Sequences & Series
Arithmetic: $S_n = \dfrac{n}{2}(2a_1 + (n-1)d)$
Geometric: $S_\infty = \dfrac{a}{1-r},\;|r|<1$
Remember: geometric series converges only when $|r| < 1$
Find $S_{20}$ for $a_1=3,\,d=4$.
$S_{20} = \dfrac{20}{2}(6+76) = 820$
Binomial Theorem
$(a+b)^n = \displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r$
General term: $T_{r+1} = \binom{n}{r}a^{n-r}b^r$
Coefficient of $x^3$ in $(2x+3)^5$.
$\binom{5}{3}(2x)^3(3)^2 = 10 \cdot 8 \cdot 9 = 720$
Complex Numbers
Modulus: $|z| = \sqrt{a^2+b^2}$; Argument: $\arg(z) = \arctan\!\left(\dfrac{b}{a}\right)$
De Moivre: $(r\,\text{cis}\,\theta)^n = r^n\,\text{cis}(n\theta)$
Polar form: $z = r(\cos\theta + i\sin\theta) = r\,e^{i\theta}$
$(1+i)^8 = ?$
$r=\sqrt{2},\;\theta=\pi/4$; $({\sqrt{2}})^8\,\text{cis}(2\pi) = 16$
Proof by Induction
Step 1: Verify base case $n=1$. Step 2: Assume true for $n=k$. Step 3: Prove for $n=k+1$.
The inductive hypothesis must be clearly stated and used.
Differentiation
Product rule: $(uv)' = u'v + uv'$; Chain rule: $\dfrac{d}{dx}f(g(x)) = f'(g(x))\cdot g'(x)$
$f(x)=x^3\ln x \Rightarrow f'(x) = 3x^2\ln x + x^2$
Integration
By parts: $\displaystyle\int u\,dv = uv - \int v\,du$
$\displaystyle\int xe^x\,dx = (x-1)e^x + C$
Vectors
$\cos\theta = \dfrac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$
Parallel: $\mathbf{a}\times\mathbf{b}=\mathbf{0}$; Perpendicular: $\mathbf{a}\cdot\mathbf{b}=0$
Statistics & Probability
$X\sim B(n,p)$: $P(X=k) = \binom{n}{k}p^k(1-p)^{n-k}$
$X\sim N(\mu,\sigma^2)$: standardise $Z = \dfrac{X-\mu}{\sigma}$
Calculus Applications
Related rates: use chain rule. Area between curves: $\displaystyle\int_a^b [f(x)-g(x)]\,dx$
Maclaurin Series & Limits
$e^x = 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots$
$\displaystyle\lim_{x\to 0}\frac{\sin kx}{mx} = \frac{k}{m}$
An arithmetic sequence has first term $a_1 = 3$ and common difference $d = 4$. What is the sum of the first 20 terms, $S_{20}$?
$S_{20} = 760$
$S_{20} = 800$
$S_{20} = 820$
$S_{20} = 840$
Using $S_n = \dfrac{n}{2}(2a_1+(n-1)d)$: $S_{20} = \dfrac{20}{2}(2\cdot3 + 19\cdot4) = 10(6+76) = 10 \times 82 = \mathbf{820}$.
A geometric series has first term $a = 6$ and common ratio $r = \dfrac{1}{2}$. Find the sum to infinity $S_\infty$.
$S_\infty = 10$
$S_\infty = 12$
$S_\infty = 18$
$S_\infty = 24$
$S_\infty = \dfrac{a}{1-r} = \dfrac{6}{1-\frac{1}{2}} = \dfrac{6}{\frac{1}{2}} = \mathbf{12}$. The series converges since $|r|=\tfrac{1}{2}<1$.
Find the coefficient of $x^3$ in the binomial expansion of $(2x+3)^5$.
$360$
$480$
$600$
$720$
The general term is $T_{r+1}=\binom{5}{r}(2x)^{5-r}(3)^r$. For $x^3$: set $5-r=3\Rightarrow r=2$. $T_3 = \binom{5}{2}(2x)^3(3)^2 = 10 \cdot 8x^3 \cdot 9 = \mathbf{720}x^3$.
Let $z = \dfrac{3+4i}{1+2i}$. What is $|z|$?
$\sqrt{3}$
$\sqrt{5}$
$\sqrt{7}$
$\sqrt{10}$
Multiply numerator and denominator by the conjugate: $z = \dfrac{(3+4i)(1-2i)}{(1+2i)(1-2i)} = \dfrac{3-6i+4i+8}{1+4} = \dfrac{11-2i}{5}$. Therefore $|z| = \dfrac{\sqrt{121+4}}{5} = \dfrac{\sqrt{125}}{5} = \dfrac{5\sqrt{5}}{5} = \mathbf{\sqrt{5}}$.
