 = 9$.

Hence 1 unit of A, 2 units of B and 2 units of C are required.

Equilibrium: set $P_D = P_S$.

$120 - 4Q = -60 + 5Q \;\Rightarrow\; 180 = 9Q \;\Rightarrow\; Q_0 = 20,\qquad P_0 = 120 - 80 = 40$

Consumer's surplus

$CS = \int_0^{20}(120 - 4Q)\,dQ - P_0Q_0 = \left[120Q - 2Q^2\right]_0^{20} - 800 = (2400 - 800) - 800 = 800$

Producer's surplus

$PS = P_0Q_0 - \int_0^{20}(-60 + 5Q)\,dQ = 800 - \left[-60Q + \tfrac52Q^2\right]_0^{20} = 800 - (-1200 + 1000) = 1000$

Total surplus

$TS = CS + PS = 800 + 1000 = 1800$

Given $C(x)=300x-10x^2+\tfrac13x^3$.

(i) Minimum marginal cost

$MC=C'(x)=300-20x+x^2$

$MC'=-20+2x=0\Rightarrow x=10$, and $MC''=2>0$, so MC is a minimum at $x=10$.

$MC_{min}=300-20(10)+10^2=300-200+100=200$

The minimum marginal cost is Rs 200.

(ii) Minimum average cost

$AC=\frac{C}{x}=300-10x+\frac{x^2}{3}$

$AC'=-10+\tfrac{2x}{3}=0\Rightarrow x=15$, and $AC''=\tfrac23>0$, so AC is a minimum at $x=15$.

$AC_{min}=300-10(15)+\frac{15^2}{3}=300-150+75=225$

The minimum average cost is Rs 225.

(iii) Output where MC = AC

$300-20x+x^2=300-10x+\frac{x^2}{3}$

$\frac{2x^2}{3}-10x=0\;\Rightarrow\;x\left(\frac{2x}{3}-10\right)=0$

So $x=0$ (rejected) or $x=15$. MC = AC at output $x=15$.

(iv) Showing MC = AC at minimum AC

At $x=15$: $MC=300-300+225=225$ and $AC=225$. So $MC=AC=225$ at $x=15$, which is exactly the output at which AC is minimum (part ii). Generally, $AC'=\dfrac{x\,MC-C}{x^2}=\dfrac{MC-AC}{x}$, so $AC'=0$ exactly when $MC=AC$. Hence marginal cost equals average cost at the minimum of average cost. $\blacksquare$

$A\Delta B$ is the symmetric difference, the elements in exactly one of $A$ and $B$:

$A\Delta B=(A-B)\cup(B-A)=\{x:(x\in A\text{ and }x\notin B)\text{ or }(x\in B\text{ and }x\notin A)\}$

Equivalently, $A\Delta B=\{x:x\in A\cup B\text{ and }x\notin A\cap B\}$.

Given $-15\le x+5\le3$. Subtracting 5 from every part:

$-15-5\le x\le3-5\;\Rightarrow\;-20\le x\le-2$

Hence proved.

Taking $x,y$ as real, compare real and imaginary parts of $x+4y=ix+y+3$.

Real parts: $x+4y=y+3$. Imaginary parts: $0=x$.

Hence $x=0$, and then $4y=y+3\Rightarrow y=1$.

So $x=0,\ y=1$.

(If the term on the left is read as $4yi$, i.e. $x+4yi=y+3+ix$, then $x=y+3$ and $4y=x$, giving $y=1,\ x=4$.)

$z=-1+i\sqrt3$ lies in the second quadrant.

$r=\sqrt{1+3}=2,\qquad\theta=\pi-\tan^{-1}\sqrt3=\pi-\frac\pi3=\frac{2\pi}3$

$z=2\left(\cos\frac{2\pi}3+i\sin\frac{2\pi}3\right)$

$x=\frac{-1\pm\sqrt{1-4}}{2}=\frac{-1\pm i\sqrt3}{2}$

So $x=-\dfrac12+\dfrac{\sqrt3}2i$ or $x=-\dfrac12-\dfrac{\sqrt3}2i$.

