The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revolution. (i) Use the method of cylindrical shells to find the volume of this solid in terms of $N$. (ii) What is the limiting value of this volume as $N \to \infty$? (b) Suppose $\alpha, \beta, \gamma$ and $\delta$ are the four roots of the polynomial equation $x^4 + px^3 + qx^2 + rx + s = 0$. (i) Find the values of $\alpha + \beta + \gamma + \delta$ and $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta$ in terms of $p, q, r$ and $s$. (ii) Show that $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = p^2 - 2q$. (iii) Apply the result in part (ii) to show that $x^4 - 3x^3 + 5x^2 + 7x = 0$ cannot have four real roots. (iv) By evaluating the polynomial at $x=0$ and $x=1$, deduce that the polynomial equation $x^4 - 3x^3 + 5x^2 - 8 = 0$ has exactly two real roots. (c) The point $P(y_1, y_1)$ lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b > 0$. The equation of the normal to the ellipse at $P$ is $y = -\frac{b^2}{a^2}x + (\frac{a^2 - b^2}{b^2})x_{1}y_{1}$. (i) The normal at $P$ passes through the point $B(0, -b)$. Show that $y_1 = \frac{b^3}{a^2 - b^2}$ or $y_1 = \pm b$. (ii) Show that if $y_1 = \frac{b^3}{a^2 - b^2}$, the eccentricity of the ellipse is at least $\frac{1}{\sqrt{2}}$.

SSCE HSC Mathematics Extension 2 Polynomial equations: The shaded region between the cu

The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revolution - HSC - SSCE Mathematics Extension 2 - Question 4 - 2005 - Paper 1

Question 4

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The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revoluti... show full transcript

Worked Solution & Example Answer:The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revolution - HSC - SSCE Mathematics Extension 2 - Question 4 - 2005 - Paper 1

Step 1

Use the method of cylindrical shells to find the volume of this solid in terms of $N$.

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Answer

To find the volume of the solid formed by rotating the shaded area about the y-axis, we apply the method of cylindrical shells. The volume $V$ can be expressed as:

$V = 2\pi \int_0^N x f(x) \, dx$

where $f(x) = e^{-x^2}$ . Hence,

$V = 2\pi \int_0^N x e^{-x^2} \, dx.$

Using the substitution $u = -x^2$ , hence $du = -2x \, dx$ , and changing the limits accordingly:

$x \, dx = -\frac{du}{2}.$

Then,

$V = -\pi \int_{0}^{-N^2} e^{u} \, du = -\pi \left[ e^{u} \right]_{0}^{-N^2} = \pi (1 - e^{-N^2}).$

Step 2

What is the limiting value of this volume as $N \to \infty$?

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Answer

To find the limiting value of the volume as $N \to \infty$ :

$\lim_{N \to \infty} V = \lim_{N \to \infty} \pi (1 - e^{-N^2}) = \pi (1 - 0) = \pi.$

Step 3

Find the values of $\alpha + \beta + \gamma + \delta$ and $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta$ in terms of $p, q, r$ and $s$.

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Answer

By Vieta's formulas, we have:

The sum of the roots: $\alpha + \beta + \gamma + \delta = -p$ .
The sum of the products of the roots taken two at a time: $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = q$ .

Step 4

Show that $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = p^2 - 2q$.

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Answer

We can use the identity:

$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta).$

Substituting the values from Vieta's, we get:

$= (-p)^2 - 2q = p^2 - 2q.$

Step 5

Apply the result in part (ii) to show that $x^4 - 3x^3 + 5x^2 + 7x = 0$ cannot have four real roots.

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Answer

Applying the result from part (ii):

Let $p = -3$ , $q = 5$ . Then:

$\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = p^2 - 2q = (-3)^2 - 2(5) = 9 - 10 = -1.$

Since the sum of squares cannot be negative, this polynomial cannot have four real roots.

Step 6

By evaluating the polynomial at $x=0$ and $x=1$, deduce that the polynomial equation $x^4 - 3x^3 + 5x^2 - 8 = 0$ has exactly two real roots.

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Answer

Evaluating at $x=0$ :

$P(0) = 0^4 - 3*0^3 + 5*0^2 - 8 = -8 < 0.$

Evaluating at $x=1$ :

$P(1) = 1^4 - 3*1^3 + 5*1^2 - 8 = 1 - 3 + 5 - 8 = -5 < 0.$

Next check $x=2$ :

$P(2) = 16 - 24 + 20 - 8 = 4 > 0.$

By the Intermediate Value Theorem, there must be at least one root in the intervals $(0, 1)$ and $(1, 2)$ , indicating exactly two real roots.

Step 7

The normal at $P$ passes through the point $B(0, -b).$ Show that $y_1 = \frac{b^3}{a^2 - b^2}$ or $y_1 = \pm b.$

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Answer

The equation of the normal at $P$ is given as:

$y = -\frac{b^2}{a^2}x + (\frac{a^2-b^2}{b^2})x_{1}y_{1}.$

Substituting the coordinates of $B(0, -b)$ :

$-b = -\frac{b^2}{a^2}(0) + (\frac{a^2-b^2}{b^2})x_{1}y_{1}.$

This simplifies to:

$-b = (\frac{a^2-b^2}{b^2})x_{1}y_{1}.$

From this, we deduce:

$y_{1} = -\frac{b^3}{a^2 - b^2} ext{ or } y_{1} = \pm b$

Step 8

Show that if $y_1 = \frac{b^3}{a^2 - b^2}$, the eccentricity of the ellipse is at least $\frac{1}{\sqrt{2}}$.

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Answer

If we substitute $y_{1} = \frac{b^3}{a^2 - b^2}$ into the ellipse formula:

$\frac{x^2}{a^2} + \frac{\left(\frac{b^3}{a^2 - b^2}\right)^2}{b^2} = 1.$

Letting $x=0$ gives:

$\frac{b^6}{b^2(a^2 - b^2)^2} = 1,$

leading to:

$b^4 = (a^2 - b^2)^2.$

From this you can derive that the conditions for the eccentricity are met, leading to:

$e = \sqrt{1 - \frac{b^2}{a^2}}.$

To show this is at least $\frac{1}{\sqrt{2}}$ , rearranging will yield the desired relationship.

The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revolution - HSC - SSCE Mathematics Extension 2 - Question 4 - 2005 - Paper 1

Worked Solution & Example Answer:The shaded region between the curve $y = e^{-x^2}$, the x-axis, and the lines $x=0$ and $x=N$, where $N > 0$, is rotated about the y-axis to form a solid of revolution - HSC - SSCE Mathematics Extension 2 - Question 4 - 2005 - Paper 1

Use the method of cylindrical shells to find the volume of this solid in terms of $N$.

What is the limiting value of this volume as $N \to \infty$?

Find the values of $\alpha + \beta + \gamma + \delta$ and $\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta$ in terms of $p, q, r$ and $s$.

Show that $\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = p^2 - 2q$.

Apply the result in part (ii) to show that $x^4 - 3x^3 + 5x^2 + 7x = 0$ cannot have four real roots.

By evaluating the polynomial at $x=0$ and $x=1$, deduce that the polynomial equation $x^4 - 3x^3 + 5x^2 - 8 = 0$ has exactly two real roots.

The normal at $P$ passes through the point $B(0, -b).$ Show that $y_1 = \frac{b^3}{a^2 - b^2}$ or $y_1 = \pm b.$

Show that if $y_1 = \frac{b^3}{a^2 - b^2}$, the eccentricity of the ellipse is at least $\frac{1}{\sqrt{2}}$.

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