Find \( \int \sin^3 x \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

The slopes of the lines are ( m_1 = 2 ) and ( m_2 = -3 ). The formula for the angle ( \theta ) between two lines is given by:

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

Substituting the values gives:

$$\tan \theta = \left| \frac{2 - (-3)}{1 + 2 \cdot (-3)} \right| = \left| \frac{5}{-5} \right| = 1$$

Thus, the angle ( \theta = \frac{\pi}{4} = 45^\circ ) since the acute angle is taken.

To solve the inequality, we first rearrange it:

$$\frac{4}{x + 3} - 1 \geq 0$$

This simplifies to:

$$\frac{4 - (x + 3)}{x + 3} \geq 0$$

or

$$\frac{1 - x}{x + 3} \geq 0$$

The critical points are ( x = 1 ) and ( x = -3 ). Testing intervals, we find:

The solution is ( x \leq 1 ) or ( x > -3 ), thus the solution is ( -3 < x \leq 1 )

To rewrite ( 5 \cos x - 12 \sin x ) in the required form, we can find ( A ) and ( \alpha ) using:

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To find ( \alpha ):

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Thus, the expression can be written as:

$13 \cos(x + \alpha)$

Using the substitution ( u = 2x - 1 \implies du = 2dx ightarrow dx = \frac{du}{2} ).

The limits change: when ( x = 1, u = 1 ) and when ( x = 2, u = 3 ).

The integral becomes:

$$\int_{1}^{3} \frac{2}{u^3} \cdot \frac{du}{2} = \int_{1}^{3} \frac{1}{u^3} \, du$$

Evaluating, we have:

$$= \left[-\frac{1}{2u^2}\right]_{1}^{3} = -\frac{1}{2(3)^2} + \frac{1}{2(1)^2} = -\frac{1}{18} + \frac{1}{2} = \frac{8}{18} = \frac{4}{9}$$

To show that ( k = 6 ) for the polynomial ( P(x) = x^3 - kx^2 + 5x + 12 ) when ( A(x) = x - 3 ) is a factor, we use the factor theorem:

Since ( A(3) = 0 ), we substitute ( x = 3 ) into ( P(x) ):

$$P(3) = 3^3 - k(3^2) + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k$$

Setting ( P(3) = 0:)

$$54 - 9k = 0 \Rightarrow 9k = 54 \Rightarrow k = 6$$

Given ( P(x) = x^3 - 6x^2 + 5x + 12 ), we can factor it as follows:

We already know that ( x = 3 ) is a zero. Performing polynomial long division to find the other factors, we find:

Resulting polynomial after division is ( x^2 - 3x - 4 = 0 ).

Factoring gives:

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Thus, the zeros of ( P(x) ) are ( x = 3, 4, -1 ).