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 - 3x $. (c) Solve the inequality $ \frac{4}{x + 3} \geq 1 $. (d) Express $ 5 \cos x - 12 \sin x $ in the form $ A \cos(x + \alpha) $, where $ 0 \leq \alpha \leq \frac{\pi}{2} $. (e) Use the substitution $ u = 2x - 1 $ to evaluate $ \int_{1}^{2} \frac{x}{(2x - 1)^3} \, dx $. (f) 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 $.

SSCE HSC Mathematics Extension 1 Polynomial functions: 11. Use a SEPARATE writing bookl

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 find the integral ( \int \sin x^2 , dx ), we can use integration by parts or refer to known integral tables. The integral does not have a simple antiderivative and generally requires numerical methods or special functions for evaluation. Thus, we state the integral as ( \int \sin x^2 , dx = F(x) + C ) where ( F(x) ) is the integral function.

Step 2

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

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Answer

The slopes of the lines are ( m_1 = 2 ) and ( m_2 = -3 ). We can find the acute angle ( \theta ) using the tangent formula:

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

Thus, ( \theta = \tan^{-1}(1) = \frac{\pi}{4} \text{ radians or } 45^{\circ}. )

Step 3

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

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Answer

To solve ( \frac{4}{x + 3} \geq 1 ):

Rearranging gives ( 4 \geq x + 3 ).
Thus, ( x \leq 1 ).
We also need to consider the critical point when the denominator is zero, which is ( x = -3 ).
Therefore, the solution 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

We can express ( 5 \cos x - 12 \sin x ) using the formula ( R \cos(x + \alpha) ) where ( R = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ) and ( \tan \alpha = \frac{-12}{5} ). Thus, the expression becomes:

$5 \cos x - 12 \sin x = 13 \cos\left(x + \tan^{-1}(-\frac{12}{5})\right).$

Step 5

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

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Answer

Using the substitution ( u = 2x - 1 ), when ( x = 1 ), ( u = 1 ) and when ( x = 2 ), ( u = 3 ). The integral becomes:

$\int_{1}^{3} \frac{(u + 1)/2}{u^3} \cdot \frac{1}{2} \, du = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^3} \, du.$

This now can be simplified and solved accordingly.

Step 6

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

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Answer

To show that ( k = 6 ), we can substitute ( x = 3 ) into ( P(x) ) since ( A(3) = 0 ):

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

For ( P(3) = 0 ), we set ( 54 - 9k = 0 ) giving ( k = 6 ).

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 \( A \cos(x + \alpha) \)

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

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

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