In a proof by mathematical induction that $\displaystyle\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$, which expression correctly represents the inductive step for $n=m+1$?
$\dfrac{m(m+1)}{2} + (m+1)$
$\dfrac{m(m+1)}{2} + m$
$\dfrac{(m+1)(m+2)}{2} - (m+1)$
$\dfrac{(m-1)m}{2} + (m+1)$
Assuming the formula holds for $n=m$: $\displaystyle\sum_{k=1}^{m+1}k = \underbrace{\dfrac{m(m+1)}{2}}_{\text{inductive hypothesis}} + (m+1)$. Factoring: $= \dfrac{m(m+1)+2(m+1)}{2} = \dfrac{(m+1)(m+2)}{2}$, which is the formula for $n=m+1$. ✓
Let $f(x) = x^3\ln x$ for $x>0$. Find $f'(x)$.
$f'(x) = 3x^2\ln x + x^2$
$f'(x) = 3x^2\ln x + x^3$
$f'(x) = 3x^2\ln x - x^2$
$f'(x) = x^2\ln x + 3x^2$
Apply the product rule with $u=x^3$, $v=\ln x$: $f'(x) = 3x^2 \cdot \ln x + x^3 \cdot \dfrac{1}{x} = \mathbf{3x^2\ln x + x^2}$.
Evaluate $\displaystyle\int xe^x\,dx$.
$e^x(x+1)+C$
$e^x(x-1)+C$
$xe^x + C$
$e^x(x^2-x)+C$
Integration by parts: let $u=x$, $dv=e^x\,dx$, so $du=dx$, $v=e^x$. $\displaystyle\int xe^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = \mathbf{e^x(x-1)+C}$.
Vectors $\mathbf{a} = \begin{pmatrix}1\\2\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}2\\-1\\2\end{pmatrix}$. Find the angle $\theta$ between them.
$\theta = \arccos\!\left(\dfrac{\sqrt{14}}{7}\right)$
$\theta = \arccos\!\left(\dfrac{1}{7}\right)$
$\theta = \arccos\!\left(\dfrac{6}{\sqrt{14}}\right)$
$\theta = \arccos\!\left(\dfrac{2}{7}\right)$
$\mathbf{a}\cdot\mathbf{b} = 2-2+6 = 6$; $|\mathbf{a}|=\sqrt{1+4+9}=\sqrt{14}$; $|\mathbf{b}|=\sqrt{4+1+4}=3$. Therefore $\cos\theta = \dfrac{6}{3\sqrt{14}} = \dfrac{2}{\sqrt{14}} = \dfrac{\sqrt{14}}{7}$. So $\theta = \arccos\!\left(\dfrac{\sqrt{14}}{7}\right)$.
Find the determinant of the matrix $A = \begin{pmatrix}2&1\\5&3\end{pmatrix}$.
$-1$
$0$
$1$
$11$
$\det(A) = (2)(3)-(1)(5) = 6-5 = \mathbf{1}$. Since $\det(A)\neq 0$, the matrix is invertible.
Write the Maclaurin series for $e^x$ up to and including the $x^3$ term.
$1+x+x^2+x^3$
$1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}$
$x+\dfrac{x^2}{2}+\dfrac{x^3}{3}$
$1+x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}$
The Maclaurin series is $e^x = \displaystyle\sum_{n=0}^{\infty}\dfrac{x^n}{n!} = 1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\cdots = \mathbf{1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}}+\cdots$
Solve $2\sin^2 x - \sin x - 1 = 0$ for $x\in[0,2\pi]$. How many solutions are there?
$2$ solutions
$3$ solutions
$4$ solutions
$1$ solution
Factor: $(2\sin x+1)(\sin x-1)=0$, giving $\sin x = -\tfrac{1}{2}$ or $\sin x = 1$. For $\sin x=1$: $x=\dfrac{\pi}{2}$ (1 solution). For $\sin x=-\tfrac{1}{2}$: $x=\dfrac{7\pi}{6},\dfrac{11\pi}{6}$ (2 solutions). Total: $\mathbf{3}$ solutions.
Solve $\log_2 x + \log_2(x-2) = 3$.
$x = 2$
$x = 3$
$x = 4$
$x = 6$
Combine: $\log_2[x(x-2)]=3 \Rightarrow x(x-2)=8 \Rightarrow x^2-2x-8=0 \Rightarrow (x-4)(x+2)=0$. So $x=4$ or $x=-2$. Since $x>0$ and $x-2>0$, we need $x>2$, so $x=\mathbf{4}$.
