1. **Problem 4:** Given two lines r and s in the plane with the following properties: - Line s forms an angle of $\frac{\pi}{3}$ with the x-axis. - Lines r and s are perpendicular and intersect at point $(1,0)$. - Line r is an asymptote to the graph of function $f$ as $x \to +\infty$. We want to determine which of the given limits is true: (A) $\lim_{x \to +\infty} \frac{f(x)}{x} = \sqrt{3}$ (B) $\lim_{x \to +\infty} (\sqrt{3}x - 3f(x)) = \sqrt{3}$ (C) $\lim_{x \to +\infty} (3f(x) - \sqrt{3}x) = \sqrt{3}$ (D) $\lim_{x \to +\infty} \frac{f(x)}{x} = -\sqrt{3}$ --- 2. **Step 1: Find the slope of line s.** Since line s forms an angle $\theta = \frac{\pi}{3}$ with the x-axis, its slope is: $$m_s = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$$ 3. **Step 2: Find the slope of line r.** Lines r and s are perpendicular, so their slopes satisfy: $$m_r \cdot m_s = -1 \implies m_r = -\frac{1}{m_s} = -\frac{1}{\sqrt{3}}$$ 4. **Step 3: Equation of line r.** Line r passes through $(1,0)$ with slope $m_r = -\frac{1}{\sqrt{3}}$: $$y - 0 = -\frac{1}{\sqrt{3}}(x - 1) \implies y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$$ 5. **Step 4: Since line r is an asymptote to $f$ as $x \to +\infty$, the graph of $f$ approaches line r.** This means: $$\lim_{x \to +\infty} \left(f(x) - \left(-\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}\right)\right) = 0$$ 6. **Step 5: Analyze the limits given.** - For (A) and (D), consider $\lim_{x \to +\infty} \frac{f(x)}{x}$. From the asymptote: $$f(x) \approx -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$$ Dividing by $x$: $$\frac{f(x)}{x} \approx -\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}x}$$ Taking limit as $x \to +\infty$: $$\lim_{x \to +\infty} \frac{f(x)}{x} = -\frac{1}{\sqrt{3}}$$ Neither (A) $\sqrt{3}$ nor (D) $-\sqrt{3}$ matches this exactly, but (D) is close except for the denominator. 7. **Step 6: Check (B) and (C).** Rewrite (B): $$\lim_{x \to +\infty} (\sqrt{3}x - 3f(x))$$ Using the asymptote approximation: $$3f(x) \approx 3\left(-\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}\right) = -\sqrt{3}x + \sqrt{3}$$ So: $$\sqrt{3}x - 3f(x) \approx \sqrt{3}x - \left(-\sqrt{3}x + \sqrt{3}\right) = \sqrt{3}x + \sqrt{3}x - \sqrt{3} = 2\sqrt{3}x - \sqrt{3}$$ As $x \to +\infty$, this goes to $+\infty$, so (B) is false. Check (C): $$3f(x) - \sqrt{3}x \approx -\sqrt{3}x + \sqrt{3} - \sqrt{3}x = -2\sqrt{3}x + \sqrt{3}$$ As $x \to +\infty$, this goes to $-\infty$, so (C) is false. 8. **Step 7: Conclusion for problem 4.** The only plausible limit is: $$\lim_{x \to +\infty} \frac{f(x)}{x} = -\frac{1}{\sqrt{3}}$$ Since none of the options exactly match $-\frac{1}{\sqrt{3}}$, but (D) is $-\sqrt{3}$, which is different, none is exactly correct. However, given the options, the closest is (D) if the problem expects the negative slope of the asymptote. --- **Problem 5.1:** Given $g$ differentiable on $\mathbb{R} \setminus \{-\frac{7}{2}\}$ with $$g'(x) = \frac{2x^2 - 12}{2x + 7}$$ Find: $$\lim_{x \to 9} \frac{x^2 - 81}{g(x) - g(9)}$$ --- 9. **Step 1: Recognize the limit form.** The limit resembles the reciprocal of the derivative of $g$ at $x=9$: $$\lim_{x \to 9} \frac{x^2 - 81}{g(x) - g(9)} = \lim_{x \to 9} \frac{(x-9)(x+9)}{g(x) - g(9)}$$ If $g$ is differentiable at 9, then: $$\lim_{x \to 9} \frac{g(x) - g(9)}{x - 9} = g'(9)$$ So: $$\lim_{x \to 9} \frac{x^2 - 81}{g(x) - g(9)} = \lim_{x \to 9} \frac{(x-9)(x+9)}{g(x) - g(9)} = \lim_{x \to 9} (x+9) \cdot \frac{x-9}{g(x) - g(9)} = (9+9) \cdot \frac{1}{g'(9)} = \frac{18}{g'(9)}$$ 10. **Step 2: Calculate $g'(9)$.** $$g'(9) = \frac{2(9)^2 - 12}{2(9) + 7} = \frac{2 \cdot 81 - 12}{18 + 7} = \frac{162 - 12}{25} = \frac{150}{25} = 6$$ 11. **Step 3: Calculate the limit.** $$\lim_{x \to 9} \frac{x^2 - 81}{g(x) - g(9)} = \frac{18}{6} = 3$$ **Answer:** (B) 3 --- **Problem 6:** Given $g$ twice differentiable on $\mathbb{R}$ and the graph of $g'$ has a relative minimum at $x=2$. Determine which statement is true: (A) $g$ is decreasing on $\mathbb{R}$. (B) $g$ is decreasing on $]-\infty, 2]$. (C) The graph of $g$ is concave up on $\mathbb{R}$. (D) The graph of $g$ is concave down on $]-\infty, 2]$. --- 12. **Step 1: Understand the meaning of $g'$ having a relative minimum at $x=2$.** - Since $g'$ has a relative minimum at $x=2$, then: $$g''(2) = 0 \quad \text{and} \quad g''(x) > 0 \text{ near } x=2$$ because the derivative of $g'$ is $g''$, and a minimum of $g'$ means $g''$ changes from negative to positive or is positive at that point. 13. **Step 2: Interpret concavity of $g$.** - $g''(x) > 0$ near $x=2$ means $g$ is concave up near $x=2$. - Since $g'$ has a minimum at $x=2$, $g''$ changes sign from negative to positive at $x=2$. - So for $x < 2$, $g''(x) < 0$ (concave down), and for $x > 2$, $g''(x) > 0$ (concave up). 14. **Step 3: Analyze the options.** - (A) $g$ is decreasing on $\mathbb{R}$: Not necessarily true, depends on $g'$ sign. - (B) $g$ is decreasing on $]-\infty, 2]$: Not necessarily true. - (C) $g$ is concave up on $\mathbb{R}$: False, since $g''$ changes sign at $x=2$. - (D) $g$ is concave down on $]-\infty, 2]$: True, since $g''(x) < 0$ for $x < 2$. **Answer:** (D) --- **Summary:** - Problem 4: Closest correct is (D) but exact slope is $-\frac{1}{\sqrt{3}}$. - Problem 5.1: Answer is (B) 3. - Problem 6: Answer is (D).