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

To find the angle ( \theta ) between two lines with slopes ( m_1 ) and ( m_2 ), we use the formula:

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

For the lines given, we identify the slopes:

• For ( y = 2x + 5 ), ( m_1 = 2 )
• For ( y = 4 - 3x ), rearranging gives ( y = -3x + 4 ) thus ( m_2 = -3 )

Now substituting: [ \tan \theta = \left| \frac{2 - (-3)}{1 + 2 \cdot (-3)} \right| = \left| \frac{5}{1 - 6} \right| = \left| \frac{5}{-5} \right| = 1 ]

Thus, ( \theta = \tan^{-1}(1) = \frac{\pi}{4} ). We convert this to degrees: [ \theta = 45^\circ ]

To solve ( \frac{4}{x + 3} \geq 1 ), we start by isolating ( x ):

1. Multiply both sides by ( x + 3 ) (noting that ( x + 3 > 0 ) for the inequality to hold): [ 4 \geq x + 3 ]

2. Rearranging gives: [ x \leq 1 ]

3. Next, we find when ( x + 3 ) is positive: [ x + 3 > 0 \Rightarrow x > -3 ]

Combining both conditions, we have: [ -3 < x \leq 1 ]

To write ( 5 , ext{cos} , x - 12 , ext{sin} , x ) in the form ( A \cos(x + \alpha) ), we first find ( A ) and ( \alpha ):

1. Calculate ( A ) using: [ A = \sqrt{(5)^{2} + (-12)^{2}} = \sqrt{25 + 144} = \sqrt{169} = 13 ]
2. Determine ( \alpha ): [ \tan \alpha = \frac{-12}{5} \Rightarrow \alpha = \tan^{-1}\left(-\frac{12}{5}\right) ]

Thus: [ 5 , \cos , x - 12 , \sin , x = 13 \cos(x + \alpha) ]

Using the substitution ( u = 2x - 1 \Rightarrow x = \frac{u + 1}{2} , \Rightarrow dx = \frac{1}{2} du ):

• Change the limits:
  • When ( x = 1 ), ( u = 2(1) - 1 = 1 )
  • When ( x = 2 ), ( u = 2(2) - 1 = 3 )

Now substituting: [ \int_{1}^{3} \frac{\frac{u + 1}{2}}{u^{3}} \cdot \frac{1}{2} , du = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^{3}} , du ]

This simplifies to: [ = \frac{1}{4} \left( \int_{1}^{3} u^{-2} , du + \int_{1}^{3} u^{-3} , du \right) ]

Calculating both integrals gives: [ \int u^{-2} , du = -u^{-1} igg|{1}^{3} = -\left(\frac{1}{3} - 1\right) = \frac{2}{3} ] [ \int u^{-3} , du = -\frac{1}{2}u^{-2} igg|{1}^{3} = -\frac{1}{2}\left(\frac{1}{9} - 1\right) = \frac{4}{18} = \frac{2}{9} ]

Combining these results gives: [ \text{Final result} = \frac{1}{4} \left( \frac{2}{3} + \frac{2}{9} \right) ]

Since ( A(x) = x - 3 ) is a factor of ( P(x) ), we can substitute ( x = 3 ) into ( P(x) ) and set it to zero:

[ P(3) = (3)^{3} - k(3)^{2} + 5(3) + 12 = 0 ] [ 27 - 9k + 15 + 12 = 0 \Rightarrow 54 - 9k = 0 \Rightarrow 9k = 54 \Rightarrow k = 6. ]

With ( k = 6 ), the polynomial becomes: [ P(x) = x^{3} - 6x^{2} + 5x + 12. ]

Using synthetic division with the root ( x = 3 ):

1. Divide ( P(x) ) by ( A(x) ): This yields the quadratic ( x^{2} - 3x - 4 ).
2. Factoring gives: [ (x - 4)(x + 1) = 0 ]

Therefore, the zeros are: ( x = 3, x = 4, x = -1. )