Check $x=0$: $f(x)=|x-\pi|(e^{|x|}-1)\sin|x|$. Near $x=0$, both $(e^{|x|}-1)$ and $\sin|x|$ have factors of $|x|$, so $f(x)\approx |x-\pi|\cdot|x|^2$ near 0, which is differentiable at 0.

Check $x=\pi$: near $\pi$, $(e^{|x|}-1)\sin|x|$ is smooth and non-zero ($e^\pi\sin\pi=0$!). Since $\sin\pi=0$, the product $(e^{|x|}-1)\sin|x|$ vanishes at $\pi$, making $f(x)=0$ near $\pi$ — wait, $\sin\pi=0$, so $f(\pi)=0$. Checking differentiability more carefully shows $f$ is differentiable at $\pi$ too.

Therefore $S=\boldsymbol{\emptyset}$.

This is a binomial experiment: $n=10$, $p=P(\text{green})=\dfrac{15}{25}=\dfrac{3}{5}$, $q=\dfrac{2}{5}$.

$\text{Variance} = npq = 10 \times \dfrac{3}{5} \times \dfrac{2}{5} = 10 \times \dfrac{6}{25} = \dfrac{60}{25} = \dfrac{12}{5}$

Using Raoult's Law: $\dfrac{P^0 - P}{P^0} = x_{\text{solute}}$

$\dfrac{185 - 183}{185} = \dfrac{1.2/M}{(1.2/M) + (100/58)}$

$\dfrac{2}{185} = x_{\text{solute}}$

Since $1.2/M \ll 100/58$, approximate: $x_{\text{solute}} \approx \dfrac{1.2/M}{100/58} = \dfrac{1.2 \times 58}{100M}$

$\dfrac{2}{185} = \dfrac{69.6}{100M}$

$M = \dfrac{69.6 \times 185}{200} = \dfrac{12876}{200} \approx \mathbf{128\ \text{g mol}^{-1}}$

Flextime is the practice of permitting employees to choose, with certain limitations, their own working hours. This helps organizations attract qualified individuals by allowing both employment and family needs to be addressed.

The Threshold Limit Value (TLV), established by the American Conference of Governmental Industrial Hygienists (ACGIH), refers to the maximum concentration of a chemical or physical agent in the workplace air to which workers can be repeatedly exposed, day after day, without adverse health effects.

Types of TLV:

• TLV-TWA: Time-Weighted Average (8-hour workday, 40-hour week)
• TLV-STEL: Short-Term Exposure Limit (15-minute exposure)
• TLV-C: Ceiling value — should never be exceeded

Understanding TLVs is essential for occupational health nurses managing industrial settings in Pakistan.