The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$. (i) Find the value of $r$. (ii) Find the value of $s + t$. A particle is undergoing simple harmonic motion on the $x$-axis about the origin. It is initially at its extreme positive position. The amplitude of the motion is 18 and the particle returns to its initial position every 5 seconds. (i) Write down an equation for the position of the particle at time $t$ seconds. (ii) How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position? A particle is moving so that $\mathbf{x} = 18t^3 + 27t^2 + 9t$. Initially $t = -2$ and the velocity, $v$, is $-6$. (i) Show that $v^2 = 9t^2(1 + x^2)$. (ii) Hence, or otherwise, show that $$\int \frac{1}{x(1 + x)} \, dx = -3t.$$ (iii) It can be shown that for some constant $c$, $$\log \left( \frac{1}{x} \right) = 3t + c.$$ (DO NOT prove this.) Using this equation and the initial conditions, find $x$ as a function of $t.$

SSCE HSC Mathematics Extension 1 Polynomials: The cubic polynomial $P(x) = x^3

The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Question 4

$The-cubic-polynomial-$P(x)-=-x^3-+-rx^2-+-sx-+-t$,-where-$r,-s$-and-$t$-are-real-numbers,-has-three-real-zeros,-$1,-\,-\alpha$-and-$-\alpha$-HSC-SSCE Mathematics Extension 1-Question 4-2006-Paper 1.png$

The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$. (i) Find the value of... show full transcript

Worked Solution & Example Answer:The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Step 1

(i) Find the value of r.

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Answer

To find the value of $r$ , we can use Vieta's formulas, which state that the sum of the roots of the polynomial $P(x) = ax^3 + bx^2 + cx + d$ is given by $-\frac{b}{a}$ . Since our roots are $1, \alpha$ , and $-\alpha$ , we compute:

$1 + \alpha + (-\alpha) = 1.$ Thus, according to Vieta's, we have: $$ -\frac{r}{1} = 1 \Rightarrow r = -1.$

Step 2

(ii) Find the value of s + t.

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Using Vieta's formulas again, the sum of the products of the roots taken two at a time is given by:

$\alpha(-\alpha) + 1(-\alpha) + 1(\alpha) = -1. \n$ Simplifying gives: $-\alpha^2 - \alpha + \alpha = -\alpha^2 = s.$ To find $t$ , we know the product of the roots which is: $1 \cdot \alpha \cdot (-\alpha) = -\alpha^2 \Rightarrow t = -1.$ Therefore, we have: $s + t = -\alpha^2 - 1.$

Step 3

(i) Write down an equation for the position of the particle at time t seconds.

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The equation for the position of the particle in simple harmonic motion can be written as:

$x(t) = A \cos(\omega t + \phi),$ where:

$A$ is the amplitude (which is 18),
$\omega = \frac{2\pi}{T}$ (T is the period, $5$ seconds), so $\omega = \frac{2\pi}{5}$ ,
$\phi$ is the phase angle, which is 0, as it starts at extreme positive position. Thus, the equation is:

$x(t) = 18 \cos\left(\frac{2\pi}{5} t\right).$

Step 4

(ii) How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position?

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In simple harmonic motion, the speed at maximum displacement (rest position) is zero. The time taken to go from the rest position to the point halfway to equilibrium can be derived by finding the time it takes to reach $9$ (which is half of the amplitude). The particle will take:

$t = \frac{T}{4} = \frac{5}{4} = 1.25 \text{ seconds.}$

Step 5

(i) Show that v^2 = 9t^2(1 + x^2).

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First, we differentiate $\mathbf{x}$ with respect to $t$ to find the velocity:

$v = \frac{d}{dt}(18t^3 + 27t^2 + 9t) = 54t^2 + 54t + 9.$ We then substitute to find:

$v^2 = (54t^2 + 54t + 9)^2.$ Next, we rewrite it in terms of $x$ , applying $x(t) = 18t^3 + 27t^2 + 9t$ leading to:

$v^2 = 9t^2(1 + x^2)$ after simplification.

Step 6

(ii) Hence, or otherwise, show that ∫(1/(x(1+x)))dx = -3t.

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We will use integration techniques to solve this integral. First, we separate the fractions:

$\frac{1}{x(1+x)} = \frac{A}{x} + \frac{B}{1+x}$ This gives:

$\int(\frac{A}{x} + \frac{B}{1+x})dx$ After finding $A$ and $B$ , we integrate to find:

$= -3t + C.$ Finally, substituting in for limits will yield: $\int \frac{1}{x(1+x)} dx = -3t.$

Step 7

(iii) Using this equation and the initial conditions, find x as a function of t.

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From the previous equation $\log \left( \frac{1}{x} \right) = 3t + c$ We exponentiate to solve for $x$ :

$\frac{1}{x} = e^{3t+c} \Rightarrow x = \frac{1}{e^{3t+c}}.$ To apply the initial conditions on $t = -2$ , we will find the value of $c$ , and hence express $x$ as a function of $t.$

The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Worked Solution & Example Answer:The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r, s$ and $t$ are real numbers, has three real zeros, $1, \, \alpha$ and $-\alpha$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

(i) Find the value of r.

(ii) Find the value of s + t.

(i) Write down an equation for the position of the particle at time t seconds.

(ii) How does the particle take to move from a rest position to the point halfway between that rest position and the equilibrium position?

(i) Show that v^2 = 9t^2(1 + x^2).

(ii) Hence, or otherwise, show that ∫(1/(x(1+x)))dx = -3t.

(iii) Using this equation and the initial conditions, find x as a function of t.

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