New Millennium Academy

If $f: A \to B$ and $g: B \to C$, the composite from $A$ to $C$ is denoted $gof(x)$, written $gf(x)$. Verify orientation of arrows in the diagram before confirming.

$2x - x = 0 \Rightarrow x = 0$

$\sin(30^\circ+60^\circ)=\sin 90^\circ = 1$

$\tan 2A = \dfrac{\frac{1}{7}+\frac{2}{3}}{1-\frac{1}{7}\cdot\frac{2}{3}} = \dfrac{17/21}{19/21} = \dfrac{17}{19}$

Slope of $x-3y+7=0$ is $\frac{1}{3}$. $\frac{3-2}{a+1}=\frac{1}{3}\Rightarrow a=2$

$\vec{a}^2=2^2+3^2=13$

$Q_1=15,\ Q_3=45\Rightarrow \frac{Q_3-Q_1}{Q_3+Q_1}=\frac{30}{60}=0.5$

Read the $y$-value the curve approaches as $x \to 4$ from the left. Confirm against the figure.

• A composite function combines two functions $f$ and $g$ so that the output of one becomes the input of the other: $(gof)(x)=g[f(x)]$.
• $y=\frac{2x-7}{3}\Rightarrow x=\frac{3y+7}{2}$, so $g^{-1}(x)=\frac{3x+7}{2}$
• $ff(x)=2(2x-3)-3=4x-9$. Set $4x-9=\frac{3x+7}{2} \Rightarrow 8x-18=3x+7 \Rightarrow x=\boxed{5}$

• A matrix has an inverse iff it is a square matrix and its determinant is non-zero ($|A|\ne 0$).
• $AB=I$: solving gives $m=\boxed{2},\ n=\boxed{-7}$ (verify: $A=\begin{bmatrix}4&7\\5&9\end{bmatrix}$, $|A|=1$, $A^{-1}=\begin{bmatrix}9&-7\\-5&4\end{bmatrix}=B$)
• Standard proof: transposing twice restores rows/columns to original position, so $(A^T)^T=A$.

• $g\circ f(a)=g(1)=p$;   $g\circ f(b)=g(2)=q$
• Mapping diagram: $a \to p$, $b \to q$ (arrows from set $\{a,b\}$ directly to $\{p,q\}$)

• Compound angles are angles formed by the algebraic sum or difference of two or more angles, e.g. $A+B$, $A-B$.
• $\cos(A+B)-\cos(A-B) = -2\sin A \sin B$
• From $\tan(A+B)=\tan 45^\circ=1$: $\tan A+\tan B+\tan A\tan B=1$. Adding 1 to both sides: $1+\tan A+\tan B+\tan A\tan B=2 \Rightarrow (1+\tan A)(1+\tan B)=2$

• $\tan(50^\circ+40^\circ)=\tan 90^\circ$, which is undefined ($\infty$).
• Since $50^\circ+40^\circ=90^\circ$, $\cos 40^\circ=\sin 50^\circ$. Then $\tan50^\circ-\tan40^\circ=\frac{\sin10^\circ}{\cos50^\circ\cos40^\circ}=\frac{\sin10^\circ}{\frac{1}{2}\cos10^\circ}=2\tan10^\circ$, hence $\tan50^\circ-2\tan10^\circ=\tan40^\circ$

• $\cos A=\frac{4}{5}$, $\sin B=\frac{1}{5\sqrt2}$. $\sin(A+B)=\frac{3}{5}\cdot\frac{7}{5\sqrt2}+\frac{4}{5}\cdot\frac{1}{5\sqrt2}=\frac{25}{25\sqrt2}=\boxed{\frac{1}{\sqrt2}}$
• $\tan A=\frac34,\ \tan B=\frac17 \Rightarrow \tan(A+B)=\frac{3/4+1/7}{1-3/28}=\frac{25/28}{25/28}=1 \Rightarrow A+B=45^\circ=\frac{\pi^c}{4}$. Gita is correct.

• It means all three points satisfy the same linear equation, i.e. the slope between any two of the three points is equal.
• Line through $A(p,0)$, $B(0,q)$: $\frac{x}{p}+\frac{y}{q}=1$. Substituting $C(5,5)$: $\frac{5}{p}+\frac{5}{q}=1 \Rightarrow \frac{1}{p}+\frac{1}{q}=\frac{1}{5}$

• Midpoint of $BC=(-2,5)$. Line through $A(2,2)$ and $(-2,5)$: slope $=-\frac34$; $4(y-2)=-3(x-2)\Rightarrow 3x+4y=14$
• Length $=\sqrt{(-2-2)^2+(5-2)^2}=\sqrt{16+9}=\boxed{5 \text{ units}}$

• $X'(1,6)$, $Y'(3,2)$, $Z'(7,4)$  (add $(2,3)$ to each vertex)
• Plot $\triangle XYZ$ and $\triangle X'Y'Z'$ on the same graph, shifted 2 right and 3 up.
• $T$ is called the translation vector (displacement vector).

