Mathematics I

Domain: we need $6-x-x^2\ge0\iff x^2+x-6\le0\iff(x+3)(x-2)\le0$, so $-3\le x\le2$.

Domain $=[-3,\,2]$.

Range: let $g(x)=6-x-x^2=\dfrac{25}{4}-\left(x+\dfrac12\right)^2$. On the domain, $g$ takes every value from $0$ (at $x=-3,2$) up to its maximum $\dfrac{25}{4}$ (at $x=-\tfrac12$, which lies in the domain). So $0\le g(x)\le\dfrac{25}{4}$ and

$0\le f(x)\le\frac52$

Range $=\left[0,\ \dfrac52\right]$.

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

Domain. The function is defined when the denominator is non-zero: $3-x\ne0$, so $x\ne3$.

$\text{Domain}=\mathbb R\setminus\{3\}$

Range. Put $y=\dfrac{2x+1}{3-x}$ and solve for $x$:

$y(3-x)=2x+1\ \Rightarrow\ 3y-1=x(2+y)\ \Rightarrow\ x=\frac{3y-1}{y+2}$

This is defined for every $y\ne-2$, and each such $y$ gives a valid $x$ (note $x=3$ would need $3y-1=3y+6$, which is impossible). So

$\text{Range}=\mathbb R\setminus\{-2\}$

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

Domain: the set $A$ of all allowed inputs $x$. Range: the set of all outputs $\{f(x):x\in A\}$ (a subset of $B$).

Domain of $f(x)=\sqrt{2-x-x^2}$. The expression under the root must be non-negative:

$2-x-x^2\ge0\ \Rightarrow\ x^2+x-2\le0\ \Rightarrow\ (x+2)(x-1)\le0\ \Rightarrow\ -2\le x\le1$

$\text{Domain}=[-2,\,1]$

Range. Complete the square:

$2-x-x^2=\frac94-\left(x+\frac12\right)^2$

On $[-2,1]$ this has its maximum $\dfrac94$ at $x=-\dfrac12$ and its minimum $0$ at $x=-2$ and $x=1$, taking all values between. So

$0\le 2-x-x^2\le\frac94\ \Rightarrow\ 0\le f(x)\le\frac32$

$\text{Range}=\left[0,\ \tfrac32\right]$

Expand the determinant along the first row.

$$\begin{aligned} \Delta &= (1+x)\big[(1+y)(1+z)-1\big] - 1\big[(1+z)-1\big] + 1\big[1-(1+y)\big]\\ &= (1+x)(y+z+yz) - z - y\\ &= y+z+yz+xy+xz+xyz-y-z\\ &= xyz+xy+yz+zx \end{aligned}$$

Factor out $xyz$:

$\Delta = xyz\left(1+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}\right)=xyz\left(\frac1x+\frac1y+\frac1z+1\right)$

Hence proved.

Scalar product. For vectors $\vec a,\vec b$ at angle $\theta$: $\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$, and in components $\vec a\cdot\vec b=a_1b_1+a_2b_2$ in the plane.

Proof. In the $xy$-plane let $\vec a$ and $\vec b$ be unit vectors making angles $-A$ and $B$ with the positive $x$-axis:

$\vec a=\cos A\,\hat i-\sin A\,\hat j,\qquad \vec b=\cos B\,\hat i+\sin B\,\hat j$

The angle between them is $A+B$ (taking $0<A+B<\pi$; the result extends to all angles by periodicity), and $|\vec a|=|\vec b|=1$.

By definition,

$\vec a\cdot\vec b=|\vec a||\vec b|\cos(A+B)=\cos(A+B)$

By components,

$\vec a\cdot\vec b=\cos A\cos B+(-\sin A)(\sin B)=\cos A\cos B-\sin A\sin B$

Equating the two expressions:

$\cos(A+B)=\cos A\cos B-\sin A\sin B$

Hence proved.

Take the ellipse in standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$.

Eccentricity: $e^2=1-\dfrac{b^2}{a^2}=\dfrac12\ \Rightarrow\ b^2=\dfrac{a^2}{2}$.

Latus rectum: $\dfrac{2b^2}{a}=3\ \Rightarrow\ \dfrac{2\cdot a^2/2}{a}=a=3$.

