Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 1 - 2006 - Paper 1

First, we notice that ( \sqrt{4 + 8} = \sqrt{12} = 2\sqrt{3} ), so the integral simplifies to:

$\int x(2\sqrt{3}) \, dx = 2\sqrt{3} \int x \, dx$

Integrating gives:

$2\sqrt{3} \cdot \frac{x^2}{2} + C = \sqrt{3} x^2 + C$

Using the standard limit ( \lim_{y \to 0} \frac{\sin y}{y} = 1 ), we rewrite the limit as follows:

$\lim_{x \to 0} \frac{\sin 5x}{3x} = \lim_{x \to 0} \frac{\sin 5x}{5x} \cdot \frac{5}{3} = \frac{5}{3} \cdot 1 = \frac{5}{3}$

We can use the sum of cubes formula, which states: [ a^3 + b^3 = (a + b)(a^2 - ab + b^2) ] Here, let ( a = \sin \theta ) and ( b = \cos \theta ):

Thus:

$\sin^3 \theta + \cos^3 \theta = (\sin \theta + \cos \theta)(\sin^2 \theta - \sin \theta \cos \theta + \cos^2 \theta)$

Using ( \sin^2 \theta + \cos^2 \theta = 1 ):

$= (\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)$

The original expression becomes:

$\frac{(\sin \theta + \cos \theta)(1 - \sin \theta \cos \theta)}{\sin \theta + \cos \theta} - 1 = 1 - \sin \theta \cos \theta - 1 = -\sin \theta \cos \theta$

To find the values of ( b ), we require that the line ( y = 12x + b ) touches the curve ( y = x^3 ) at some point. This occurs when they intersect at exactly one point, which requires that:

1. The two functions are equal at some point: ( 12x + b = x^3 )
2. Their slopes are equal at the same point: The slope of the line is 12, while the slope of the curve at ( x ) is given by ( 3x^2 ).

Setting these equal, we find:

$3x^2 = 12 \implies x^2 = 4 \implies x = 2 \text{ or } x = -2$

We substitute these values back into the first equation to solve for ( b ):

For ( x = 2 ): [ 12(2) + b = (2)^3 \implies 24 + b = 8 \implies b = 8 - 24 = -16 ]

For ( x = -2 ): [ 12(-2) + b = (-2)^3 \implies -24 + b = -8 \implies b = -8 + 24 = 16 ]

Thus, the values of ( b ) that make the line tangent to the curve are ( b = -16 ) and ( b = 16 ).