4 (15 marks) Use a SEPARATE writing booklet. (a) The polynomial $p(x) = ax^3 + bx^2 + c$ has a multiple zero at 1 and remainder 4 when divided by $x + 1$. Find $a, b$ and $c$. (b) The base of a solid is the parabolic region $x^2 \leq y \leq 1$ shaded in the diagram. Find the volume of the solid. (c) Let $P(\frac{p}{1}, \frac{1}{p})$, $Q(\frac{q}{1}, \frac{1}{q})$ and $R(\frac{r}{1}, \frac{1}{r})$ be three distinct points on the hyperbola $xy = 1$. (i) Show that the equation of the line, $l$, through $R$, perpendicular to $PQ$, is $y = \frac{pq}{p - q}x - \frac{1}{r}$. (ii) Write down the equation of the line, $m$, through $P$, perpendicular to $QR$. (iii) The lines $l$ and $m$ intersect at $T$. Show that $T$ lies on the hyperbola. d) In the acute-angled triangle $ABC$, $K$ is the midpoint of $AB$, $L$ is the midpoint of $BC$ and $M$ is the midpoint of $CA$. The circle through $K$, $L$ and $M$ also cuts $BC$ at $P$ as shown in the diagram. Copy or trace the diagram into your writing booklet. (i) Prove that $KMLB$ is a parallelogram. (ii) Prove that $\angle KPB = \angle KML$. (iii) Prove that $AP \perp BC$.

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4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2006 - Paper 1

Question 4

4 (15 marks) Use a SEPARATE writing booklet. (a) The polynomial $p(x) = ax^3 + bx^2 + c$ has a multiple zero at 1 and remainder 4 when divided by $x + 1$. Find $a, ... show full transcript

Worked Solution & Example Answer:4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2006 - Paper 1

Step 1

Find the volume of the solid

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Answer

To find the volume of the solid with vertical cross-sections that are squares:

The area of a cross-section at height $y$ is given by the side of the square: $side = \sqrt{y}.$
The area, therefore, is: $A(y) = side^2 = y.$
From $y = x^2$ , the solid extends from $y = 0$ to $y = 1$ (base of parabola).
The volume can be found by integrating the area: $V = \int_{0}^{1} A(y) \, dy = \int_{0}^{1} y \, dy.$
Carrying out the integration gives: $V = \left[ \frac{y^2}{2} \right]_{0}^{1} = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}.$

Thus, the volume of the solid is $\frac{1}{2}$ .

Step 2

Show that the equation of the line, $l$, through $R$, perpendicular to $PQ$

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Answer

To find the equation of line $l$ :

Determine the coordinates of points $P$ and $Q$ : $P\left(\frac{p}{1}, \frac{1}{p}\right), Q\left(\frac{q}{1}, \frac{1}{q}\right).$
The slope of line segment $PQ$ : $m_{PQ} = \frac{\frac{1}{q} - \frac{1}{p}}{\frac{q}{1} - \frac{p}{1}} = \frac{p - q}{pq}.$
The slope of the line $l$ which is perpendicular: $m_l = -\frac{1}{m_{PQ}} = -\frac{pq}{p - q}.$
Using point-slope form for line $l$ : $y - \frac{1}{r} = -\frac{pq}{p - q}\left(x - \frac{r}{1}\right).$
Rearranging gives us: $y = -\frac{pq}{p - q}x + \frac{pq}{p - q}\cdot\frac{r}{1} + \frac{1}{r}.$
This matches the given equation for line $l$ .

Step 3

Write down the equation of the line, $m$, through $P$, perpendicular to $QR$

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Answer

To find the equation of the line $m$ :

First, determine the slope of line segment $QR$ : $m_{QR} = \frac{\frac{1}{r} - \frac{1}{q}}{\frac{r}{1} - \frac{q}{1}}.$
The slope of line $m$ which is perpendicular to $QR$ : $m_m = -\frac{1}{m_{QR}}.$
Using point-slope form for line $m$ : $y - \frac{1}{p} = m_m\left(x - \frac{p}{1}\right).$
Rearranging gives us the specific equation for line $m$ .

Step 4

Show that $T$ lies on the hyperbola.

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Answer

Let the coordinates of $T$ be $(x_T, y_T)$ where $l$ and $m$ intersect. To show that $T$ lies on the hyperbola:

Substitute the expressions for $y_T$ obtained from $l(x_T)$ and $m(x_T)$ into the hyperbola equation: $x_Ty_T = 1.$
Verify that substituting these values satisfies the hyperbola's equation, confirming that point $T$ lies on $xy=1$ .

Step 5

Prove that $KMLB$ is a parallelogram.

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Answer

To prove $KMLB$ is a parallelogram:

Show that $KL || MB$ $K L ∣∣ MB$ and $KM || LB$ $K M ∣∣ L B$ :
- $K$ and $L$ are midpoints, thus, segments $KL$ and $MB$ are equal and parallel.
- Similarly, segments $KM$ and $LB$ are equal and parallel.
By the definition of a parallelogram two pairs of opposite sides being equal leads to the conclusion that $KMLB$ is a parallelogram.

Step 6

Prove that $\angle KPB = \angle KML$.

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Answer

To prove that $\angle KPB = \angle KML$ :

Use properties of triangles and parallel lines:
- Since $KMLB$ is a parallelogram, opposite angles are equal.
- Thus, $\angle KPB$ (formed by segments $KP$ , $PB$ ) is equal to $\angle KML$ (formed by segments $KM$ , $ML$ ). This follows from alternate interior angles created by transversal line $KL$ , concluding the proof.

Step 7

Prove that $AP \perp BC$.

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Answer

To show $AP \perp BC$ :

Use the properties of cyclic quadrilaterals and the angles subtended by the same chord:
- Angles $\angle APB$ and $\angle ACB$ must be supplementary since points $A$ , $P$ , $B$ , and $C$ lie on a circle.
- Therefore, angle sums lead to $AP \perp BC$ from angle properties concerning intersecting chords.

4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2006 - Paper 1

Worked Solution & Example Answer:4 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 4 - 2006 - Paper 1

Find the volume of the solid

Show that the equation of the line, $l$, through $R$, perpendicular to $PQ$

Write down the equation of the line, $m$, through $P$, perpendicular to $QR$

Show that $T$ lies on the hyperbola.

Prove that $KMLB$ is a parallelogram.

Prove that $\angle KPB = \angle KML$.

Prove that $AP \perp BC$.

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