Option C is correct.

• Asset not capitalised → Non-current assets understated.
• Full cost charged as expense → expenses increase → loss increases → Loss overstated.

Let $z=e^{i\theta}=\cos\theta+i\sin\theta$. Then $\bar{z}=e^{-i\theta}$.

$\frac{1+z}{1+\bar{z}}=\frac{1+e^{i\theta}}{1+e^{-i\theta}}=\frac{e^{i\theta/2}(e^{-i\theta/2}+e^{i\theta/2})}{e^{-i\theta/2}(e^{i\theta/2}+e^{-i\theta/2})}=\frac{e^{i\theta/2}}{e^{-i\theta/2}}=e^{i\theta}$

So $\arg\!\left(\frac{1+z}{1+\bar{z}}\right)=\theta$

Option B ($29,200) is correct.

X's salary = $550 ÷ 5% = $11,000; Y's salary = $450 ÷ 5% = $9,000

Residual profit = $94,000 − $3,000 + $1,000 − $11,000 − $9,000 = $73,000 (Interest on drawings added back: $550+$450=$1,000)

Y's share = 2/5 × $73,000 = $29,200

From the condition $|f(x)-f(y)|\le 2|x-y|^{3/2}$, divide both sides by $|x-y|$:

$\left|\frac{f(x)-f(y)}{x-y}\right|\le 2|x-y|^{1/2}$

Taking $y\to x$: $|f'(x)|\le 2\cdot0=0$, so $f'(x)=0$ for all $x$.

Hence $f$ is constant. Since $f(0)=1$, we have $f(x)=1$ for all $x$.

$\int_0^1 f^2(x)\,dx = \int_0^1 1\,dx = \mathbf{1}$

• Across a period: decreases left to right (increasing Z*)
• Down a group: increases (new shells added)

The time period of a pendulum is $T = 2\pi\sqrt{\frac{L}{g}}$, so $L \propto T^2$.

• Initial length: $L = kT^2$
• New length: $L + \Delta L = k T_M^2$
• Strain: $\frac{\Delta L}{L} = \frac{T_M^2 - T^2}{T^2} = (\frac{T_M}{T})^2 - 1$
• From $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{Mg/A}{\Delta L/L}$
• Therefore, $\frac{1}{Y} = \frac{\Delta L/L}{Mg/A} = [(\frac{T_M}{T})^2 - 1]\frac{A}{Mg}$