1. **State the problem:** We are given the function $f(x) = \ln(13x^2 + 5)$ and need to find the points of inflection $Q_1$ and $Q_2$ where the concavity changes. 2. **Find the first derivative:** $$f'(x) = \frac{d}{dx} \ln(13x^2 + 5) = \frac{1}{13x^2 + 5} \times 26x = \frac{26x}{13x^2 + 5}$$ 3. **Find the second derivative to determine concavity:** Using the quotient rule, $$f''(x) = \frac{(26)(13x^2 + 5) - 26x(26x)}{(13x^2 + 5)^2} = \frac{26(13x^2 + 5) - 676x^2}{(13x^2 + 5)^2}$$ Simplify numerator: $$26 \times 13x^2 = 338x^2, \quad 338x^2 + 130 - 676x^2 = -338x^2 + 130$$ So, $$f''(x) = \frac{-338x^2 + 130}{(13x^2 + 5)^2}$$ 4. **Find points of inflection by setting $f''(x) = 0$:** $$-338x^2 + 130 = 0 \implies 338x^2 = 130 \implies x^2 = \frac{130}{338} = \frac{65}{169}$$ $$x = \pm \sqrt{\frac{65}{169}} = \pm \frac{\sqrt{65}}{13}$$ Thus, $$Q_1 = -\frac{\sqrt{65}}{13}, \quad Q_2 = \frac{\sqrt{65}}{13}$$ 5. **Determine concavity between $Q_1$ and $Q_2$:** Choose $x=0$ between $Q_1$ and $Q_2$: $$f''(0) = \frac{-338(0)^2 + 130}{(5)^2} = \frac{130}{25} = 5.2 > 0$$ Concave up on $(Q_1, Q_2)$ so $Q_3 = 1$. 6. **Calculate $Q$:** $$Q = \ln\left(3 + |Q_1| + 2|Q_2| + 3|Q_3|\right) = \ln\left(3 + \frac{\sqrt{65}}{13} + 2 \times \frac{\sqrt{65}}{13} + 3 \times 1\right)$$ $$= \ln\left(3 + 3 \times \frac{\sqrt{65}}{13} + 3\right) = \ln\left(6 + \frac{3\sqrt{65}}{13}\right)$$ 7. **Calculate $T$:** $$T = 5 \sin^2(100Q)$$ Since $\sin^2$ ranges from 0 to 1, $T$ ranges from 0 to 5. 8. **Summary:** - Points of inflection: $Q_1 = -\frac{\sqrt{65}}{13}$, $Q_2 = \frac{\sqrt{65}}{13}$ - Concavity between $Q_1$ and $Q_2$ is up, so $Q_3 = 1$ - $Q = \ln\left(6 + \frac{3\sqrt{65}}{13}\right)$ - $T = 5 \sin^2(100Q)$ with $0 \leq T \leq 5$ This completes the solution for the first problem regarding points of inflection and related values.