1. **State the problem:** We have a function $$f(x) = \frac{x^2}{3 + 3x}$$ and its second derivative is given by $$f''(x) = \frac{Q_1}{(3 + 3x)^3}$$ where $$Q_1$$ is a constant. We need to find $$Q_1$$, then compute $$Q = \ln(3 + |Q_1|)$$, and finally evaluate $$T = 5 \sin^2(100Q)$$ to determine which interval $$T$$ lies in. 2. **Find the first derivative $$f'(x)$$:** Use the quotient rule: $$f(x) = \frac{x^2}{3(1+x)}$$ Let $$u = x^2$$ and $$v = 3 + 3x = 3(1+x)$$. Then, $$f'(x) = \frac{u'v - uv'}{v^2} = \frac{2x(3+3x) - x^2(3)}{(3+3x)^2}$$ Simplify numerator: $$2x(3+3x) - 3x^2 = 6x + 6x^2 - 3x^2 = 6x + 3x^2$$ So, $$f'(x) = \frac{6x + 3x^2}{(3+3x)^2} = \frac{3x(2 + x)}{9(1+x)^2} = \frac{x(2+x)}{3(1+x)^2}$$ 3. **Find the second derivative $$f''(x)$$:** Differentiate $$f'(x) = \frac{x(2+x)}{3(1+x)^2}$$. Rewrite numerator: $$x(2+x) = 2x + x^2$$. Use quotient rule again with: $$u = 2x + x^2, \quad v = 3(1+x)^2$$ Calculate derivatives: $$u' = 2 + 2x$$ $$v' = 3 \cdot 2(1+x) = 6(1+x)$$ Then, $$f''(x) = \frac{u'v - uv'}{v^2} = \frac{(2+2x)3(1+x)^2 - (2x + x^2)6(1+x)}{[3(1+x)^2]^2}$$ Simplify numerator: $$3(1+x)^2(2+2x) - 6(1+x)(2x + x^2)$$ Factor terms: $$3(1+x)^2 2(1+x) - 6(1+x) x(2 + x) = 6(1+x)^3 - 6(1+x) x(2 + x)$$ Expand: $$6(1+x)^3 - 6(1+x) x(2 + x)$$ Calculate each term: $$6(1 + 3x + 3x^2 + x^3) - 6(1+x)(2x + x^2)$$ $$= 6 + 18x + 18x^2 + 6x^3 - 6(2x + x^2 + 2x^2 + x^3)$$ $$= 6 + 18x + 18x^2 + 6x^3 - 6(2x + 3x^2 + x^3)$$ $$= 6 + 18x + 18x^2 + 6x^3 - (12x + 18x^2 + 6x^3)$$ $$= 6 + (18x - 12x) + (18x^2 - 18x^2) + (6x^3 - 6x^3) = 6 + 6x$$ Denominator: $$[3(1+x)^2]^2 = 9(1+x)^4$$ So, $$f''(x) = \frac{6 + 6x}{9(1+x)^4} = \frac{6(1+x)}{9(1+x)^4} = \frac{6}{9(1+x)^3} = \frac{2}{3(1+x)^3}$$ 4. **Match with given form:** Given $$f''(x) = \frac{Q_1}{(3 + 3x)^3} = \frac{Q_1}{27(1+x)^3}$$. From our calculation, $$f''(x) = \frac{2}{3(1+x)^3} = \frac{18}{27(1+x)^3}$$ So, $$\frac{Q_1}{27(1+x)^3} = \frac{18}{27(1+x)^3} \implies Q_1 = 18$$ 5. **Calculate $$Q = \ln(3 + |Q_1|)$$:** $$Q = \ln(3 + 18) = \ln(21)$$ 6. **Calculate $$T = 5 \sin^2(100Q)$$:** Since $$Q = \ln(21)$$, compute: $$100Q = 100 \ln(21)$$ Note that $$\sin^2(\theta)$$ is always between 0 and 1. Therefore, $$0 \leq T = 5 \sin^2(100Q) \leq 5$$ To find the exact interval, consider the value modulo $$\pi$$ because sine is periodic with period $$\pi$$ for $$\sin^2$$. Calculate: $$100Q \bmod \pi = 100 \ln(21) \bmod \pi$$ Approximate values: $$\ln(21) \approx 3.044522$$ $$100Q \approx 304.4522$$ $$\pi \approx 3.141593$$ Number of full $$\pi$$ periods in $$100Q$$: $$n = \frac{304.4522}{3.141593} \approx 96.9$$ Fractional part: $$0.9 \times \pi \approx 2.827433$$ So, $$\sin^2(100Q) = \sin^2(2.827433)$$ Calculate $$\sin(2.827433)$$: $$\sin(2.827433) \approx 0.3090$$ Square it: $$0.3090^2 = 0.0955$$ Finally, $$T = 5 \times 0.0955 = 0.4775$$ 7. **Determine the interval for $$T$$:** $$0 \leq T = 0.4775 < 1$$ So, the correct choice is (A) $$0 \leq T < 1$$. **Final answer:** $$T \in [0,1)$$, choice (A).