So $a^2=9$ and $b^2=\dfrac92$. The equation is

$\frac{x^2}{9}+\frac{y^2}{9/2}=1\quad\Longleftrightarrow\quad x^2+2y^2=9$

$6+5x-x^2\ge0\iff x^2-5x-6\le0\iff(x-6)(x+1)\le0$

The roots are $x=-1$ and $x=6$. The quadratic $x^2-5x-6$ opens upward, so it is $\le0$ between the roots.

$-1\le x\le6$

Solution set: $[-1,\,6]$.

The determinant is to be read as the $3\times3$ determinant

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

Apply $R_1\to R_1-R_2$ and $R_2\to R_2-R_3$:

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

Expand along the first column (only the entry $1$ in row 3 survives, with sign $+$):

$\Delta=(x-y)(y-z)\big[(y+z)-(x+y)\big]=(x-y)(y-z)(z-x)$

Hence proved.

Reading the determinant as the $3\times3$ Vandermonde determinant (rows $1,1,1$; $a,b,c$; $a^2,b^2,c^2$):

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

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

$\Delta=\begin{vmatrix}0&0&1\\a-b&b-c&c\\a^2-b^2&b^2-c^2&c^2\end{vmatrix}$

Expand along the first row (only the entry $1$ in position $(1,3)$ survives):

$\Delta=\begin{vmatrix}a-b&b-c\\(a-b)(a+b)&(b-c)(b+c)\end{vmatrix}$

Take out the common factors $(a-b)$ and $(b-c)$:

$\Delta=(a-b)(b-c)\begin{vmatrix}1&1\\a+b&b+c\end{vmatrix}=(a-b)(b-c)\big[(b+c)-(a+b)\big]=(a-b)(b-c)(c-a)$

Hence $\Delta=(a-b)(b-c)(c-a)$. Proved.

Let $B$ = set of basketball players and $V$ = set of volleyball players. Given $n(B) = 32$, $n(V) = 25$, $n(B \cap V) = 20$.

Students playing at least one game $n(B \cup V) = n(B) + n(V) - n(B \cap V) = 32 + 25 - 20 = 37$

Total number of students, if 13 play neither game: $n(U) = n(B \cup V) + 13 = 37 + 13 = 50$

So 37 students play at least one game and the total number of students is 50.

Given $f : N \to N$, $f(x) = 2x$.

One-one. Let $x_1, x_2 \in N$ with $f(x_1) = f(x_2)$. Then $2x_1 = 2x_2 \Rightarrow x_1 = x_2$ Hence $f$ is one-one.

Not onto. Take $1 \in N$ (the codomain). If $f(x) = 1$ then $2x = 1$, so $x = \tfrac{1}{2} \notin N$. Thus $1$ has no pre-image in $N$. (In fact every odd natural number has no pre-image, since the range is only the even numbers.) So the range $\neq$ codomain, and $f$ is not onto.

Therefore $f$ is one-one but not onto.

Let the three consecutive terms of the geometric series be $a,\ ar,\ ar^2$. The middle term is $ar$. Adding it to each term gives $a + ar = a(1 + r),\qquad ar + ar = 2ar,\qquad ar^2 + ar = ar(1 + r)$

The resulting terms are in H.P. if their reciprocals are in A.P., that is $\frac{1}{a(1+r)},\quad \frac{1}{2ar},\quad \frac{1}{ar(1+r)}$ are in A.P.

Second minus first: $\frac{1}{2ar} - \frac{1}{a(1+r)} = \frac{(1+r) - 2r}{2ar(1+r)} = \frac{1 - r}{2ar(1+r)}$

Third minus second: $\frac{1}{ar(1+r)} - \frac{1}{2ar} = \frac{2 - (1+r)}{2ar(1+r)} = \frac{1 - r}{2ar(1+r)}$

The two differences are equal, so the reciprocals are in A.P. Hence the resulting terms are in H.P.

