1. Problem: Find the sum $S$ and product $P$ of all primitive 10th roots of unity. Step 1: The 10th roots of unity are the complex solutions to $z^{10} = 1$. Step 2: The primitive 10th roots of unity are those with order exactly 10, so $z^{k} \neq 1$ for $1 \leq k \leq 9$. Step 3: The number of primitive 10th roots is $\phi(10) = 4$, where $\phi$ is Euler's totient function. Step 4: The sum of primitive $n$th roots of unity equals the Möbius function $\mu(n)$. Here, $S = \mu(10)$. Step 5: Since $10 = 2 \times 5$, product of two distinct primes, $\mu(10) = 1$ if the number has even number of distinct primes, otherwise $-1$. 2 primes implies $\mu(10) = 1 * (-1)^2 = 1$. Actually, the sum of primitive $n$th roots of unity is $\mu(n)$, but $\mu(10) = 1$ means $S = 1$. Step 6: The product of the primitive $n$th roots of unity is $(-1)^{\phi(n)/2}$. Since $\phi(10) = 4$, $$P = (-1)^{4/2} = (-1)^2 = 1.$$ Answer: $S=1$ and $P=1$ which corresponds to option (E). 2. Problem: Determine for which metric $(\mathbb{R}, d)$ is NOT complete. Step 1: Recall that $(\mathbb{R}, |x - y|)$ is complete. Step 2: For $d(x,y) = |\arctan x - \arctan y|$, since arctan is continuous and increasing mapping $\mathbb{R}$ onto $(-\pi/2, \pi/2)$, $(\mathbb{R}, d)$ is isometric to $((-\pi/2, \pi/2), |\cdot|)$ which is NOT complete as an open interval in $\mathbb{R}$. Step 3: For $d(x,y) = |x^3 - y^3|$, since $f(x) = x^3$ is a homeomorphism (continuous with continuous inverse) from $\mathbb{R}$ to $\mathbb{R}$, completeness is preserved, so complete. Step 4: For $d(x,y) = |\sqrt{x} - \sqrt{y}|$, this is only defined for $x,y \geq 0$. If domain intended $\mathbb{R}$, sqrt is not defined for negatives, so metric not valid over all $\mathbb{R}$. Assuming domain $[0, \infty)$, $( [0, \infty), d )$ is isometric to $([0, \infty), |x-y|)$ which is complete as a closed half-line. Step 5: Discrete metric $d(x,y) =0$ if $x=y$, else $1$, is complete because every Cauchy sequence is eventually constant. Answer: The ONLY NOT complete metric space is (B). 3. Problem: Compute $$\int_0^{\pi} \frac{\sin(100x)}{\sin x} dx$$. Step 1: Use the identity for Dirichlet kernel: $$\frac{\sin (nx)}{\sin x} = \sum_{k=0}^{n-1} \cos((n-1-2k)x)$$ So, $$\frac{\sin(100x)}{\sin x} = \sum_{k=0}^{99} \cos((99 - 2k) x)$$ Step 2: The integral becomes $$\int_0^{\pi} \sum_{k=0}^{99} \cos((99 - 2k) x) dx = \sum_{k=0}^{99} \int_0^{\pi} \cos(m_k x) dx$$ where $m_k = 99 - 2k$ is an odd integer. Step 3: Integral of cosine: $$\int_0^{\pi} \cos(mx) dx = \frac{\sin(m\pi)}{m} - 0 = 0$$ because $\sin(m\pi) = 0$ for all integer $m$. Step 4: Except when $m=0$, but as $m=99-2k$ and $k$ integer, $m$ never zero (99 is odd). Step 5: However, known formula: $$\int_0^{\pi} \frac{\sin(nx)}{\sin x} dx = n\pi$$ when $n$ is integer. Wait, this formula is incorrect. Correct formula according to the Dirichlet kernel integral is: $$\int_0^{\pi} \frac{\sin((2n+1)x)}{\sin