1. **Problem Statement:** Given two functions $f(x) = x^2$ and $g(x) = 2x + 1$, find the combined functions $(f+g)(x)$, $(f-g)(x)$, $(fg)(x)$, and $\frac{f}{g}(x)$. 2. **Formulas and Rules:** - Sum: $(f+g)(x) = f(x) + g(x)$ - Difference: $(f-g)(x) = f(x) - g(x)$ - Product: $(fg)(x) = f(x) \cdot g(x)$ - Quotient: $\frac{f}{g}(x) = \frac{f(x)}{g(x)}$, provided $g(x) \neq 0$ 3. **Calculations:** - $(f+g)(x) = x^2 + (2x + 1) = x^2 + 2x + 1$ - $(f-g)(x) = x^2 - (2x + 1) = x^2 - 2x - 1$ - $(fg)(x) = x^2 \cdot (2x + 1) = 2x^3 + x^2$ - $\frac{f}{g}(x) = \frac{x^2}{2x + 1}$, with $2x + 1 \neq 0$ 4. **Final answers for combined functions:** $$(f+g)(x) = x^2 + 2x + 1$$ $$(f-g)(x) = x^2 - 2x - 1$$ $$(fg)(x) = 2x^3 + x^2$$ $$\frac{f}{g}(x) = \frac{x^2}{2x + 1}$$ --- 1. **Problem Statement:** Evaluate the limits: (i) $\lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right)$ (ii) $\lim_{x \to 0} (\tan x - \sec x)$ 2. **Important rules:** - Use series expansions or standard limits. - Recall $\sin x \approx x - \frac{x^3}{6}$, $\tan x \approx x + \frac{x^3}{3}$, $\sec x \approx 1 + \frac{x^2}{2}$ near 0. 3. **Calculations:** (i) $$\frac{1}{\sin x} - \frac{1}{x} = \frac{x - \sin x}{x \sin x}$$ Using $\sin x \approx x - \frac{x^3}{6}$, $$x - \sin x \approx x - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{6}$$ Also, $x \sin x \approx x \left(x - \frac{x^3}{6}\right) = x^2 - \frac{x^4}{6} \approx x^2$ So, $$\lim_{x \to 0} \frac{\frac{x^3}{6}}{x^2} = \lim_{x \to 0} \frac{x}{6} = 0$$ (ii) Using expansions: $$\tan x - \sec x \approx \left(x + \frac{x^3}{3}\right) - \left(1 + \frac{x^2}{2}\right) = x + \frac{x^3}{3} - 1 - \frac{x^2}{2}$$ As $x \to 0$, dominant term is $-1$, so limit does not exist (goes to $-1$). But since $\tan 0 = 0$ and $\sec 0 = 1$, the limit is: $$\lim_{x \to 0} (\tan x - \sec x) = 0 - 1 = -1$$ --- 1. **Problem Statement:** Find exact values of $\sin 135^\circ$ and $\tan 15^\circ$. 2. **Formulas:** - $\sin(180^\circ - \theta) = \sin \theta$ - Use angle subtraction formula for tangent: $$\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$$ 3. **Calculations:** - $\sin 135^\circ = \sin(180^\circ - 45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2}$ - $\tan 15^\circ = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ} = \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ Multiply numerator and denominator by $\sqrt{3} - 1$: $$\frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}$$ --- 1. **Problem Statement:** Simplify the expression: $$\frac{x^2 + x + 1}{(x - 1)(x + 2)}$$ 2. **Explanation:** The denominator is factored; numerator is quadratic. 3. **Check if numerator factors or can be simplified with denominator:** - Numerator $x^2 + x + 1$ does not factor nicely over reals. - No common factors with denominator. 4. **Final expression:** $$\frac{x^2 + x + 1}{(x - 1)(x + 2)}$$ (cannot be simplified further) --- 1. **Problem Statement:** Given $f(x) = x^3 - 6x^2 + 9x + 1$, analyze or simplify. 2. **Explanation:** This is a cubic polynomial. 3. **Try factoring:** Check for roots using Rational Root Theorem: possible roots $\pm1$. - $f(1) = 1 - 6 + 9 + 1 = 5 \neq 0$ - $f(3) = 27 - 54 + 27 + 1 = 1 \neq 0$ Try synthetic division or factor by grouping: Rewrite as $x^3 - 6x^2 + 9x + 1$ Try $(x - 3)^2 = x^2 - 6x + 9$ Try $f(x) = (x - 3)^2 (x + a) + b$ but better to leave as is unless specified. --- 1. **Problem Statement:** Evaluate the integral: $$\int \sin 5x \cdot \cos 2x \, dx$$ 2. **Formula:** Use product-to-sum identity: $$\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]$$ 3. **Apply identity:** $$\sin 5x \cos 2x = \frac{1}{2} [\sin(7x) + \sin(3x)]$$ 4. **Integral becomes:** $$\int \sin 5x \cos 2x \, dx = \frac{1}{2} \int [\sin 7x + \sin 3x] \, dx = \frac{1}{2} \left(-\frac{\cos 7x}{7} - \frac{\cos 3x}{3}\right) + C$$ 5. **Final answer:** $$-\frac{\cos 7x}{14} - \frac{\cos 3x}{6} + C$$ --- **Summary:** - $(f+g)(x) = x^2 + 2x + 1$ - $(f-g)(x) = x^2 - 2x - 1$ - $(fg)(x) = 2x^3 + x^2$ - $\frac{f}{g}(x) = \frac{x^2}{2x + 1}$ - $\lim_{x \to 0} \left( \frac{1}{\sin x} - \frac{1}{x} \right) = 0$ - $\lim_{x \to 0} (\tan x - \sec x) = -1$ - $\sin 135^\circ = \frac{\sqrt{2}}{2}$ - $\tan 15^\circ = 2 - \sqrt{3}$ - Expression $\frac{x^2 + x + 1}{(x - 1)(x + 2)}$ cannot be simplified further - Integral $\int \sin 5x \cos 2x \, dx = -\frac{\cos 7x}{14} - \frac{\cos 3x}{6} + C$