a) Express $\frac{3 - i}{2 + i}$ in the form $x + iy$, where $x$ and $y$ are real numbers - HSC - SSCE Mathematics Extension 2 - Question 11 - 2022 - Paper 1

To evaluate the integral $\sin^2 2x \cos 2x \, dx$, we can use the substitution method. Let:

Thus:

To write the complex number $-\sqrt{3} + i$ in exponential form, we find the modulus and argument:

The modulus is: $r = | -\sqrt{3} + i | = \sqrt{(-\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = 2.$

The argument is: $\theta = \tan^{-1}\left(\frac{1}{-\sqrt{3}}\right) = \frac{5\pi}{6}.$

Thus, the exponential form is: $2 \text{e}^{i \frac{5\pi}{6}}.$

Using the exponential form found previously: $(-\sqrt{3} + i)^{10} = (2 \text{e}^{i \frac{5\pi}{6}})^{10} = 2^{10} \text{e}^{i \frac{25\pi}{3}}$

We must reduce the argument: $\frac{25\pi}{3} = 8\pi + \frac{1\pi}{3} = \frac{\pi}{3}.$

Therefore: $(-\sqrt{3} + i)^{10} = 1024 \text{e}^{i \frac{\pi}{3}} = 1024 \left(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}\right) = 1024 \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) = 512 + 512i.$

Let ( \vec{a} = B - A = (0, 2, -1) - (1, -1, 2) = (-1, 3, -3) ) and ( \vec{b} = C - B = (2, 1, 1) - (0, 2, -1) = (2, -1, 2) ).

Calculating the dot product: $\vec{a} \cdot \vec{b} = (-1)(2) + (3)(-1) + (-3)(2) = -2 - 3 - 6 = -11.$

Finding magnitudes: $|\vec{a}| = \sqrt{(-1)^2 + 3^2 + (-3)^2} = \sqrt{1 + 9 + 9} = \sqrt{19}$ $|\vec{b}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = 3.$

Using the cosine rule: $\cos \angle ABC = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-11}{\sqrt{19} \cdot 3}.$

Thus: $\angle ABC = \cos^{-1}\left(\frac{-11}{3\sqrt{19}}\right).$

Since line $\ell_2$ is parallel to $\ell_1$ and passes through point $A(-6, 5)$, we can use the slope from $\ell_1$ to find the equation of $\ell_2$. The direction vector of $\ell_1$ is $(3, 2)$, giving a slope: $m = \frac{2}{3}.$

Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 5 = \frac{2}{3}(x + 6).$

This simplifies to: $y - 5 = \frac{2}{3}x + 4 \Rightarrow y = \frac{2}{3}x + 9.$

Using the Weierstrass substitution: $dx = \frac{2 \, dt}{1 + t^2},$ $\cos x = \frac{1 - t^2}{1 + t^2}, \, \sin x = \frac{2t}{1 + t^2}.$

Substituting in the integral: $\int \frac{dx}{1 + \cos x - \sin x} = \int \frac{\frac{2}{1+t^2} dt}{1 + \frac{1 - t^2}{1 + t^2} - \frac{2t}{1 + t^2}}$

This simplifies to: $\int \frac{2 \, dt}{3 + t^2 - 2t}.$

To find anti-derivatives, you can complete the square, yielding the final result.