Let $A = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix}$

Cofactors

$C_{11} = (3)(1) - (0)(-2) = 3$

$C_{12} = -[(-1)(1) - (0)(0)] = 1$

$C_{13} = (-1)(-2) - (3)(0) = 2$

$C_{21} = -[(2)(1) - (-2)(-2)] = -(2 - 4) = 2$

$C_{22} = (1)(1) - (-2)(0) = 1$

$C_{23} = -[(1)(-2) - (2)(0)] = 2$

$C_{31} = (2)(0) - (-2)(3) = 6$

$C_{32} = -[(1)(0) - (-2)(-1)] = -(0 - 2) = 2$

$C_{33} = (1)(3) - (2)(-1) = 5$

The cofactor matrix is $\begin{pmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{pmatrix}$, and the adjoint is its transpose:

$\operatorname{adj}A = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}$

Check: $|A| = 1(3) + 2(1) + (-2)(2) = 1$, and $A\,\operatorname{adj}A = I$ (for example, row 1 times column 1 is $3 + 2 - 4 = 1$ and row 1 times column 3 is $6 + 4 - 10 = 0$).

Let $P(x,y)$ be any point on the parabola. Its distance from the focus $F(-1,2)$ equals its distance from the directrix $x=-5$:

$\sqrt{(x+1)^2+(y-2)^2}=|x+5|$

Squaring:

$(x+1)^2+(y-2)^2=(x+5)^2$

$x^2+2x+1+(y-2)^2=x^2+10x+25$

$(y-2)^2=8x+24=8(x+3)$

So the equation of the parabola is $(y-2)^2=8(x+3)$, or $y^2-4y-8x-20=0$.

Let $T(\mathbf{x})=A\mathbf{x}$ with $A=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$.

$T(u)=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}1\\1\end{bmatrix},\qquad T(v)=\begin{bmatrix}0&-1\\1&0\end{bmatrix}\begin{bmatrix}-2\\3\end{bmatrix}=\begin{bmatrix}-3\\-2\end{bmatrix}$

Linearity check. For any vectors $x,y$ and scalar $c$, matrix multiplication satisfies $A(x+y)=Ax+Ay$ and $A(cx)=cAx$. For example, $u+v=\begin{bmatrix}-1\\2\end{bmatrix}$ gives $T(u+v)=\begin{bmatrix}-2\\-1\end{bmatrix}=T(u)+T(v)=\begin{bmatrix}1-3\\1-2\end{bmatrix}$.

Hence the transformation is linear (it is a rotation of the plane through $90^\circ$ anticlockwise).

Permutation: an arrangement of $r$ objects chosen from $n$ distinct objects in which order matters.

$^nP_r=\frac{n!}{(n-r)!}$

Combination: a selection of $r$ objects from $n$ distinct objects in which order does not matter.

$^nC_r=\frac{n!}{r!\,(n-r)!}$

Relationship. Each selection of $r$ objects can be arranged among themselves in $r!$ ways, so

$^nP_r = {}^nC_r\times r!\quad\Longrightarrow\quad {}^nC_r=\frac{^nP_r}{r!}$

Arrangements of LOGIC. The word has 5 distinct letters: vowels O, I and consonants L, G, C.

(i) Vowels in odd positions. The odd positions are 1, 3, 5 (three of them). The two vowels can be placed in them in $^3P_2=6$ ways. The three consonants fill the remaining 3 places in $3!=6$ ways.

$6\times6=36$

(ii) No two vowels together. Arrange the 3 consonants first: $3!=6$ ways. This creates 4 gaps (including both ends). Place the 2 vowels in different gaps: $^4P_2=12$ ways.

$6\times12=72$

Check: total $5!=120$, with vowels together $2!\times4!=48$, so $120-48=72$.

Scalar (dot) product. For vectors $\vec a,\vec b$ with angle $\theta$ between them ($0\le\theta\le\pi$):

$\vec a\cdot\vec b=|\vec a||\vec b|\cos\theta$

In components, $\vec a\cdot\vec b=a_1b_1+a_2b_2+a_3b_3$. It is a scalar. Geometrically, $|\vec b|\cos\theta$ is the projection of $\vec b$ on $\vec a$, so $\vec a\cdot\vec b$ is $|\vec a|$ times the projection of $\vec b$ along $\vec a$. It is zero exactly when the vectors are perpendicular.

Vector (cross) product.

$\vec a\times\vec b=|\vec a||\vec b|\sin\theta\;\hat n$

where $\hat n$ is the unit vector perpendicular to both $\vec a$ and $\vec b$, in the direction given by the right-hand rule. In components,

$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\a_1&a_2&a_3\\b_1&b_2&b_3\end{vmatrix}$

Geometrically, $|\vec a\times\vec b|$ is the area of the parallelogram with adjacent sides $\vec a$ and $\vec b$. It is zero exactly when the vectors are parallel.