For $x>0$, $f(x)=\dfrac{x}{x}=1$; for $x<0$, $f(x)=\dfrac{-x}{x}=-1$.

At every point $a\neq0$, $f$ is constant near $a$, so $\lim_{x\to a}f(x)=f(a)$ and $f$ is continuous.

At $x=0$: $\lim_{x\to0^+}f(x)=1$ and $\lim_{x\to0^-}f(x)=-1$. The limits differ, so $\lim_{x\to0}f(x)$ does not exist (and $f(0)$ is undefined). Hence $f$ is continuous at all $x\neq0$ and has a jump discontinuity at $x=0$.

With $x_0=3$: $P_0=20-2(3)-2(9)=-4$.

$\text{C.S.}=\int_0^{3}(20-2x-2x^2)\,dx-x_0P_0=\left[20x-x^2-\frac{2x^3}{3}\right]_0^3-3(-4)$

$=(60-9-18)+12=33+12=45$

Consumer's surplus $=45$ (note that the price at $x_0=3$ comes out negative, so this is a purely formal evaluation of the formula).

Form the determinant of the vectors:

$\begin{vmatrix}2&3&1\\3&2&3\\-1&-2&-3\end{vmatrix}=2(-6+6)-3(-9+3)+1(-6+2)=0+18-4=14$

Since the determinant is $14\neq0$, the vectors are linearly independent.

Reflection in the line $y=x$ interchanges the coordinates, so the image of $(x,y)$ is $(y,x)$.

Integrate by parts with $u=\log x$, $dv=x\,dx$:

$\int x\log x\,dx=\frac{x^2}{2}\log x-\int\frac{x^2}{2}\cdot\frac1x\,dx=\frac{x^2}{2}\log x-\frac{x^2}{4}+C$

(a) Let $T$ and $C$ be the tea and coffee drinkers. $n(T\cup C)=100$, $n(T)=70$, $n(C)=50$.

$n(T\cap C)=70+50-100=20$

Tea only $=70-20=50$ students; coffee only $=50-20=30$ students.

(b) $|2x-5|\le7\iff-7\le2x-5\le7\iff-2\le2x\le12\iff-1\le x\le6$.

So $x$ lies in the interval $[-1,6]$.

Let $\sqrt{-1-i\sqrt3}=a+ib$. Squaring: $a^2-b^2=-1$ and $2ab=-\sqrt3$.

Modulus: $|z|=\sqrt{1+3}=2$, so $a^2+b^2=2$.

Hence $a^2=\dfrac{2-1}{2}=\dfrac12$ and $b^2=\dfrac{2+1}{2}=\dfrac32$. Since $ab<0$, $a$ and $b$ have opposite signs:

$\sqrt{-1-i\sqrt3}=\pm\frac{1-i\sqrt3}{\sqrt2}$

Check: $(1-i\sqrt3)^2=-2-2i\sqrt3$, and dividing by 2 gives $-1-i\sqrt3$.

$\vec{AB}=(1-1,\,2-1,\,3-1)=(0,1,2)$ and $\vec{AC}=(2-1,\,3-1,\,1-1)=(1,2,0)$.

$\vec{AB}\times\vec{AC}=\begin{vmatrix}\vec i&\vec j&\vec k\\0&1&2\\1&2&0\end{vmatrix}=(0-4)\vec i-(0-2)\vec j+(0-1)\vec k=-4\vec i+2\vec j-\vec k$

$|\vec{AB}\times\vec{AC}|=\sqrt{16+4+1}=\sqrt{21}$

$\text{Area}=\tfrac12|\vec{AB}\times\vec{AC}|=\frac{\sqrt{21}}{2}\approx2.29\text{ square units}$

Expand along the first row:

$(x+1)\big[(x+2)(x+3)-6\big]-1\big[2(x+3)-6\big]+1\big[6-3(x+2)\big]=0$

$(x+1)(x^2+5x)-2x-3x=0$

$x^3+6x^2+5x-5x=0\;\Rightarrow\;x^2(x+6)=0$

Therefore $x=0$ (twice) or $x=-6$.