Mathematics

(c) $[-3,-2]$

$A-B$ is the part of $A$ not in $B$: the points of $(-3,2]$ that are not in $(-2,3)$. These are the points from $-3$ up to and including $-2$. Exactly this set is $(-3,-2]$, since $-3\notin A$; option (c) is the closest listed.

(d) $2n\pi+\frac{5\pi}{6}$

$\sin x>0$ and $\cos x<0$ place $x$ in the second quadrant. The reference angle with $\sin=\frac12$ is $\frac\pi6$, so $x=\pi-\frac\pi6=\frac{5\pi}{6}$. Adding the period gives $2n\pi+\frac{5\pi}{6}$.

(a) $\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}$

For $A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$, $\operatorname{adj}A=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}$. Here $a=2,b=3,c=5,d=-2$, giving $\begin{bmatrix}-2&-3\\-5&2\end{bmatrix}$.

(a) 1 or 4

Equal roots need discriminant zero:

$4(k+2)^2-4(9k)=0\Rightarrow k^2+4k+4=9k\Rightarrow k^2-5k+4=0$

So $(k-1)(k-4)=0$ and $k=1$ or $4$.

(c) greater than 1

For a hyperbola, $e=\sqrt{1+\frac{b^2}{a^2}}>1$. (An ellipse has $e<1$ and a parabola has $e=1$.)

(b) 0

Let $\alpha=\omega$, $\beta=\omega^2$. Then $\alpha^4=\omega^4=\omega$, $\beta^4=\omega^8=\omega^2$, and $\alpha\beta=\omega^3=1$.

$\omega+\omega^2+1=0$

(b) $\frac\pi3$

$\sin\frac{2\pi}{3}=\frac{\sqrt3}{2}$. The principal value of $\sin^{-1}$ lies in $[-\frac\pi2,\frac\pi2]$, so $\sin^{-1}\frac{\sqrt3}{2}=\frac\pi3$.