• $\vec{m}\cdot\vec{n}=0$, i.e. $m_1n_1+m_2n_2=0$
• $3(6)+m(-2)=0 \Rightarrow m=\boxed{9}$
• With $\vec{m}=3\vec{i}+9\vec{j}$: $\cos\theta=\frac{3}{\sqrt{90}}=\frac{1}{\sqrt{10}} \Rightarrow \theta\approx 71.57^\circ$. Claim of $70^\circ$ is incorrect; correct angle $\approx 71.57^\circ$ ($71^\circ34'$).

• Coefficient of Q.D. $=\dfrac{Q_3-Q_1}{Q_3+Q_1}$
• $0.71=\dfrac{65-Q_1}{65+Q_1} \Rightarrow Q_1 \approx \boxed{11 \text{ years}}$
• A smaller coefficient of Q.D. means less dispersion, so the camp with coefficient 0.30 is more consistent.

• $f$ is continuous at $x=a$ if $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$
• $\lim_{x\to4^-}f(x)=2(4)+1=9$;   $\lim_{x\to4^+}f(x)=4+5=9$
• $\lim_{x\to4}f(x)=9$, but $f(4)=7$, so $\lim_{x\to4}f(x)\ne f(4)$
• Not continuous at $x=4$. Redefine $f(4)=9$ instead of 7 to make it continuous.

• $A'(5,3)$, $B'(8,5)$, $C'(9,4)$  (add $(3,2)$ to each vertex)
• $P=$ midpoint of $BC=(5.5,\,2.5)$. Slope of $AP=\dfrac{2.5-1}{5.5-2}=\boxed{\frac{3}{7}}$
• $\vec{BA}=(-3,-2)$, $\vec{BC}=(1,-1)$. $\cos(\angle ABC)=\dfrac{-1}{\sqrt{13}\sqrt2}=\dfrac{-1}{\sqrt{26}} \Rightarrow \angle ABC\approx \boxed{101.31^\circ}$ ($101^\circ19'$)
• $|\vec a+\vec b|=|\vec a-\vec b| \Rightarrow \vec a \cdot \vec b = 0 \Rightarrow \vec a \perp \vec b$ (squaring both sides and simplifying gives $4\vec a\cdot\vec b=0$)

• $f\circ g(\theta)=\dfrac{\cos18^\circ-\sin18^\circ}{\cos18^\circ+\sin18^\circ}=\dfrac{1-\tan18^\circ}{1+\tan18^\circ}=\tan(45^\circ-18^\circ)=\boxed{\tan27^\circ}$
• As derived above, $m=\tan27^\circ$ (proved via the tangent subtraction identity).
• Since $m=\tan27^\circ$, taking $\tan^{-1}$ of both sides: $\tan^{-1}(m)=27^\circ$. Yes, correct.
• $m_1=\dfrac{\cos18^\circ+\sin18^\circ}{\cos18^\circ-\sin18^\circ}=\tan(45^\circ+18^\circ)=\tan63^\circ=\dfrac{1}{\tan27^\circ}$. So $m\cdot m_1=\tan27^\circ\cdot\dfrac{1}{\tan27^\circ}=\boxed{1}$. Yes.

• $g\circ f(x)=6x+11$. Check $(0,11)$ ✓. $g\circ f(2)=23 \Rightarrow q=\boxed{23}$. $6p+11=-1 \Rightarrow p=\boxed{-2}$
• $\cos(45^\circ+30^\circ)=\cos45^\circ\cos30^\circ-\sin45^\circ\sin30^\circ=\dfrac{\sqrt6-\sqrt2}{4}$
• Midpoint of $A(2,-1)$ and $B(8,5)$ is $(5,2)=C$. Yes, $C$ lies exactly on segment $AB$ (in fact at its midpoint).

• $N=50$. $Q_1$ class $=24\text{-}28$: $Q_1=24+\frac{12.5-6}{10}(4)=26.6$. $Q_3$ class $=32\text{-}36$: $Q_3=32+\frac{37.5-30}{12}(4)=34.5$. Q.D.$=\dfrac{34.5-26.6}{2}=\boxed{3.95\text{ kg}}$
• L.H.L.$=2(3.95)+1=8.9$;   R.H.L.$=3(3.95)-2.95=8.9$
• Since L.H.L. $=$ R.H.L. $=f(\text{Q.D.})=8.9$, $f(x)$ is continuous at $x=$ Q.D.
• Check $(2, 3.95)$: $1.975(2)+3.95=3.95+3.95=7.9$. Yes, the point lies on the same line as the two warehouses.