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

The slopes of the lines are ( m_1 = 2 ) and ( 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 values, [ \tan \theta = \left| \frac{2 - (-3)}{1 + (2)(-3)} \right| = \left| \frac{5}{-5} \right| = 1 ] Thus, ( \theta = 45^{\circ} ) (or ( \frac{\pi}{4} ) radians).

To solve the inequality: [ \frac{4}{x + 3} \geq 1 ] Multiply both sides by ( x + 3 ) (note the sign): [ 4 \geq x + 3 ] Rearranging gives: [ x \leq 1 ] However, we also have to ensure that ( x + 3 > 0 ) or ( x > -3 ). Thus the solution is: [ -3 < x \leq 1 ]

Using the identity for the cosine of a sum: [ A \cos(x + \alpha) = A (\cos x \cos \alpha - \sin x \sin \alpha) ] This gives: [ A \cos \alpha = 5, , A \sin \alpha = -12 ] To find ( A ): [ A = \sqrt{(5)^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13 ] Then, we can find ( \alpha ) using: [ \cos \alpha = \frac{5}{13}, , \sin \alpha = \frac{-12}{13} ] Thus: [ 5 \cos x - 12 \sin x = 13 \cos(x + \alpha) ] where ( \alpha = \tan^{-1}\left( \frac{-12}{5} \right) ).

Using the substitution ( u = 2x - 1 ), we have ( du = 2 , dx ) or ( dx = \frac{du}{2} ). 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. ]

Since ( A(x) = x - 3 ) is a factor of ( P(x) ), it must hold that ( P(3) = 0 ). Calculating: [ P(3) = 3^3 - k(3^2) + 5(3) + 12 = 0 ]
This simplifies to: [ 27 - 9k + 15 + 12 = 0 ]
Solving gives: [ 54 - 9k = 0 \Rightarrow k = 6. ]

The polynomial becomes: [ P(x) = x^3 - 6x^2 + 5x + 12. ] To find the zeros, we can use the Rational Root Theorem or synthetic division. After determining roots, we can factor the polynomial and solve for the remaining zeros. Assuming we find a root at ( x = 3 ), we factor it and solve the resulting quadratic to find other zeros.