Find $$\int \sin x^2 \, dx$$ - HSC - SSCE Mathematics Extension 1 - Question 11 - 2015 - Paper 1

To find the acute angle ( \theta ) between the two lines, we first find their slopes. The slope of the first line is ( m_1 = 2 ) and of the second line is ( m_2 = -3 ). The formula for the angle ( \theta ) between two lines is given by:

$\tan \theta = \frac{|m_1 - m_2|}{1 + m_1m_2} = \frac{|2 - (-3)|}{1 + 2(-3)}$

Calculating this, we find:

$\tan \theta = \frac{5}{1 - 6} = -5, \quad \Rightarrow \theta \approx \tan^{-1}(-5) \text{ (considering the positive angle)} \approx 1.3734 \text{ radians}$.

To solve the inequality, we first rewrite it:

$\frac{4}{x + 3} - 1 \geq 0 \Rightarrow \frac{4 - (x + 3)}{x + 3} \geq 0$.

This simplifies to:

$\frac{1 - x}{x + 3} \geq 0$.

Finding the critical points, we set the numerator and denominator to zero, giving us (x = 1) and (x = -3). Testing intervals around these points will show where the inequality holds.

To express in the required form, use the identities:

$A = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = 13$,

and ( \tan \alpha = \frac{-12}{5} \Rightarrow \alpha = \tan^{-1}\left(-\frac{12}{5}\right). $$.

Thus, the expression becomes:

$5\cos x - 12\sin x = 13 \cos(x + \alpha).$

With the substitution, ( u = 2x - 1 \Rightarrow du = 2 , dx \Rightarrow dx = \frac{du}{2} ). The limits change accordingly: when ( x = 1 \Rightarrow u = 1 ); when ( x = 2 \Rightarrow u = 3 ). Then the integral becomes:

$$ \int_{1}^{3} \frac{2}{u^3} \cdot \frac{du}{2} = \int_{1}^{3} \frac{1}{u^3} , du = \left[ -\frac{1}{2u^2} \right]_{1}^{3} = -\frac{1}{2(3^2)} + \frac{1}{2(1^2)} = \frac{1}{2} - \frac{1}{18} = \frac{9}{18} - \frac{1}{18} = \frac{8}{18} = \frac{4}{9}.$

For divisibility, ( P(x) = (x - 3)(x^2 + bx + c) ). Expanding and matching coefficients with ( P(x) = x^3 - kx^2 + 5x + 12 ), we find:

[\Rightarrow -k = -3 + b] [\Rightarrow 5 = -3b + c] [\Rightarrow 12 = -3c]. By solving the equations, we can find that ( k = 6 ).

With ( k = 6), the polynomial becomes ( P(x) = x^3 - 6x^2 + 5x + 12). We can use synthetic division or the Rational Root Theorem to find other roots. Once we find one root, we can factor and find remaining zeros.