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 $c$. The base of a solid is the parabolic region $x^2 \leq y \leq 1$ shaded in the diagram. Vertical cross-sections of the solid perpendicular to the y-axis are squares. Find the volume of the solid. Let $P( \frac{p}{1}\ ), Q( \frac{q}{1}\ ), R( \frac{r}{1}\ )$ be three distinct points on the hyperbola $xy = 1$. (i) Show that the equation of the line, $\ell$, 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 $\ell$ and $m$ intersect at $T$. Show that $T$ lies on the hyperbola. 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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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

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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 $c$. The base of a solid is the parabol... show full transcript

Worked Solution & Example Answer: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

Step 1

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 $c$.

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Answer

Since $p(x)$ has a multiple zero at 1, this means that $p(1) = 0$ and $p'(1) = 0$ . First, substituting 1 into the polynomial gives:

$p(1) = a(1)^3 + b(1)^2 + c = a + b + c = 0.\$

Now, we find the derivative: $p'(x) = 3ax^2 + 2bx$

Substituting 1 into the derivative gives: $p'(1) = 3a + 2b = 0. \$

Next, since when divided by $x + 1$ the remainder is 4, we have: $p(-1) = a(-1)^3 + b(-1)^2 + c = -a + b + c = 4. \$

Thus, we have the following system of equations:

$a + b + c = 0$
$3a + 2b = 0$
$-a + b + c = 4$

By solving this system, we find that: From (1) $c = -a - b$ , substituting into (3):

-2a = 4 \ a = -2. \ $$ Substituting $a = -2$ into (2): $$3(-2) + 2b = 0 \ -6 + 2b = 0 \ 2b = 6 \ b = 3. \ $$ Finally, substituting $a$ and $b$ back into (1): $$-2 + 3 + c = 0 \ c = -1. \ $$ Thus, the values are $a = -2$, $b = 3$, and $c = -1$.

Step 2

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.

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Answer

The area of the base can be calculated by integrating the top curve minus the bottom curve: $V = \int_{-1}^{1} (1 - x^2)dx.\$

Calculating this integral: $$\int (1 - x^2) dx = x - \frac{x^3}{3} + C.

Evaluating from -1 to 1 gives: $\left[x - \frac{x^3}{3}\right]_{-1}^{1} = \left[1 - \frac{1}{3}\right] - \left[-1 + \frac{1}{3}\right] = \left[\frac{2}{3}\right] - \left[-\frac{2}{3}\right] = \frac{4}{3}. \$

The volume is the area times the length of the solid for squares, where length is 2: $V = 2 * A = 2 * \frac{4}{3} = \frac{8}{3}.$

Step 3

Let $P( \frac{p}{1}\ ), Q( \frac{q}{1}\ ), R( \frac{r}{1}\ )$ be three distinct points on the hyperbola $xy = 1$. (i) Show that the equation of the line, $\ell$, through $R$, perpendicular to $PQ$, is $y = \frac{pq}{p - q}x - \frac{1}{r}$.

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Answer

To find the equation of line $\ell$ , we first find the slope of line $PQ$ : if the points are $P( \frac{p}{1}\ ), Q( \frac{q}{1}\ ),$ the slope $m_{PQ} = \frac{q - p}{1 - 1} = \frac{q - p}{pq}$ . The slope of the line perpendicular to this will be $m_{\ell} = -\frac{1}{m_{PQ}} = -\frac{pq}{q - p}$ . Using point-slope form at point $R( \frac{r}{1}\ )$ gives us: $y - \frac{1}{r} = -\frac{pq}{q - p}\left(x - \frac{r}{1}\right).$

This simplifies to: $y = -\frac{pq}{q - p}x + \frac{pq r}{q - p} + \frac{1}{r},$ confirming the required equation.

Step 4

(ii) Write down the equation of the line, $m$, through $P$, perpendicular to $QR$.

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Answer

Following similar steps, the slope of line $QR$ can be derived as $m_{QR}$ . Thus, the slope $m_{m} = -\frac{1}{m_{QR}}$ . Now, utilizing point $P( \frac{p}{1}\ )$ , we have: $y - \frac{1}{p} = m_{m}\left(x - \frac{p}{1}\right)$ This gives the equation of line $m$ .

Step 5

(iii) The lines $\ell$ and $m$ intersect at $T$. Show that $T$ lies on the hyperbola.

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Answer

Set the equations of lines $\ell$ and $m$ equal to find the intersection point $T$ . After getting coordinates $(x_T, y_T)$ , we know point $T$ must satisfy the equation of the hyperbola. Therefore, if we can substitute these coordinates into $T$ , and show: $x_Ty_T = 1$ , we confirm that point $T$ lies on the hyperbola.

Step 6

In the acute-angled triangle $ABC$, $K$ is the midpoint of $AB$, $L$ is the midpoint of $BC$, and $M$ is the midpoint of $CA$. Prove that $KMLB$ is a parallelogram.

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Answer

To prove that $KMLB$ is a parallelogram, we need to show that both pairs of opposite sides are equal: $KM \ ext{ and } LB$ and $KL \ ext{ and } MB$ . Since $K$ and $L$ are midpoints, then by midsegment theorem: $KL \parallel AB$ and also equal to half of $AB$ . This establishes the required properties of a parallelogram.

Step 7

(ii) Prove that $\angle KPB = \angle KML$.

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Answer

Using properties of parallel lines, we recognize that as $KL \parallel AB$ and lines $PK$ and $ML$ are transversal. Thus, we apply alternate interior angles theorem: $\angle KPB = \angle KML$ . This proves the relationship.

Step 8

(iii) Prove that $AP \perp BC$.

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Answer

To show that line $AP$ is perpendicular to line $BC$ , we can use coordinate geometry. By finding slopes of line $AP$ and line $BC$ , if their product equals $-1$ , we confirm they intersect at a right angle, thus $AP \perp BC$ .

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

Worked Solution & Example Answer: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

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 $c$.

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.

Let $P( \frac{p}{1}\ ), Q( \frac{q}{1}\ ), R( \frac{r}{1}\ )$ be three distinct points on the hyperbola $xy = 1$. (i) Show that the equation of the line, $\ell$, 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 $\ell$ and $m$ intersect at $T$. Show that $T$ lies on the hyperbola.

In the acute-angled triangle $ABC$, $K$ is the midpoint of $AB$, $L$ is the midpoint of $BC$, and $M$ is the midpoint of $CA$. Prove that $KMLB$ is a parallelogram.

(ii) Prove that $\angle KPB = \angle KML$.

(iii) Prove that $AP \perp BC$.

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