9. (a) an aircraft took off from an airstrip with an average speed of 35 km/h on a bearing of 015° for 2 hours. it then changed course and flew on a bearing of 100° with an average speed of 22 km/h for another 2 1/2 hours. find its: i. distance from the starting point to two decimal places ii. bearing from the airstrip to the nearest degree (b) differentiate (1 + x)(1 + 2x^2)/x with respect to x. (12 marks) 10. (a) using a ruler and a pair of compasses only, construct: i. △pqr such that |pq| = 8 cm, rpq = 120° and pqr = 30° ii. the locus l1 of points equidistant from p and q iii. the locus l2 of points equidistant from pq and pr which passes through triangle pqr (b) label the point where l1 and l2 intersect as x and measure |qx|. (12 marks) 11. (a) copy and complete table 2.2 for y = 2x^2 - 4x - 3. table 2.2: table of values x -2 -1 0 1 2 3 4 5 6 y 3 -3 45 (b) using a scale of 2 cm to 1 unit on x - axis and 2 cm to 5 units on y - axis, draw the graph of y = 2x^2 - 4x - 3 for -2 ≤ x ≤ 6. (c) use your graph to find the: i. roots of the equation 2x^2 - 4x - 3 = 4 correct to 1 decimal place ii. gradient of the curve y = 2x^2 - 4x - 3 at x = 4 (12 marks)

Question

Accepted Answer

### 9. (a)
# Explanation:
## Step1: Calculate the distances of each - part of the journey
The distance formula is $d = vt$.
For the first part of the journey: $v_1 = 35$ km/h and $t_1=2$ h, so $d_1=v_1t_1 = 35	imes2=70$ km.
For the second - part of the journey: $v_2 = 22$ km/h and $t_2 = 2.5$ h, so $d_2=v_2t_2=22	imes2.5 = 55$ km.
## Step2: Find the angle between the two paths
The angle between the two paths $	heta=100 - 15=85^{\circ}$.
## Step3: Use the cosine - law to find the distance from the starting point
The cosine - law states that $c^{2}=a^{2}+b^{2}-2ab\cos C$. Let $a = 70$, $b = 55$, and $C = 85^{\circ}$.
$\cos(85^{\circ})\approx0.0872$.
$c^{2}=70^{2}+55^{2}-2	imes70	imes55	imes0.0872$
$c^{2}=4900 + 3025-2	imes70	imes55	imes0.0872$
$c^{2}=4900+3025 - 671.44$
$c^{2}=7253.56$
$c=\sqrt{7253.56}\approx85.17$ km.
## Step4: Use the sine - law to find the bearing
Let $\alpha$ be the angle opposite to the side of length $55$ km in the triangle formed by the two paths and the line from the starting point to the end - point.
By the sine - law, $\frac{\sin\alpha}{55}=\frac{\sin(85^{\circ})}{85.17}$
$\sin\alpha=\frac{55	imes\sin(85^{\circ})}{85.17}$
$\sin\alpha=\frac{55	imes0.9962}{85.17}\approx0.645$
$\alpha=\sin^{-1}(0.645)\approx40^{\circ}$
The bearing from the airstrip is $15^{\circ}+\alpha\approx15 + 40=55^{\circ}$

# Answer:
i. $85.17$ km
ii. $55^{\circ}$

### 9. (b)
# Explanation:
## Step1: Expand the numerator
$(1 + x)(1 + 2x^{2})=1+2x^{2}+x + 2x^{3}$
So, $\frac{(1 + x)(1 + 2x^{2})}{x}=\frac{1+2x^{2}+x + 2x^{3}}{x}=\frac{1}{x}+2x + 1+2x^{2}$
## Step2: Differentiate term - by - term
The derivative of $\frac{1}{x}=x^{-1}$ is $-x^{-2}=-\frac{1}{x^{2}}$ using the power rule $\frac{d}{dx}(x^{n})=nx^{n - 1}$.
The derivative of $2x$ is $2$.
The derivative of the constant $1$ is $0$.
The derivative of $2x^{2}$ is $4x$.
The derivative of $\frac{(1 + x)(1 + 2x^{2})}{x}$ is $-\frac{1}{x^{2}}+2 + 4x$

