Question

One of the points of intersection of the curves $y = 1 + 3x - 2x^2$ and $y = \frac{1}{x}$ is $\left( \frac{1}{2}, 2 \right)$. Let the area of the region enclosed by these curves be $\frac{1}{24} \left( \ell \sqrt{5} + m \right) - n \log_e \left( 1 + \sqrt{5} \right),$ where $\ell, m, n \in \mathbb{N}$. Then $\ell + m + n$ is equal to:

• A. 32
• B. 30
• C. 29
• D. 31

Answer 1

Answer: (B)

30

Answer 2

$A = \int_{1/2}^{1+\sqrt{5}} \left(1 + 3x - 2x^2 - \frac{1}{x}\right) \, dx$
$A = \left[ x + \frac{3x^2}{2} - \frac{2x^3}{3} - \ln x \right]_{1/2}^{1+\sqrt{5}}$
$A = \left( \left(1 + \sqrt{5}\right) + \frac{3\left(1+\sqrt{5}\right)^2}{2} - \frac{2\left(1+\sqrt{5}\right)^3}{3} - \ln\left(1+\sqrt{5}\right) \right) - \left( \frac{1}{2} + \frac{3\left(1/2\right)^2}{2} - \frac{2\left(1/2\right)^3}{3} - \ln\left(\frac{1}{2}\right) \right)$

Simplify step-by-step:

$A = (1 + \sqrt{5}) + \frac{3}{2}(1+\sqrt{5})^2 - \frac{2}{3}(1+\sqrt{5})^3 - \ln(1+\sqrt{5}) - \left( \frac{1}{2} + \frac{3}{8} - \frac{1}{12} + \ln 2 \right)$
$A = \frac{1}{2} + \sqrt{5} + \frac{3}{2}(1 + 2\sqrt{5} + 5) - \frac{2}{3}(1 + 3\sqrt{5} + 3(5) + 5\sqrt{5}) - \ln(1+\sqrt{5}) - \frac{1}{2} - \frac{3}{8} + \frac{1}{12} - \ln 2$

Combine and simplify further:
$A = \sqrt{5}\left(1 + \frac{3}{2} - 2\right) + \frac{15}{8} - \frac{4}{3} + \frac{1}{12} - \ln(1+\sqrt{5})$
$A = \frac{14\sqrt{5}}{24} + \frac{15}{24} - \ln(1+\sqrt{5})$

Final answer: $A = \frac{14\sqrt{5}}{24} + \frac{15}{24} - \ln(1+\sqrt{5})$