11. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

To find the acute angle between the lines, we need their slopes:

1. Slope of ( y = 2x + 5 ) is ( m_1 = 2 ) and slope of ( y = 4 - 3x ) is ( m_2 = -3 ).

2. Use the formula for the angle ( \theta ) between two lines: [ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{-5} \right| = 1 ]

3. Hence, ( \theta = \tan^{-1}(1) = \frac{\pi}{4} ) radians. The acute angle is thus: [ \frac{\pi}{4} \text{ radians} = 45^\circ ]

To solve this inequality:

1. Rearrange the inequality: [ \frac{4}{x + 3} - 1 \geq 0 ]

2. Combine the fractions: [ \frac{4 - (x + 3)}{x + 3} \geq 0 \implies \frac{1 - x}{x + 3} \geq 0 ]

3. Identify critical points by setting the numerator and denominator equal to zero:

   • Numerator: ( 1 - x = 0 \implies x = 1 )
   • Denominator: ( x + 3 = 0 \implies x = -3 )

4. Determine the intervals: ( (-\infty, -3), (-3, 1), (1, \infty) ). Test each interval:

   • Choose ( x = -4 ): Positive
   • Choose ( x = 0 ): Negative
   • Choose ( x = 2 ): Positive

5. The solution is: [ x \in (-\infty, -3) \cup [1, \infty) \text{ (including critical points where valid)} ]

To express ( 5 \cos x - 12 \sin x ) in the required form:

1. Identify ( A ) and ( \alpha ): [ A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ]

2. Find ( \alpha ) using: [ \tan \alpha = \frac{-12}{5} \implies \alpha = \tan^{-1}\left(-\frac{12}{5}\right) ] (Ensure ( 0 \leq \alpha \leq \frac{\pi}{2} ).

3. The expression is: [ 5 \cos x - 12 \sin x = 13 \cos(x + \alpha) ]

For the integral, apply the substitution:

1. If ( u = 2x - 1 ), then ( du = 2 , dx ), hence ( dx = \frac{du}{2} ).
2. Change limits for ( x = 1 ) to ( u = 1 ) and for ( x = 2 ) to ( u = 3 ): [ \int_{1}^{2} \frac{x}{(2x-1)^3} , dx = \int_{1}^{3} \frac{\frac{u+1}{2}}{u^3} \frac{du}{2} = \frac{1}{4} \int_{1}^{3} \frac{u+1}{u^3} , du ]
3. Simplify the integral: [ = \frac{1}{4} \int_{1}^{3} \left( u^{-2} + u^{-3} \right) , du = \frac{1}{4} \left[ -u^{-1} + \frac{1}{2} u^{-2} \right]_{1}^{3} ]
4. Calculate: [ = \frac{1}{4} \left( -\frac{1}{3} + \frac{1}{2 \cdot 9} + 1 - \frac{1}{2} \right) = \frac{1}{4} \left( -\frac{1}{3} + \frac{1}{18} \right) ]

Since ( P(x) ) is divisible by ( A(x) ), the polynomial value at the root must equal zero:

1. Substitute ( x = 3 ) into ( P(x) = x^3 - kx^2 + 5x + 12 ): [ P(3) = 27 - 9k + 15 + 12 = 0 ]
2. Thus, [ 54 - 9k = 0 \implies k = 6 ]

Now that we have ( k = 6 ), the polynomial is:

1. Substitute: [ P(x) = x^3 - 6x^2 + 5x + 12 ]
2. To find zeros, we can use the rational root theorem and synthetic division or factor it:
3. By factorization or using synthetic division: [ P(x) = (x - 3)(x^2 - 3) = (x - 3)(x - \sqrt{3})(x + \sqrt{3}) ]
4. Hence:
   • Real zeros are: ( x = 3, x = \sqrt{3}, x = -\sqrt{3} ).