Write (1 + \sqrt{5})^3 in the form a + b\sqrt{5}, where a and b are integers - HSC - SSCE Mathematics Extension 1 - Question 1 - 2007 - Paper 1

To find the coordinates of point P(x, y) dividing the segment AB in the ratio ( m:n = 2:3 ), we use the section formula:

$x = \frac{mx_2 + nx_1}{m+n} \quad \text{and} \quad y = \frac{my_2 + ny_1}{m+n}$

Substituting A(4, 5) and B(19, -5):

$x = \frac{2 \cdot 19 + 3 \cdot 4}{2 + 3} = \frac{38 + 12}{5} = \frac{50}{5} = 10$

$y = \frac{2 \cdot (-5) + 3 \cdot 5}{2 + 3} = \frac{-10 + 15}{5} = \frac{5}{5} = 1$

Thus, point P is (10, 1).

To differentiate ( \tan^{-1}(x^4) ), we apply the chain rule:

$\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1+u^2} \cdot \frac{du}{dx}, \text{ where } u = x^4.$

Now we find ( \frac{du}{dx} = 4x^3 ).

Thus, the derivative becomes:

$\frac{d}{dx} \tan^{-1}(x^4) = \frac{1}{1+(x^4)^2} \cdot 4x^3 = \frac{4x^3}{1+x^8}.$

First, rewrite the equation of the line:

$y = \frac{3}{2}.$

Next, we find the slope of the line, ( m_1 = 0 ) since it's horizontal.

For the curve ( y = x^3 + 1 ):

The derivative is:

$\frac{dy}{dx} = 3x^2.$

Evaluating at the point (1, 2):

$m_2 = 3(1^2) = 3.$

Now, using the formula for the angle ( \theta ) between the two lines:

$\tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right|.$

Substituting the values:

$\tan(\theta) = \left| \frac{3 - 0}{1 + 0 \cdot 3} \right| = 3.$

The angle is:

$\theta = \tan^{-1}(3),$

and the exact value in radians is ( \tan^{-1}(3) ).

Using the substitution ( u = 25 - x^2 ), we find:

$du = -2x \, dx \Rightarrow dx = \frac{du}{-2x}.$

For the limits: When ( x = -3 ), ( u = 25 - 9 = 16 ) When ( x = 4 ), ( u = 25 - 16 = 9. ) Using these limits, the integral becomes:

$\int_{16}^{9} \frac{2x}{\sqrt{u}} \cdot \frac{du}{-2x} = -\int_{16}^{9} \frac{1}{\sqrt{u}} du.$

Now reversing the limits gives:

$\int_{9}^{16} u^{-1/2} du = \left[ 2u^{1/2} \right]_{9}^{16} = 2(4) - 2(3) = 8 - 6 = 2.$

Thus, the answer is 2.