1.1 The picture below shows the curved flight path of an aircraft - NSC Technical Mathematics - Question 1 - 2019 - Paper 1

Setting $p(x)$ to zero: $2 \left( x - \frac{2}{9} \right) \left( x + \frac{2}{9} \right) = 0$ This gives us two equations to solve:

1. $x - \frac{2}{9} = 0 \Rightarrow x = \frac{2}{9}$
2. $x + \frac{2}{9} = 0 \Rightarrow x = -\frac{2}{9}$

Rearranging the equation gives us: $(3x-5)(x+2) + 13 = 0$ Expanding: $3x^2 + 6x - 5x - 10 + 13 = 0$ Which simplifies to: $3x^2 + x + 3 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: Here, $a=3$, $b=1$, $c=3$: $x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3}\Rightarrow x = \frac{-1 \pm \sqrt{1 - 36}}{6}\Rightarrow x = \frac{-1 \pm \sqrt{-35}}{6}$ Final results can be expressed as: $x = \frac{-1}{6} \pm \frac{i \sqrt{35}}{6}$

To solve the inequality, we first find the critical points:

1. $4-x=0 \Rightarrow x=4$
2. $x+3=0 \Rightarrow x=-3$ These points divide the number line into intervals: $(-\infty, -3)$, $(-3, 4)$, and $(4, \infty)$. We test each interval:

• For $x < -3$, choose $x=-4$, $(4-(-4))(-4+3)=8(-1) < 0$ (valid)
• For $-3 < x < 4$, choose $x=0$, $(4-0)(0+3)=4(3) > 0$ (not valid)
• For $x > 4$, choose $x=5$, $(4-5)(5+3)=-1(8) < 0$ (valid) Thus, the solution set is: $x \in (-\infty, -3) \cup (4, \infty)$

Substituting $y$ into the second equation: $x^2 - x(3x - 8) + (3x - 8)^2 = 39$ Expanding: $x^2 - (3x^2 - 8x) + (9x^2 - 48x + 64) = 39$ Combining terms leads to: $7x^2 - 40x + 25 - 39 = 0$ This simplifies to: $7x^2 - 40x - 14 = 0$ Using the quadratic formula: $x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4 \cdot 7 \cdot (-14)}}{2 \cdot 7}\Rightarrow x = \frac{40 \pm \sqrt{1600 + 392}}{14}\Rightarrow x = \frac{40 \pm \sqrt{1992}}{14}$ The value of $y$ can be calculated by substituting $x$ back into $y=3x-8$.

Starting with the equation: $V = I \cdot Z$ Dividing both sides by $Z$: $I = \frac{V}{Z}$

Using the earlier derived formula: $I = \frac{V}{Z} = \frac{7i}{3 - i}$ To simplify, multiply the numerator and denominator by the complex conjugate of the denominator: $I = \frac{(7i)(3+i)}{(3-i)(3+i)} = \frac{21i + 7i^2}{9 + 1} = \frac{21i - 7}{10} = -\frac{7}{10} + \frac{21}{10}i$

To simplify the mathematical expression:

1. Start with $101, x_{11}$ which is equivalent to $1 + 0 + 1 = 2$ (by summing individual digits).
2. Then, calculate $5 \times 3 = 15. Thus, combined we get:$2 + 15 = 17.$