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11. Find \( \int \sin x^2 \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1
Question 11
11. Find \( \int \sin x^2 \, dx \). (b) Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x \). (c) Solve the inequality \( ... show full transcript
Worked Solution & Example Answer:11. Find \( \int \sin x^2 \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1
Step 1
Find \( \int \sin x^2 \, dx \)
Answer
To solve this integral, we will use a substitution method. Start by noting that ( \sin x^2 ) does not have a standard antiderivative; hence, we can either look for a series expansion or numerical methods to approximate the integral. The integral can be computed using special functions if necessary, or numerically for practical applications.
Step 2
Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x \)
Answer
First, find the slopes of both lines: ( 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| ] Plugging in the values: [ \tan \theta = \left| \frac{2 - (-3)}{1 + 2(-3)} \right| = \left| \frac{5}{-5} \right| = 1 ] Thus, ( \theta = \frac{\pi}{4} ) radians or ( 45^{\circ} ).
Step 3
Solve the inequality \( \frac{4}{x + 3} \geq 1 \)
Answer
To solve the inequality:
- Start by isolating the fraction: [ \frac{4}{x + 3} - 1 \geq 0 ]
- Rewrite it as: [ \frac{4 - (x + 3)}{x + 3} \geq 0 ] This simplifies to: [ \frac{1 - x}{x + 3} \geq 0 ]
- Determine the critical points from ( 1 - x = 0 ) and ( x + 3 = 0 ), yielding ( x = 1 ) and ( x = -3 ).
- Test intervals for the sign of the inequality. The solution is ( -3 < x \leq 1 ).
Step 4
Express \( 5 \cos x - 12 \sin x \) in the form \( A \cos (x + \alpha) \)
Answer
To express in the form ( A \cos (x + \alpha) ):
- Recognize that: [ A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ].
- Find ( \alpha ): [ \tan \alpha = \frac{-12}{5} \rightarrow \alpha = \tan^{-1} \left(-\frac{12}{5}\right) ]. Thus, the expression can be rewritten as ( 13 \cos (x + \alpha) ).
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 ), we have:
- Find differential: ( du = 2 , dx ) or ( dx = \frac{du}{2} ).
- Change limits: when ( x = 1 ), ( u = 1 ) and when ( x = 2 ), ( u = 3 ).
- Substitute into the integral: [ \int_{1}^{3} \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 through partial fractions.
Step 6
Given that \( P(x) \) is divisible by \( A(x) \), show that \( k = 6 \)
Answer
To show this, apply polynomial long division or substitutive evaluation: Substituting ( x = 3 ) into ( P(x) ),
- ( P(3) = 3^3 - k(3^2) + 5(3) + 12 = 27 - 9k + 15 + 12 = 0 )
- This yields: ( 54 - 9k = 0 ) leading to ( k = 6 ) as the condition for divisibility.
Step 7
Find all the zeros of \( P(x) \) when \( k = 6 \)
Answer
Substituting ( k = 6 ) into ( P(x) ):
- ( P(x) = x^3 - 6x^2 + 5x + 12 = 0 ).
- Employ synthetic division or the Rational Root Theorem to identify potential zeros.
- After testing, you find the roots: ( x = 3, x = -2, and x = 2 ) by solving ( (x - 3)(x + 2)(x - 2) = 0 ).