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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
Step 2
Calculate the size of the acute angle between the lines \( y = 2x + 5 \) and \( y = 4 - 3x \)
Answer
To calculate the angle between the two lines, we first find their slopes:
- Slope of line 1 (( m_1 )): 2
- Slope of line 2 (( m_2 )): -3
Applying the formula for the tangent of the angle between two lines:
[ \tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right| ]
we have:
[ \tan \theta = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{-5} \right| = 1 ]
Hence, ( \theta = \frac{\pi}{4} ) radians or 45 degrees.
Step 3
Solve the inequality \( \frac{4}{x + 3} \geq 1 \)
Answer
First, let's rewrite the inequality:
[ \frac{4}{x + 3} - 1 \geq 0 ]
This simplifies to:
[ \frac{4 - (x + 3)}{x + 3} \geq 0 ]
which can also be expressed as:
[ \frac{1 - x}{x + 3} \geq 0 ]
The critical points occur at:
- ( x = 1 )
- ( x = -3 )
By testing intervals around these points, we find the solution set:
( -3 < x \leq 1 ).
Step 4
Express \( 5 \cos x - 12 \sin x \) in the form \( A \cos(x + \alpha) \) where \( 0 \leq \alpha \leq \frac{\pi}{2} \)
Answer
To express in the required form, we can use the identity:
[ A = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ]
Next, to find ( \alpha ):
[ \tan \alpha = \frac{-12}{5} \quad \Rightarrow \quad \alpha = \tan^{-1}\left(\frac{-12}{5}\right) ]
Thus, we can write:
[ 5 \cos x - 12 \sin x = 13 \cos\left(x + \alpha \right) ].
Step 5
Use the substitution \( u = 2x - 1 \) to evaluate \( \int_{1}^{2} \frac{x}{(2x - 1)^3} \, dx \)
Answer
Substituting ( u = 2x - 1 ) yields: ( x = \frac{u + 1}{2} ).
When ( x = 1 ), then ( u = 1 ) and when ( x = 2 ), ( u = 3 ).
So the integral will change to:
[ \int_{1}^{3} \frac{\frac{u + 1}{2}}{u^3} \cdot \frac{1}{2} du = \frac{1}{4} \int_{1}^{3} \frac{u + 1}{u^3} , du = \frac{1}{4} \int_{1}^{3} \left( \frac{1}{u^2} + \frac{1}{u^3} \right) , du ]
Evaluating the integral gives us the final result.
Step 6
Given that \( P(x) \) is divisible by \( A(x) \), show that \( k = 6 \).
Answer
Since ( A(x) = x - 3 ) is a factor of ( P(x) ), applying the Factor Theorem:
Substituting ( x = 3 ) gives:
[ P(3) = 3^3 - k(3^2) + 5(3) + 12 = 0 ]
Solving this for ( k ):
[ 27 - 9k + 15 + 12 = 0 \Rightarrow 54 - 9k = 0 \Rightarrow k = 6. ]
Step 7