Find $ \int \sin x^2 \, dx $. (a) Calculate the size of the acute angle between the lines $ y = 2x + 5 $ and $ y = 4 - 3x $. (b) Solve the inequality $ \frac{4}{x + 3} \geq 1 $. (c) Express $ 5 \cos x - 12 \sin x $ in the form $ A \cos(x + \alpha) $, where $ 0 \leq \alpha \leq \frac{\pi}{2} $. d) Use the substitution $ u = 2x - 1 $ to evaluate $ \int_{1}^{2} \frac{2}{(2x-1)^3} \, dx $. (e) Consider the polynomials $ P(x) = x^3 - kx^2 + 5x + 12 $ and $ A(x) = x - 3 $. (i) Given that $ P(x) $ is divisible by $ A(x) $, show that $ k = 6 $. (ii) Find all the zeros of $ P(x) $ when $ k = 6 $.

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Find $ \int \sin x^2 \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

Question 11

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Find $ \int \sin x^2 \, dx $. (a) Calculate the size of the acute angle between the lines $ y = 2x + 5 $ and $ y = 4 - 3x $. (b) Solve the inequality \( \fra... show full transcript

Worked Solution & Example Answer:Find $ \int \sin x^2 \, dx $ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

Step 1

Find $ \int \sin x^2 \, dx $

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Answer

The integral ( \int \sin x^2 , dx ) can be approached using integration by substitution or numerical methods, as it does not have a standard elementary form.

Step 2

Calculate the size of the acute angle between the lines $ y = 2x + 5 $ and $ y = 4 - 3x $

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Answer

First, identify the slopes of the lines. The slope of the first line is ( m_1 = 2 ) and the second line is ( m_2 = -3 ). We use the formula for the tangent of the angle ( \theta ) between two lines: [ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|. ] Substituting the values: [ \tan \theta = \left| \frac{2 - (-3)}{1 + 2(-3)} \right| =\left| \frac{5}{-5} \right| = 1. ] Thus, ( \theta = 45^{\circ} ).

Step 3

Solve the inequality $ \frac{4}{x + 3} \geq 1 $

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Answer

First, rearranging gives us: [ 4 \geq x + 3 ]
This simplifies to:
[ x \leq 1 ]
Next, consider the domain where ( x + 3 > 0 ), which gives ( x > -3 ).
Thus, the solution set is: [ -3 < x \leq 1. ]

Step 4

Express $ 5 \cos x - 12 \sin x $ in the form $ A \cos(x + \alpha) $

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Answer

To express this in the desired form, we first identify ( A = \sqrt{5^2 + (-12)^2} = 13 ) and use the identity: [ A \cos(x + \alpha) = A \left( \cos x \cos \alpha - \sin x \sin \alpha \right). ] We identify ( \cos \alpha = \frac{5}{13} ) and ( \sin \alpha = \frac{-12}{13} ).
Thus, ( \alpha = \tan^{-1}\left(-\frac{12}{5}\right) ) within the specified range.

Step 5

Use the substitution $ u = 2x - 1 $ to evaluate $ \int_{1}^{2} \frac{2}{(2x-1)^3} \, dx $

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Answer

Using the substitution ( u = 2x - 1 ), we have ( du = 2 , dx ) or ( dx = \frac{du}{2} ).
The limits change as follows:

When ( x = 1 ), ( u = 1).
When ( x = 2 ), ( u = 3). Thus, the integral transforms to: [ \int_{1}^{3} \frac{2}{u^3} \cdot \frac{du}{2} = \int_{1}^{3} \frac{1}{u^3} , du = \left[-\frac{1}{2u^2}\right]_{1}^{3}. ] Evaluating gives: [ -\frac{1}{2(3^2)} + \frac{1}{2(1^2)} = -\frac{1}{18} + \frac{1}{2} = \frac{8}{18} = \frac{4}{9}. ]

Step 6

Given that $ P(x) $ is divisible by $ A(x) $, show that $ k = 6 $

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Answer

Since ( A(3) = 0 ), we calculate ( P(3) ): [ P(3) = 3^3 - k(3^2) + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k. ] For ( P(3) = 0 ): [ 54 - 9k = 0 ]
This implies ( 9k = 54 ) so ( k = 6. ]

Step 7

Find all the zeros of $ P(x) $ when $ k = 6 $

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Answer

Substituting ( k = 6 ), we have: [ P(x) = x^3 - 6x^2 + 5x + 12. ] Using synthetic division for ( A(x) = x - 3 ) confirms that 3 is a root. Now factor ( P(x) ) as: [ P(x) = (x - 3)(x^2 - 3x - 4). ] Solving ( x^2 - 3x - 4 = 0 ) gives: [ x = \frac{3 \pm \sqrt{(-3)^2 - 4(1)(-4)}}{2(1)} = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}. ] Thus, the roots are ( x = 4 ) and ( x = -1 ). So, the zeros are ( x = 3, 4, -1. ]

Find \( \int \sin x^2 \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

Worked Solution & Example Answer:Find \( \int \sin x^2 \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

Find \( \int \sin x^2 \, dx \)

Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x \)

Solve the inequality \( \frac{4}{x + 3} \geq 1 \)

Express \( 5 \cos x - 12 \sin x \) in the form \( A \cos(x + \alpha) \)

Use the substitution \( u = 2x - 1 \) to evaluate \( \int_{1}^{2} \frac{2}{(2x-1)^3} \, dx \)

Given that \( P(x) \) is divisible by \( A(x) \), show that \( k = 6 \)

Find all the zeros of \( P(x) \) when \( k = 6 \)

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