# Answer:
$-\frac{1}{x^{2}}+2 + 4x$

### 10. (a)
# Explanation:
## Step1: Construct $	riangle PQR$
1. Draw a line segment $PQ = 8$ cm.
2. At point $P$, use a protractor to construct an angle $\angle RPQ=120^{\circ}$.
3. At point $Q$, use a protractor to construct an angle $\angle PQR = 30^{\circ}$. The intersection of the two rays is point $R$.
## Step2: Construct the locus $L_1$
The locus of points equidistant from $P$ and $Q$ is the perpendicular bisector of the line segment $PQ$. Use a compass to construct the perpendicular bisector.
## Step3: Construct the locus $L_2$
The locus of points equidistant from $\overline{PQ}$ and $\overline{PR}$ is the angle - bisector of $\angle QPR$. Use a compass to construct the angle - bisector of $\angle QPR$ which passes through $	riangle PQR$.

### 10. (b)
# Explanation:
Label the intersection of $L_1$ and $L_2$ as $X$. Then, use a ruler to measure the length of $|QX|$.

### 11. (a)
# Explanation:
## Step1: Substitute $x$ values into $y = 2x^{2}-4x - 3$
When $x=-2$:
$y=2	imes(-2)^{2}-4	imes(-2)-3=2	imes4 + 8-3=8 + 8-3 = 13$
When $x = 0$:
$y=2	imes0^{2}-4	imes0-3=-3$
When $x = 1$:
$y=2	imes1^{2}-4	imes1-3=2-4-3=-5$
When $x = 3$:
$y=2	imes3^{2}-4	imes3-3=2	imes9-12 - 3=18-12-3 = 3$
When $x = 4$:
$y=2	imes4^{2}-4	imes4-3=2	imes16-16-3=32-16-3 = 13$
When $x = 5$:
$y=2	imes5^{2}-4	imes5-3=2	imes25-20-3=50-20-3 = 27$

# Answer:
|$x$|$-2$|$-1$|$0$|$1$|$2$|$3$|$4$|$5$|$6$|
|----|----|----|----|----|----|----|----|----|----|
|$y$|$13$|$3$|$-3$|$-5$|$-3$|$3$|$13$|$27$|$45$|

### 11. (b)
# Explanation:
1. Mark the $x$ - values from $-2$ to $6$ on the $x$ - axis with a scale of $2$ cm to $1$ unit.
2. Mark the $y$ - values on the $y$ - axis with a scale of $2$ cm to $5$ units.
3. Plot the points $(-2,13),(-1,3),(0,-3),(1,-5),(2,-3),(3,3),(4,13),(5,27),(6,45)$ and draw a smooth curve through them.

### 11. (c)
# Explanation:
## Step1: Find the roots of $2x^{2}-4x - 3=4$
Rewrite the equation as $2x^{2}-4x - 7 = 0$. On the graph of $y = 2x^{2}-4x - 3$, find the $x$ - values where $y = 4$.
By looking at the graph, the roots are approximately $x\approx - 1.2$ and $x\approx3.2$
## Step2: Find the gradient at $x = 4$
The gradient of the curve $y = 2x^{2}-4x - 3$ at $x = 4$ can be approximated by drawing a tangent to the curve at $x = 4$ and finding the slope of the tangent.
The slope of the tangent at $x = 4$ is approximately $12$ (by drawing the tangent and using the rise - over - run method on the graph).

# Answer:
i. $x\approx - 1.2,x\approx3.2$
ii. $12$

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QUESTION IMAGE

Question

Explanation:

9. (a)

Step1: Calculate the distances of each - part of the journey

Step2: Find the angle between the two paths

Step3: Use the cosine - law to find the distance from the starting point

Step4: Use the sine - law to find the bearing

Step1: Expand the numerator

Step2: Differentiate term - by - term

Step1: Construct $\triangle PQR$

Step2: Construct the locus $L_1$

Step3: Construct the locus $L_2$

10. (b)

Answer:

9. (b)