Roots: $x=\frac{-2\pm\sqrt{4-8}}{2}=-1\pm i$. So $\alpha=-1+i, \beta=-1-i$.

In polar form: $|\alpha|=\sqrt{2}$, $\arg(\alpha)=\frac{3\pi}{4}$. So $\alpha=\sqrt{2}\,e^{i3\pi/4}$.

$\alpha^{15}=(\sqrt{2})^{15}e^{i\cdot45\pi/4}=2^{15/2}e^{i\pi/4}$ (since $45\pi/4=11\pi+\pi/4$, so $e^{i45\pi/4}=e^{i\pi/4}\cdot(-1)^{11}=-e^{i\pi/4}$... careful: $45/4=11.25$, $11\pi+\pi/4$, $e^{i(11\pi+\pi/4)}=e^{i\pi}\cdot e^{i\pi/4} \cdot (-1)^{10}... $)

More cleanly: $\alpha^{15}+\beta^{15}=2\,\text{Re}(\alpha^{15})=2(\sqrt{2})^{15}\cos\!\left(\frac{45\pi}{4}\right)$. Since $\cos(45\pi/4)=\cos(\pi/4+11\pi)=-\cos(\pi/4)=-\frac{1}{\sqrt{2}}$: $=2\cdot2^{15/2}\cdot(-\frac{1}{\sqrt{2}})=-2^{15/2+1-1/2}=-2^8=-\mathbf{256}$

Rationalise the denominator by multiplying by $\dfrac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$:

Denominator becomes $2-(1+\cos x)=1-\cos x$.

$\lim_{x\to0}\dfrac{\sin^2x(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$

Use $\dfrac{\sin^2x}{1-\cos x}=\dfrac{1-\cos^2x}{1-\cos x}=1+\cos x$:

$=\lim_{x\to0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})=(1+1)(\sqrt{2}+\sqrt{2})=2\cdot2\sqrt{2}=\mathbf{4\sqrt{2}}$

Wait: $2\times2\sqrt{2}=4\sqrt{2}$... but option A is 4. Let me recheck: $(2)(\sqrt{2}+\sqrt{2})=2\cdot2\sqrt{2}=4\sqrt{2}$. Answer: $4\sqrt{2}$ is not listed as option A (which is 4). The correct answer is $\mathbf{4\sqrt{2}}$ (index 1).

The Somogyi Effect (post-hypoglycemic hyperglycemia):

• Nocturnal hypoglycemia occurs (often unnoticed during sleep)
• Counter-regulatory hormones (glucagon, cortisol, epinephrine, GH) are released
• Causes rebound morning hyperglycemia

Differentiation from Dawn Phenomenon:

Checking 3 AM blood glucose helps distinguish these two conditions — important for Charge Nurse assessment.

In a break-even chart, the gap between Total Revenue and Total Cost represents Profit (if revenue is higher) or Loss (if cost is higher).