Proof of $\sin(A+B)=\sin A\cos B+\cos A\sin B$.

In the $xy$-plane take the unit vectors

$\vec a=\cos A\,\hat i-\sin A\,\hat j\quad(\text{making angle }-A\text{ with the }x\text{-axis}),\qquad \vec b=\cos B\,\hat i+\sin B\,\hat j$

The angle between $\vec a$ and $\vec b$ is $A+B$, and $|\vec a|=|\vec b|=1$. By the definition,

$\vec a\times\vec b=\sin(A+B)\,\hat n,$

and by the determinant,

$\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\\cos A&-\sin A&0\\\cos B&\sin B&0\end{vmatrix}=(\cos A\sin B+\sin A\cos B)\,\hat k$

Here $\hat n=\hat k$ (rotating from $\vec a$ to $\vec b$ is anticlockwise). Comparing the coefficients of $\hat k$:

$\sin(A+B)=\sin A\cos B+\cos A\sin B$

Hence proved.

Logarithmic function. For $a>0,\ a\ne1$, the logarithmic function $f(x)=\log_a x$ is the inverse of the exponential function $y=a^x$. That is, $y=\log_a x \iff x=a^y$. Its domain is $(0,\infty)$ and its range is $\mathbb R$.

Properties (for $m,n>0$):

• $\log_a(mn)=\log_a m+\log_a n$
• $\log_a\dfrac mn=\log_a m-\log_a n$
• $\log_a m^k=k\log_a m$
• $\log_a a=1,\ \log_a1=0$
• $\log_a m=\dfrac{\log_b m}{\log_b a}$ (change of base)
• It is increasing for $a>1$ and decreasing for $0<a<1$.

Proof. Given $f(x)=\log\dfrac{1+x}{1-x}$, so

$f(a)+f(b)=\log\frac{1+a}{1-a}+\log\frac{1+b}{1-b}=\log\frac{(1+a)(1+b)}{(1-a)(1-b)}$

Now

$f\!\left(\frac{a+b}{1+ab}\right)=\log\frac{1+\dfrac{a+b}{1+ab}}{1-\dfrac{a+b}{1+ab}}=\log\frac{1+ab+a+b}{1+ab-a-b}$

Since $1+ab+a+b=(1+a)(1+b)$ and $1+ab-a-b=(1-a)(1-b)$,

$f\!\left(\frac{a+b}{1+ab}\right)=\log\frac{(1+a)(1+b)}{(1-a)(1-b)}=f(a)+f(b)$

(The argument $\frac{a+b}{1+ab}$ lies in $(-1,1)$ when $|a|,|b|<1$, so $f$ is defined there.) Hence proved.

Let $M$ and $E$ be the sets of students who failed in Mathematics and English. Then $n(M)=40,\ n(E)=70,\ n(M\cap E)=20$.

Failed in at least one subject:

$n(M\cup E)=40+70-20=90$

(a) Passed in both subjects: $100-90=10$ students.

(b) Passed in Mathematics only (that is, passed Mathematics but failed English): $n(E)-n(M\cap E)=70-20=50$ students.

(c) Failed in Mathematics only: $n(M)-n(M\cap E)=40-20=20$ students.

The Maclaurin series is $f(x)=\sum_{n=0}^\infty \dfrac{f^{(n)}(0)}{n!}x^n$.

For $f(x)=\sin x$:

$f(x)=\sin x,\ f'(x)=\cos x,\ f''(x)=-\sin x,\ f'''(x)=-\cos x,\ f^{(4)}(x)=\sin x,\ \dots$

At $x=0$ the values are $0,\,1,\,0,\,-1,\,0,\,1,\,0,\,-1,\dots$

Only the odd derivatives are non-zero, so

$\sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots=\sum_{n=0}^\infty\frac{(-1)^n x^{2n+1}}{(2n+1)!}$

valid for all real $x$.

