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11. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1
Question 11
11. Use a SEPARATE writing booklet. (a) Find \( \int \sin x^2 \, dx. \) (b) Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 -... show full transcript
Worked Solution & Example Answer:11. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1
Step 1
Step 2
Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x. \)
Answer
First, we find the gradients (slopes) of the given lines. For ( y = 2x + 5 ), the gradient is ( m_1 = 2 ). For ( y = 4 - 3x ), the gradient is ( m_2 = -3 ). The angle ( \theta ) between two lines can be calculated using the formula:
[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ]
Substituting in the values: [ \tan \theta = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{-5} \right| = 1 ]
Thus, ( \theta = 45^\circ ) (or ( \frac{\pi}{4} ) radians).
Step 3
Solve the inequality \( \frac{4}{x + 3} \geq 1. \)
Answer
To solve ( \frac{4}{x + 3} \geq 1 ), we first rewrite the inequality:
[ 4 \geq x + 3 \Rightarrow x \leq 1. ]
Additionally, we note that the expression is not defined at ( x = -3 ). Thus, the solution set is ( x \leq 1 ) and ( x \neq -3 ).
Step 4
Express \( 5 \cos x - 12 \sin x \) in the form \( A \cos(x + \alpha) \), where \( 0 \leq \alpha \leq \frac{\pi}{2}. \)
Answer
We can express this in the form ( A \cos(x + \alpha) ) using the identities of sine and cosine:
[ A = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ]
Next, we find ( \tan \alpha = \frac{-12}{5} ). Thus, ( \alpha = \tan^{-1}(-\frac{12}{5}) ).
Step 5
Use the substitution \( u = 2x - 1 \) to evaluate \( \int_{1}^{2} \frac{x}{(2x - 1)^2} \, dx. \)
Answer
With the substitution ( u = 2x - 1 ), it follows that ( du = 2dx ) or ( dx = \frac{du}{2} ). The bounds change as follows: when ( x = 1 ), ( u = 1 ) and when ( x = 2 ), ( u = 3 ). Thus, the integral becomes:
[ \int_{1}^{3} \frac{\frac{u + 1}{2}}{u^2} \frac{du}{2} = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^2} du. ]
This can be simplified and evaluated step by step, leading to the final result.
Step 6
Given that \( P(x) \) is divisible by \( A(x) \), show that \( k = 6. \)
Answer
To show that ( k = 6 ), we substitute ( x = 3 ) into ( P(x) ). Since ( P(3) = 0 ) due to the divisibility:
[ P(3) = 3^3 - k(3^2) + 5(3) + 12 = 0 ]
This simplifies to: [ 27 - 9k + 15 + 12 = 0 \Rightarrow -9k + 54 = 0 \Rightarrow k = 6. ]
Step 7
Find all the zeros of \( P(x) \) when \( k = 6. \)
Answer
Substituting ( k = 6 ) gives: [ P(x) = x^3 - 6x^2 + 5x + 12. ]
Using synthetic division or the Rational Root Theorem, we find that ( x = 3 ) is a root. Therefore, we can factor ( P(x) ) as: [ P(x) = (x - 3)(x^2 - 3x - 4). ]
Solving ( x^2 - 3x - 4 = 0 ) yields: [ x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}, ]
resulting in the zeros ( x = 4 ) and ( x = -1. ) Thus, the complete roots are ( x = 3, 4, -1. )