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Find \( \int \sin x^2 \, dx \) - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1
Question 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 \( \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 \( \int \sin x^2 \, dx \)
Answer
To compute this integral, we recognize that there isn't a standard elementary form for ( \int , ext{sin} , x^2 , dx ). We may express it in terms of special functions or use numerical methods. A common approach involves using a substitution or series expansion, but for exact values, employing numerical integration techniques would yield the best results.
Step 2
Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x \)
Answer
First, we find the slopes of the lines.
For the line ( y = 2x + 5 ), the slope ( m_1 = 2 ).
For the line ( y = 4 - 3x ), the slope ( m_2 = -3 ).
To find the acute angle ( \theta ) between the two lines, we use the formula:
[
\tan(\theta) = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right|
]
Substituting in the values:
[
\tan(\theta) = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{1 - 6} \right| = \left| \frac{5}{-5} \right| = 1
]
Thus, ( \theta = \frac{\pi}{4} ) or 45 degrees.
Step 3
Solve the inequality \( \frac{4}{x + 3} \geq 1 \)
Answer
To solve the inequality, we start by manipulating it.
[\frac{4}{x + 3} - 1 \geq 0]
This simplifies to:
[\frac{4 - (x + 3)}{x + 3} \geq 0]
So,
[\frac{1 - x}{x + 3} \geq 0]
The critical values are ( x = 1 ) and ( x = -3 ). We analyze the sign of the expression in the intervals determined by these points. It is positive when ( x \leq 1 ) and ( x > -3 ). Hence, the solution set is: ( -3 < x \leq 1 ).
Step 4
Express \( 5 \, ext{cos} \, x - 12 \, ext{sin} \, x \) in the form \( A \, ext{cos}(x + \alpha) \)
Answer
To express it in that form, we use the identity: [ A \cos(x + \alpha) = A \cos x \cos \alpha - A \sin x \sin \alpha ] Letting ( A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13 ), we find ( \cos \alpha = \frac{5}{13} ) and ( \sin \alpha = \frac{-12}{13} ). Thus, the angle ( \alpha) can be determined, maintaining that ( A ext{ and } \alpha ext{ meet the given conditions.} )
Step 5
Use the substitution \( u = 2x - 1 \) to evaluate \( \int \frac{2}{(2x - 1)^3} \, dx \)
Answer
Applying the substitution ( u = 2x - 1 ), hence ( du = 2 , dx ) or ( dx = \frac{du}{2} ). Thus, the integral becomes: [\int \frac{2}{u^3} \cdot \frac{du}{2} = \int \frac{1}{u^3} , du = -\frac{1}{2u^2} + C = -\frac{1}{2(2x - 1)^2} + C]
Step 6
Given that \( P(x) \) is divisible by \( A(x) \), show that \( k = 6 \)
Answer
Since ( P(x) ) is divisible by ( A(x) = x - 3 ), we set ( P(3) = 0 ). Thus, [ P(3) = 3^3 - k(3)^2 + 5(3) + 12 = 0 ] Calculating gives: [ 27 - 9k + 15 + 12 = 0 ] Simplifying yields: [ 54 - 9k = 0 \Rightarrow k = 6. ]
Step 7
Find all the zeros of \( P(x) \) when \( k = 6 \)
Answer
Substituting ( k = 6 ), we have: [ P(x) = x^3 - 6x^2 + 5x + 12 ] By trying possible rational roots (via the Rational Root Theorem), we check: ( P(-3) = 0 ). Performing synthetic division, we get: [ P(x) = (x + 3)(x^2 - 9) ] Factoring out gives: ( P(x) = (x + 3)(x - 3)(x + 3) ), yielding the zeros: ( x = -3, 3 ).