1. **Problem:** Find the points of discontinuity for the function $f(x) = \frac{5}{x^2 - 7x + 12}$. 2. **Formula and rules:** The function is discontinuous where the denominator is zero because division by zero is undefined. 3. **Find zeros of the denominator:** $$x^2 - 7x + 12 = 0$$ Factor the quadratic: $$ (x - 3)(x - 4) = 0 $$ So, $x = 3$ or $x = 4$. 4. **Conclusion:** The function is discontinuous at $x=3$ and $x=4$. 5. **Answer:** The set of discontinuities is $\{3, 4\}$ which corresponds to option b). 2. **Problem:** Calculate the limit $$\lim_{x \to 0} \frac{\sin 9x - \sin 7x^2}{3x}$$ 3. **Formula and rules:** Use the limit property $\lim_{x \to 0} \frac{\sin ax}{x} = a$ and linearity of limits. 4. **Evaluate:** Rewrite numerator using limit properties: $$\lim_{x \to 0} \frac{\sin 9x}{3x} - \lim_{x \to 0} \frac{\sin 7x^2}{3x}$$ First term: $$\lim_{x \to 0} \frac{\sin 9x}{3x} = \lim_{x \to 0} \frac{\sin 9x}{9x} \cdot \frac{9x}{3x} = 1 \cdot 3 = 3$$ Second term: $$\lim_{x \to 0} \frac{\sin 7x^2}{3x} = \lim_{x \to 0} \frac{\sin 7x^2}{x^2} \cdot \frac{x^2}{3x} = \lim_{x \to 0} \frac{\sin 7x^2}{7x^2} \cdot 7 \cdot \frac{x}{3} = 1 \cdot 7 \cdot 0 = 0$$ 5. **Result:** $$3 - 0 = 3$$ 6. **Answer:** Option a). 3. **Problem:** Find the fourth derivative of $y = \ln 3$. 4. **Explanation:** $\ln 3$ is a constant because 3 is a constant number. 5. **Derivatives:** The derivative of a constant is zero, so all derivatives including the fourth derivative are zero. 6. **Answer:** Option d) 0. 4. **Problem:** Given $$\int_3^7 f(x) dx = -5$$ Find $$\int_7^3 f(x) dx$$ 5. **Rule:** Reversing the limits of integration changes the sign: $$\int_a^b f(x) dx = -\int_b^a f(x) dx$$ 6. **Apply:** $$\int_7^3 f(x) dx = -\int_3^7 f(x) dx = -(-5) = 5$$ 7. **Answer:** Option b). 5. **Problem:** Given $$f(x) = x^4 + 3x^2 + 7$$ with $$f'(x) = 4x^3 + 6x$$ $$f''(x) = 12x^2 + 6 > 0$$ Determine the concavity of $f$. 6. **Explanation:** Since $f''(x) = 12x^2 + 6$ is always positive for all real $x$ (because $12x^2 \geq 0$ and $6 > 0$), the function is concave up everywhere. 7. **Answer:** Option d) مقعرة لأعلى على ℝ (concave up on all real numbers).