For the vectors $oldsymbol{u} = oldsymbol{i} - oldsymbol{j}$ and $oldsymbol{v} = 2oldsymbol{i} + oldsymbol{j}$, evaluate each of the following - HSC - SSCE Mathematics Extension 1 - Question 11 - 2022 - Paper 1

To evaluate the dot product $\boldsymbol{u} \bullet \boldsymbol{v}$:

$\boldsymbol{u} \bullet \boldsymbol{v} = (\boldsymbol{i} - \boldsymbol{j}) \bullet (2\boldsymbol{i} + \boldsymbol{j})$

Calculating:

$= (1)(2) + (-1)(1)$

Simplifying gives:

$= 2 - 1 = 1$

Using the substitution $u = x^2 + 4$, we differentiate to find:

$du = 2x \, dx \Rightarrow dx = \frac{du}{2x}$

Changing the limits of integration:

• When $x = 0$, $u = 4$.
• When $x = 1$, $u = 5$.

Now substituting into the integral:

$\int_0^1 \frac{x}{\sqrt{x^2 + 4}} \, dx = \int_4^5 \frac{x}{\sqrt{u}} \cdot \frac{du}{2x} = \int_4^5 \frac{1}{2\sqrt{u}} \, du$

Integrating gives:

$= \frac{1}{2} [2\sqrt{u}]_4^5 = [\sqrt{u}]_4^5 = \sqrt{5} - 2$

Using the binomial theorem:

$(1 - \frac{x}{2})^n = \sum_{k=0}^{n} \binom{n}{k} (-\frac{x}{2})^k$

For $x^2$ (where $k = 2$):

$\text{Coefficient of } x^2 = \binom{8}{2} \left(-\frac{1}{2}\right)^2 = \frac{8 \cdot 7}{2} \cdot \frac{1}{4} = 14$

For $x^3$ (where $k = 3$):

$\text{Coefficient of } x^3 = \binom{8}{3} \left(-\frac{1}{2}\right)^3 = \frac{8 \cdot 7 \cdot 6}{6} \cdot -\frac{1}{8} = -7$

To find if the vectors are perpendicular, we set:

$\boldsymbol{u} \bullet \boldsymbol{v} = 0$

Calculating the dot product:

$\boldsymbol{u} \bullet \boldsymbol{v} = \left(\frac{a}{2}\right)\left(\frac{a - 7}{4a - 1}\right) + \left(\frac{a - 7}{4a - 1}\right)\left(\frac{a}{2}\right) = 0$

This simplifies to:

$\frac{a(a-7)}{(4a-1)} = 0$

Hence, either $a = 0$ or $a = 7$. Testing each, we find:

• $a = -2$ leads to valid solutions where $\boldsymbol{u} \bullet \boldsymbol{v} = 0$ . Therefore, $a = -2$ or $a = 7$.

To express the equation in the desired form, we find $R$ and $\alpha$:

We set:

$R = \sqrt{(\sqrt{3})^2 + (-3)^2} = \sqrt{3 + 9} = \sqrt{12} = 2\sqrt{3}$

Using $\tan(\alpha) = \frac{\text{Coefficient of } \cos}{\text{Coefficient of } \sin} = \frac{-3}{\sqrt{3}} ightarrow \alpha = -\frac{\pi}{3}.$

Thus, we have:

$\sqrt{3}\sin(x) - 3\cos(x) = 2\sqrt{3}\sin\left(x - \frac{\pi}{3}\right)$

Starting from the inequality:

$\frac{x}{2 - x} - 5 \geq 0$

This can be rewritten as:

$\frac{x - 5(2 - x)}{2 - x} \geq 0$

Simplifying gives:

$\frac{6x - 10}{2 - x} \geq 0$

This implies critical values are $x = 5/3$ and $x = 2$. Testing intervals around these critical points gives:

The solution is $x \leq \frac{5}{3}$ or $x > 2$.