x} dx = \pi$$ Step 6: Alternatively, recognize that numeric answer in options is 50 or 50\pi. Indeed, the Dirichlet kernel integral: $$D_{n}(x) = \frac{\sin((n + 1/2) x)}{\sin(x/2)}$$ not matching exactly. Step 7: Another approach -- integral is $\pi$ if $n$ is odd. More precisely: $$\int_0^{\pi} \frac{\sin(mx)}{\sin x} dx = \pi$$ if $m$ is an integer. Step 8: Here, $m=100$ even, the integral is $\pi$ if odd, zero if even? The integral evaluates to $\pi$ if $n$ odd and zero if $n$ even? Step 9: Known result: \int_0^{\pi} \frac{\sin(nx)}{\sin x} dx = \pi$ if $n$ odd, 0 if $n$ even. Since 100 is even, answer is 0. Step 10: So answer is (A) 0. 4. Problem: Given $f,g$ continuously differentiable on $\mathbb{R}$ with $f(0)=g(0)=0$, and $g'(0) \neq 0$, determine which statements I, II, III are true. I. $f/g$ can be extended continuously near 0. Step 1: If $g'(0) \neq 0$, by the definition of differentiability and since $g(0)=0$, near 0, $g(x) \approx g'(0) x$ linearly. Step 2: For $f/g$ to extend continuously at 0, limit $\lim_{x \to 0} f(x)/g(x)$ must exist. Step 3: Using L'Hôpital's rule, $$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \frac{f'(0)}{g'(0)}$$ exists or finite. Thus $f/g$ can be extended continuously at 0 by defining $$ \frac{f}{g}(0) = \frac{f'(0)}{g'(0)} $$ So I is TRUE. II. $(f^2 - f)/(2g - g^3)$ can be extended continuously at 0. Step 4: Evaluate numerator and denominator at 0: $$f(0)^2 - f(0) = 0 - 0 = 0, $$ $$2 g(0) - g(0)^3 = 0 - 0 = 0.$$ Step 5: Use L'Hôpital's rule or factor near 0: Step 6: Numerator near 0: $$f(x)^2 - f(x) = f(x)(f(x)-1)$$ Since $f(0) = 0$. Denominator $$2g(x) - g(x)^3 o 0$$ Step 7: Assessment requires more assumptions, but $g'(0) \neq 0$ helps. Using expansions: $$f(x) = f'(0)x + o(x), \quad g(x) = g'(0)x + o(x).$$ Numerator: $$f(x)^2 - f(x) = (f'(0)x)^2 - f'(0) x + o(x) = f'(0)^2 x^2 - f'(0) x + o(x)$$ Denominator: $$2 g'(0) x - (g'(0) x)^3 + o(x) = 2 g'(0) x - (g'(0)^3 x^3) + o(x)$$ Step 8: The dominant terms as $x \to 0$: Numerator ~ $-f'(0) x$, denominator ~ $2 g'(0) x$. So limit is $$\lim_{x \to 0} \frac{f(x)^2 - f(x)}{2g(x) - g(x)^3} = \frac{- f'(0)}{2 g'(0)}$$ finite. Hence, II is TRUE. III. $f/g$ can be extended to a differentiable function near 0. Step 9: To be differentiable at 0, the extended function must have derivative. Step 10: From the previous steps, extended $f/g$ at 0 is continuous. Step 11: To check differentiability, check $$ \lim_{x \to 0} \frac{f(x)/g(x) - f'(0)/g'(0)}{x} $$ exists in general, unless additional conditions are placed on $f, g$. Step 12: Without further assumptions, differentiability can't be guaranteed. Therefore, III is not necessarily TRUE. Final answers: (1) Sum and product of primitive 10th roots of unity: (E) $S=1$, $P=1$ (2) Metric space $(\mathbb{R}, d)$ NOT complete for metric (B). (3) Integral evaluates to (A) 0. (4) Only statements I and II hold: (C) I and II only.