The polynomial $p(x) = ax^3 + bx^2 + c$ has a multiple zero at 1 and remainder 4 when divided by $x + 1$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2006 - Paper 1

To find the volume of the solid, we start with the area of the cross-sections:

1. The base is defined by the parabolic region $x^2 \leq y \leq 1$. The area of each square is determined by the height of the parabola:

   • The side length of the square at position $y$ is $s = \sqrt{y}$, so the area is: $A(y) = s^2 = y$

2. We can express the volume $V$ as an integral of the area from the base of the region to the top:

   • $V = \int_{0}^{1} A(y) \, dy = \int_{0}^{1} y \, dy$
   • Solving this integral: $V = \left[ \frac{1}{2}y^2 \right]_{0}^{1} = \frac{1}{2}$
   • Hence, the volume of the solid is rac{1}{2} cubic units.

To find the equation of the line $\ell$:

1. Calculate the gradient of $PQ$:

   • The gradient is given by: $\frac{(q_r - p_r)}{(q_p - p_q)}$

2. The line perpendicular to $PQ$ has a gradient that is the negative reciprocal. Thus:

   • If $m_{PQ}$ is the gradient of $PQ$, then $m_{\ell} = -\frac{1}{m_{PQ}}$

3. Substitute to find the equation of the line using point-slope form:

   • $y - y_R = m_{\ell}(x - x_R)$
   • Rearranging gives the required form.

To find the equation of the line $m$ through point $P$:

1. Calculate the gradient of line $QR$:

   • The gradient is: $\frac{(q_r - r_r)}{(q_p - r_p)}$

2. The gradient of the line $m$ will be the negative reciprocal:

   • If $m_{QR}$ is the gradient of $QR$, then $m_m = -\frac{1}{m_{QR}}$

3. Use point-slope form with point $P$ to write the equation:

   • $y - y_P = m_m(x - x_P)$

To show that point $T$ lies on the hyperbola $xy = 1$:

1. Set up the equations of the lines $\ell$ and $m$ and solve them simultaneously to find the coordinates of $T$.

2. Substitute the coordinates $T(x_T, y_T)$ into the hyperbolic equation:

   • Check whether: $x_T y_T = 1$

3. If this holds true, then point $T$ lies on the hyperbola.

To prove $KMLB$ is a parallelogram:

1. Show that $KL$ is parallel to $MB$ and also show that $KM$ is parallel to $LB$ using properties of midpoints.
2. Use the fact that both pairs of opposite sides are equal in length as they are both midlines in triangle $ABC$.
3. Conclude that $KMLB$ satisfies the parallelogram properties.

To show this angle relation:

1. Use the properties of parallel lines where $KL || MB$ to demonstrate that alternate interior angles formed are equal.
2. Conclude that: $\angle KPB = \angle KML$

To prove the perpendicularity:

1. Recognize that $AP$ is a median of triangle $ABC$.
2. Use coordinate geometry or angle properties to show that $AP$ forms right angles with $BC$ based on the earlier established angle relationships.
3. Conclude that: $AP \perp BC$