1. Evaluate $\lim_{\theta \to \infty} \frac{1 + \cos \theta}{2\theta^2}$ using the Squeeze-Theorem. Since $-1 \leq \cos \theta \leq 1$, we have: $$\frac{0}{2\theta^2} \leq \frac{1+\cos \theta}{2\theta^2} \leq \frac{2}{2\theta^2} \implies 0 \leq \frac{1+\cos \theta}{2\theta^2} \leq \frac{1}{\theta^2}$$ As $\theta \to \infty$, both bounds tend to $0$, so by Squeeze-Theorem, limit is: $$0$$ 2. Calculate volume of solid formed by rotating $y = e^x$, $0 \leq x \leq 2$ about x-axis. Volume via disk method: $$ V = \pi \int_0^2 (e^x)^2 dx = \pi \int_0^2 e^{2x} dx = \pi \left[ \frac{e^{2x}}{2} \right]_0^2 = \frac{\pi}{2}(e^4 - 1) $$ 3. Evaluate indefinite integral $\int \frac{1}{x} \cos(\ln x) dx$. Let $t=\ln x$, then $dt = \frac{1}{x} dx$ so: $$\int \frac{1}{x} \cos(\ln x) dx = \int \cos t dt = \sin t + C = \sin(\ln x) + C$$ 4. Given $z = \cos \theta + i \sin \theta$, use De Moivre's Theorem to show: $$z^n + z^{-n} = 2 \cos n\theta$$ for $n \in \mathbb{N}$. By De Moivre's Theorem: $$z^n = \cos n\theta + i \sin n\theta$$ $$z^{-n} = \cos(-n\theta) + i \sin(-n\theta) = \cos n\theta - i \sin n\theta$$ Add: $$z^n + z^{-n} = 2 \cos n\theta$$ 5. Using $$(z + z^{-1})^4 = z^4 + 4 z^2 + 6 + 4 z^{-2} + z^{-4}$$ and result from (4), deduce: $$\cos^4 \theta = \frac{1}{8} \cos 4\theta + \frac{1}{2} \cos 2\theta + \frac{3}{8}$$ Note $z = e^{i\theta}$, so: $$z + z^{-1} = 2 \cos \theta$$ Raise to power 4: $$(2 \cos \theta)^4 = (z + z^{-1})^4 = z^4 + 4 z^2 + 6 + 4 z^{-2} + z^{-4}$$ Divide both sides by 16: $$\cos^4 \theta = \frac{1}{16} (z^4 + 4 z^2 + 6 + 4 z^{-2} + z^{-4})$$ Using (4): $$z^n + z^{-n} = 2 \cos n\theta$$ So: $$\cos^4 \theta = \frac{1}{16}(2 \cos 4\theta + 8 \cos 2\theta + 6) = \frac{1}{8} \cos 4\theta + \frac{1}{2} \cos 2\theta + \frac{3}{8}$$ Evaluate the integral: $$\int_{\pi/4}^{\pi/2} \cos^4 \theta d\theta = \int_{\pi/4}^{\pi/2} \left( \frac{1}{8} \cos 4\theta + \frac{1}{2} \cos 2\theta + \frac{3}{8} \right) d\theta$$ Integrate term-wise: $$= \frac{1}{8} \int_{\pi/4}^{\pi/2} \cos 4\theta d\theta + \frac{1}{2} \int_{\pi/4}^{\pi/2} \cos 2\theta d\theta + \frac{3}{8} \int_{\pi/4}^{\pi/2} d\theta$$ Calculate each: $$\int \cos k\theta d\theta = \frac{\sin k\theta}{k}$$ So: $$= \frac{1}{8} \left[ \frac{\sin 4\theta}{4} \right]_{\pi/4}^{\pi/2} + \frac{1}{2} \left[ \frac{\sin 2\theta}{2} \right]_{\pi/4}^{\pi/2} + \frac{3}{8} \left[ \theta \right]_{\pi/4}^{\pi/2}$$ Calculate values: $$\sin 4(\pi/2) = \sin 2\pi = 0, \quad \sin 4(\pi/4) = \sin \pi = 0$$ $$\sin 2(\pi/2) = \sin \pi = 0, \quad \sin 2(\pi/4) = \sin \frac{\pi}{2} = 1$$ $$\theta \text{ from } \pi/4 \text{ to } \pi/2 \Rightarrow \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$$ Hence: $$= \frac{1}{8} \cdot 0 + \frac{1}{2} \cdot \left(0 - \frac{1}{2} \right) + \frac{3}{8} \cdot \frac{\pi}{4} = -\frac{1}{4} + \frac{3\pi}{32} = \frac{3\pi}{32} - \frac{1}{4}$$ 6. (a) Given $z = x + iy$, $z^2 = p + iq$ with $x,y,p,q \in \mathbb{R}$, prove: $$ (x + y)^2 = \sqrt{p^2 + q^2} + q $$ Calculate $z^2$: $$z^2 = (x + iy)^2 = x^2 - y^2 + 2 i x y$$ So: $$p = x^2 - y^2, \quad q = 2 x y$$ Calculate: $$p^2 + q^2 = (x^2 - y^2)^2 + (2 x y)^2 = x^4 - 2 x^2 y^2 + y^4 + 4 x^2 y^2 = x^4 + 2 x^2 y^2 + y^4 = (x^2 + y^2)^2$$ Therefore: $$\sqrt{p^2 + q^2} = x^2 + y^2$$ Add $q$: $$\sqrt{p^2 + q^2} + q = x^2 + y^2 + 2 x y = (x + y)^2$$ (b) Find quadratic curve with same tangent and normal as $$y = x + \frac{3}{x^2}$$ at $Q$, $x=1$. Compute derivative: $$y' = 1 - \frac{6}{x^3}$$ At $x=1$: $$y'(1) = 1 - 6 = -5$$ Equation of tangent at $x=1$: $$y = y(1) + y'(1)(x - 1)$$ $$y(1) = 1 + 3 = 4$$ So: $$y = 4 - 5(x - 1) = -5x + 9$$ Normal slope: $$m_{normal} = -\frac{1}{y'} = \frac{1}{5}$$ Equation normal at $x=1$: $$y - 4 = \frac{1}{5} (x - 1) \implies y = \frac{1}{5} x + \frac{19}{5}$$ Try quadratic curve: $$y = a x^2 + b x + c$$ Derivative: $$y' = 2 a x + b$$ At $x=1$: $$a + b + c = 4$$ $$2 a + b = -5$$ Normal slope at 1 is $-1/(2a + b)$ must be $1/5$, so: $$-\frac{1}{2a + b} = \frac{1}{5} \implies 2a + b = -5$$ (same as derivative slope condition) So solve: $$\begin{cases} a + b + c = 4 \\ 2a + b = -5 \end{cases}$$ Choose $a = 0$ to simplify: $$b = -5, \quad c = 4 - 0 - (-5) = 9$$ Quadratic: $$y = -5 x + 9$$ This is linear, contradicts quadratic requirement, so let $a = 1$: $$2(1) + b = -5 \implies b = -7$$ $$1 -7 + c = 4 \implies c = 10$$ So quadratic is: $$y = x^2 - 7 x + 10$$ (c) Prove: $$\int_{-\alpha}^{\alpha} \frac{f(x)}{1 + b x^2} dx = \int_0^{\alpha} f(x) dx, \quad b \neq 0$$ Assuming $f$ is even or odd, check parity of integrand: Denominator $1 + b x^2$ even function. If $f$ is even: $$\text{integrand } = \frac{f(x)}{1 + b x^2}$$ is even, integral is double from 0 to $\alpha$ not equal to RHS. If $f$ is odd: Integral from $-\alpha$ to $\alpha$ is zero. RHS integral is from 0 to $\alpha$ of odd function, not zero. Alternatively, consider substitution: Let $I = \int_{-\alpha}^{\alpha} \frac{f(x)}{1 + b x^2} dx$ By substitution $x = -t$: $$I = \int_{\alpha}^{-\alpha} \frac{f(-t)}{1 + b t^2} (-dt) = \int_{-\alpha}^{\alpha} \frac{f(-x)}{1 + b x^2} dx$$ So: $$I = \frac{1}{2} \int_{-\alpha}^{\alpha} \frac{f(x) + f(-x)}{1 + b x^2} dx$$ If assume $f$ is odd, simplifies to 0. But the given equality is true if: $$f(x) = \frac{f(x)}{1 + b x^2}, \quad \text{for } x \geq 0$$ Which implies $1 + b x^2 = 1$ or $b=0$, contradicting $b \neq 0$. Hence problem likely contains typo or additional condition. "content" returns all parts step by step as above.