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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

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

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Answer

To solve this integral, we use the substitution: let ( u = x^2 ). Then, ( du = 2x , dx ), thus ( dx = \frac{du}{2x} = \frac{du}{2\sqrt{u}} ). Therefore, the integral becomes:

$\int \sin(u) \frac{du}{2\sqrt{u}}$ The exact form depends on further integration techniques or tables.

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:

For ( y = 2x + 5 ), the slope ( m_1 = 2 ).
For ( y = 4 - 3x ), the slope ( m_2 = -3 ).

The angle ( \theta ) between the lines can be calculated using:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| = \left| \frac{2 - (-3)}{1 + 2(-3)} \right| = \left| \frac{5}{-5} \right| = 1$

Thus, ( \theta = \tan^{-1}(1) = \frac{\pi}{4} ) radians.

Step 3

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

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Answer

To solve this inequality, first rewrite it as:

$\frac{4}{x+3} - 1 \geq 0 \Rightarrow \frac{4 - (x + 3)}{x + 3} \geq 0 \Rightarrow \frac{1 - x}{x + 3} \geq 0$

Next, find the critical points by setting the numerator and denominator equal to zero:

( 1 - x = 0 ) gives ( x = 1 )
( x + 3 = 0 ) gives ( x = -3 )

Using a number line, test intervals to find the solution where the inequality holds. The answer is ( -3 < x \leq 1 ) as the critical points are included or excluded based on the signs.

Step 4

Express $ 5 \cos x - 12 \sin x $ in the form $ A \cos(x + \alpha) $, where $ 0 \leq \alpha \leq \frac{\pi}{2} $.

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Answer

To express this in the desired form, identify ( A ) and ( \alpha ) using the formula:

$A = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$

Next, determine ( \alpha ) using:

$A \cos \alpha = 5 \quad \text{and} \quad A \sin \alpha = -12.$ Thus, solving gives:

$\cos \alpha = \frac{5}{13} \quad \text{and} \quad \sin \alpha = -\frac{12}{13}.$ We can find ( \alpha ) using the inverse tangent, but adjust based on the quadrant.

Step 5

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

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Answer

With the substitution ( u = 2x - 1 ), we have ( du = 2 , dx ) hence ( dx = \frac{du}{2} ). Now substitute and simplify:

$\int \frac{2}{u^3} \frac{du}{2} = \int \frac{1}{u^3} \, du = -\frac{1}{2u^2} + C.$ Substituting back for ( u ) yields the final answer: ( -\frac{1}{2(2x-1)^2} + C ).

Step 6

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

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Answer

Since ( A(x) = x - 3 ) is a factor of ( P(x) ), we know that ( P(3) = 0 ). Substitute ( x = 3 ):

$P(3) = 3^3 - k(3^2) + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k.$

Setting this equal to zero gives:

$54 - 9k = 0 \Rightarrow 9k = 54 \Rightarrow k = 6.$

Step 7

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

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Answer

With ( k = 6 ), we have:

$P(x) = x^3 - 6x^2 + 5x + 12.$ Using polynomial long division or synthetic division with ( A(x) = x - 3 ), we obtain:

$P(x) = (x - 3)(x^2 - 3x - 4).$ Now, factor the quadratic by finding the roots:

$x^2 - 3x - 4 = 0 \Rightarrow (x - 4)(x + 1) = 0 \Rightarrow x = 4, -1.$ The zeros are therefore: ( x = 3, 4, -1. $$

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

Worked Solution & Example Answer:11. Use a SEPARATE writing booklet - 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 \( A \cos(x + \alpha) \), where \( 0 \leq \alpha \leq \frac{\pi}{2} \).

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

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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