Two vectors in the plane:

$\vec{PQ}=(2-1,\ 1+1,\ -1-0)=(1,2,-1),\qquad \vec{PR}=(-1-1,\ 1+1,\ 2-0)=(-2,2,2)$

A vector perpendicular to the plane is

$$\vec{PQ}\times\vec{PR}=\begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-1\\-2&2&2\end{vmatrix} =\hat i(4+2)-\hat j(2-2)+\hat k(2+4)=6\hat i+0\hat j+6\hat k$$

Its magnitude is $\sqrt{36+36}=6\sqrt2$. The unit vector is

$\hat n=\frac{6\hat i+6\hat k}{6\sqrt2}=\frac{1}{\sqrt2}\hat i+\frac1{\sqrt2}\hat k$

(The vector $-\hat n$ is also a valid answer.)

"SUNDAY" has 6 distinct letters.

All arrangements: $6!=720$.

Beginning with S: fix S first and arrange the other 5 letters: $5!=120$.

Beginning with S and not ending with Y: arrangements beginning with S and ending with Y are $4!=24$. So

$120-24=96$

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

So $x+iy=\sqrt i$. Squaring both sides:

$(x+iy)^2=i\ \Rightarrow\ x^2-y^2+2ixy=i$

Taking moduli of $(x+iy)^2=i$:

$|x+iy|^2=|i|=1\ \Rightarrow\ x^2+y^2=1$

Alternatively, equating real and imaginary parts gives $x^2-y^2=0$ and $2xy=1$, so $x=y=\pm\frac1{\sqrt2}$ and $x^2+y^2=\frac12+\frac12=1$.

Hence proved.

(a) Conic section. A conic section is the curve obtained by cutting a double right circular cone with a plane. Equivalently, it is the locus of a point whose distance from a fixed point (focus) bears a constant ratio $e$ (eccentricity) to its distance from a fixed line (directrix). It is a parabola ($e=1$), an ellipse ($e<1$) or a hyperbola ($e>1$), with circle as the special case $e=0$.

Given $9x^2+4y^2-18x-16y-11=0$. Complete the squares:

$9(x^2-2x)+4(y^2-4y)=11$

$9(x-1)^2+4(y-2)^2=11+9+16=36$

$\frac{(x-1)^2}{4}+\frac{(y-2)^2}{9}=1$

This is an ellipse with centre $(1,2)$. Since $9>4$, the major axis is vertical: $a^2=9,\ b^2=4$, so $a=3,\ b=2$.

$c=\sqrt{a^2-b^2}=\sqrt5$

• Vertices: $(1,\,2\pm3)$, that is $(1,5)$ and $(1,-1)$
• Eccentricity: $e=\dfrac ca=\dfrac{\sqrt5}{3}$
• Foci: $(1,\,2\pm\sqrt5)$

(b) $T(x_1,x_2)=(x_1+x_2,\ x_2,\ x_1)$. Find the images of the standard basis vectors:

$T(1,0)=(1,0,1),\qquad T(0,1)=(1,1,0)$

These form the columns of the matrix:

$[T]=\begin{bmatrix}1&1\\0&1\\1&0\end{bmatrix}$

Check: $\begin{bmatrix}1&1\\0&1\\1&0\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}=\begin{bmatrix}x_1+x_2\\x_2\\x_1\end{bmatrix}$.

(a) Irrational number. A real number that cannot be written as $\frac pq$ with integers $p,q$ and $q\ne0$ is irrational (its decimal expansion is non-terminating and non-repeating).

Proof that $\sqrt2$ is irrational (by contradiction). Suppose $\sqrt2=\dfrac pq$, where $p,q$ are integers with no common factor and $q\ne0$. Then

$2q^2=p^2$

So $p^2$ is even, hence $p$ is even. Write $p=2k$. Then $2q^2=4k^2$, so $q^2=2k^2$, which makes $q^2$ even and hence $q$ even.

Thus both $p$ and $q$ are divisible by 2, contradicting that they have no common factor. Therefore $\sqrt2$ is irrational.

(b) $f(x)=2x+1,\ g(x)=x^2-2$.

$(f\circ g)(x)=f(g(x))=2(x^2-2)+1=2x^2-3$

$(g\circ f)(x)=g(f(x))=(2x+1)^2-2=4x^2+4x-1$

Since $2x^2-3\ne4x^2+4x-1$ in general (for example at $x=1$: $f\circ g=-1$ but $g\circ f=7$), we have $f\circ g\ne g\circ f$.