A random variable $X\sim N(65,\,100)$. Find $P(X<70)$ to 4 decimal places.
$0.6554$
$0.6915$
$0.7257$
$0.5398$
Standardise: $Z = \dfrac{70-65}{10} = 0.5$. From standard normal tables, $P(Z<0.5) = \mathbf{0.6915}$. Note: $\sigma^2=100$ means $\sigma=10$.
A circular oil spill grows so that its radius increases at $2\,\text{cm/s}$. When $r=5\,\text{cm}$, find the rate of increase of the area $\dfrac{dA}{dt}$ in $\text{cm}^2/\text{s}$.
$10\pi$
$20\pi$
$25\pi$
$40\pi$
$A = \pi r^2 \Rightarrow \dfrac{dA}{dr} = 2\pi r$. By chain rule: $\dfrac{dA}{dt} = 2\pi r \cdot \dfrac{dr}{dt} = 2\pi(5)(2) = \mathbf{20\pi}\,\text{cm}^2/\text{s}$.
Find the area of the region enclosed between $y=x^2$ and $y=x+2$.
$\dfrac{7}{2}$
$4$
$\dfrac{9}{2}$
$5$
Intersections: $x^2=x+2 \Rightarrow x^2-x-2=0 \Rightarrow (x-2)(x+1)=0$, so $x=-1,\,2$. Area $= \displaystyle\int_{-1}^{2}[(x+2)-x^2]\,dx = \left[\dfrac{x^2}{2}+2x-\dfrac{x^3}{3}\right]_{-1}^{2} = \left(2+4-\dfrac{8}{3}\right)-\left(\dfrac{1}{2}-2+\dfrac{1}{3}\right) = \dfrac{10}{3}+\dfrac{7}{6} = \mathbf{\dfrac{9}{2}}$.
A random variable $X\sim B(10,\,0.3)$. Find $P(X=3)$ to 4 decimal places.
$0.2001$
$0.2335$
$0.2668$
$0.3000$
$P(X=3) = \binom{10}{3}(0.3)^3(0.7)^7 = 120 \times 0.027 \times 0.0823543 = 120 \times 0.002223567 \approx \mathbf{0.2668}$.
Using De Moivre's theorem, find $(1+i)^8$.
$8$
$16$
$-16$
$16i$
Write $1+i = \sqrt{2}\,\text{cis}\!\left(\dfrac{\pi}{4}\right)$. Then $(1+i)^8 = (\sqrt{2})^8\,\text{cis}(2\pi) = 2^4 \cdot 1 = \mathbf{16}$.
Solve the differential equation $\dfrac{dy}{dx} = 2xy$, given that $y(0) = 3$.
$y = 3e^{2x}$
$y = 3e^{x^2}$
$y = e^{x^2+3}$
$y = 3e^{x^2+1}$
Separate variables: $\dfrac{dy}{y} = 2x\,dx$. Integrate both sides: $\ln|y| = x^2 + C$, so $y = Ae^{x^2}$. Using $y(0)=3$: $3 = Ae^0 \Rightarrow A=3$. Thus $y = \mathbf{3e^{x^2}}$.
Two lines are given by $L_1: \mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + t\begin{pmatrix}1\\1\\0\end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix}2\\3\\3\end{pmatrix} + s\begin{pmatrix}2\\1\\0\end{pmatrix}$. Find their intersection point.
$(2, 3, 3)$
$(3, 4, 3)$
$(1, 2, 3)$
$(0, 1, 3)$
Setting equal: $1+t=2+2s$ and $2+t=3+s$. Subtracting: $-1=-1-s \Rightarrow s=0$. Then $t=1$. Intersection: $\mathbf{r}=\begin{pmatrix}2\\3\\3\end{pmatrix}$, i.e., the point $\mathbf{(2,3,3)}$. Verify: $L_2$ at $s=0$ gives $(2,3,3)$. ✓
Evaluate $\displaystyle\lim_{x \to 0} \frac{\sin 3x}{2x}$.
$0$
$1$
$\dfrac{3}{2}$
$3$
Write $\dfrac{\sin 3x}{2x} = \dfrac{3}{2}\cdot\dfrac{\sin 3x}{3x}$. As $x\to 0$, $\dfrac{\sin 3x}{3x}\to 1$ (standard limit). Therefore the limit $= \dfrac{3}{2}\cdot 1 = \mathbf{\dfrac{3}{2}}$.