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 $$ \int_{1}^{2} \frac{\sqrt{2}}{(2x - 1)^2} \, dx $$. Consider the polynomials $$ P(x) = x^3 - kx^2 + 5x + 12 $$ and $$ A(x) = x - 3 $$. 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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Find $$ \int \sin x^2 \, dx $$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

Question 11

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Find $$ \int \sin x^2 \, dx $$. Calculate the size of the acute angle between the lines $$ y = 2x + 5 $$ and $$ y = 4 - 3x $$. Solve the inequality $$ \frac{4}{x ... 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

To find the integral, we use integration by parts or a suitable substitution. Using the relationship $\int \sin x^2 \, dx$ does not have a simple elementary form, but can be evaluated using numerical methods or special functions.

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, determine the slopes of the lines: The slope of the first line $m_1 = 2$ and for the second line, reorganizing gives $m_2 = 3$ .

The formula for the angle $\theta$ between two lines is given by:

$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

Substituting the slopes:

$\tan \theta = \left| \frac{2 - (-3)}{1 + 2 \cdot (-3)} \right| = \left| \frac{5}{1 - 6} \right| = \frac{5}{5} = 1$ .

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

Step 3

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

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Answer

To solve the inequality:

Rearrange it: $\frac{4}{x + 3} - 1 \geq 0$
Combine the terms: $\frac{4 - (x + 3)}{x + 3} \geq 0$ $\frac{1 - x}{x + 3} \geq 0$ .
Find the critical points: $x = 1$ and $x = -3$ .
Test the intervals divided by these points:
- For $x < -3$ , the result is positive.
- For $-3 < x < 1$ , the result is negative.
- For $x > 1$ , the result is positive.

Thus, the solution set is $x \leq -3$ or $x \geq 1$ .

Step 4

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

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Answer

To express $5 \cos x - 12 \sin x$ in the required form, we identify:

Calculate $A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
Determine $\tan \alpha = \frac{-12}{5} \Rightarrow \alpha = \tan^{-1}\left(-\frac{12}{5}\right)$ .
Hence, we have: $5 \cos x - 12 \sin x = 13 \cos(x + \alpha)$ .

Step 5

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

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Answer

Applying the substitution:

If $u = 2x - 1$ , then $du = 2 \, dx$ or $dx = \frac{du}{2}$ .
The limits of integration change as follows:
- When $x = 1$ , $u = 1$ .
- When $x = 2$ , $u = 3$ .
Substitute and evaluate: $\int_{1}^{3} \frac{\sqrt{2}}{u^2} \cdot \frac{du}{2} = \frac{\sqrt{2}}{2} \int_{1}^{3} u^{-2} \, du = \frac{\sqrt{2}}{2} [-u^{-1}]_{1}^{3}$ $= \frac{\sqrt{2}}{2} \left( -\frac{1}{3} + 1 \right) = \frac{\sqrt{2}}{2} \left( \frac{2}{3} \right) = \frac{\sqrt{2}}{3}$ .

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 $P(x)$ , substituting $x = 3$ into the polynomial should yield 0:

$P(3) = (3)^3 - k(3)^2 + 5(3) + 12 = 27 - 9k + 15 + 12 = 54 - 9k = 0$ .

This gives us:

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

Step 7

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

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Answer

With $k = 6$ , the polynomial becomes:

$P(x) = x^3 - 6x^2 + 5x + 12$ .

Using the Rational Root Theorem and testing possible roots, we can find:

Testing $x = -3$ : $P(-3) = (-3)^3 - 6(-3)^2 + 5(-3) + 12 = -27 - 54 - 15 + 12 = -84$ (not a root).
Testing $x = 1$ : $P(1) = 1 - 6 + 5 + 12 = 12$ (not a root).
Testing $x = 2$ : $P(2) = 8 - 24 + 10 + 12 = 6$ (not a root).
Testing $x = 3$ : $P(3) = 27 - 54 + 15 + 12 = 0$ (a root).

Factoring out $(x - 3)$ gives:

$P(x) = (x - 3)(x^2 - 3x - 4)$ .

Factoring the quadratic, we find:

$x^2 - 3x - 4 = (x - 4)(x + 1)$ .

Thus, the zeros of $P(x)$ 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 $$ \int_{1}^{2} \frac{\sqrt{2}}{(2x - 1)^2} \, dx $$

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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