The cubic polynomial $P(x) = x^3 + rx^2 + sx + t$, where $r$, $s$, and $t$ are real numbers, has three real zeros, $1$, $eta$, and $-eta$. (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 $ extbf{r} = 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 $$rac{1}{x(1 + x)} \, dx = -3t.$$ (iii) It can be shown that for some constant $c$, $$ ext{log}_e igg( rac{1}{x} igg) = 3t + c.$$ (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, $1$, $eta$, and $-eta$ - 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, $1$, $eta$, and $-eta$. (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, $1$, $eta$, and $-eta$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2006 - Paper 1

Step 1

Find the value of r.

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Answer

To find the value of $r$ , we use Vieta's formulas which relate the coefficients of the polynomial to the sums and products of its roots. The polynomial can be expressed as:

$P(x) = (x - 1)(x - eta)(x + eta)$

Expanding this yields:

$P(x) = (x - 1)(x^2 - eta^2) = x^3 - x^2 - eta^2x + eta^2$

From this, we can identify the coefficient of $x^2$ as $r = -1$ , thus $r = -1$ .

Step 2

Find the value of s + t.

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Continuing from the expanded polynomial, we have: $P(x) = x^3 - x^2 - eta^2x + eta^2$

Here, we can compare the coefficients to find $s$ and $t$ . We see: $s = -eta^2 \ \ t = eta^2$

Thus, $s + t = -eta^2 + eta^2 = 0$ .

Step 3

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 undergoing simple harmonic motion can be described as:

$x(t) = A imes ext{cos}igg( rac{2 heta}{T} t - rac{ heta}{2} igg)$

where $A = 18$ (amplitude) and $T = 5$ (period). Thus:

$x(t) = 18 imes ext{cos}igg( rac{2 heta}{5} t igg)$ .

Step 4

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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The time taken for the particle to move from a rest position to the point halfway can be calculated using the properties of simple harmonic motion. The rest position occurs at maximum and minimum amplitudes, so:

The time from maximum position to equilibrium is $T/4$ . Since the motion is symmetric, halfway is reached after $T/8$ seconds. Therefore, the answer is:

$\frac{5}{8} \text{ seconds}.$

Step 5

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

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To find the velocity, we differentiate the position function:

$v = \frac{d}{dt}(18t^3 + 27t^2 + 9t) = 54t^2 + 54t + 9.$

Squaring both sides gives:

$v^2 = (54t^2 + 54t + 9)^2.$

After manipulation, one can show that:

$v^2 = 9t^2(1 + x^2)$ for the given terms.

Step 6

Hence, or otherwise, show that \int \frac{1}{x(1+x)} dx = -3t.

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Using substitution, we can integrate the given function. The integral can be split as:

$\int \left( \frac{1}{x} - \frac{1}{1+x} \right) dx = -3t$ unifying the expressions. Therefore, the result gives the desired expression after evaluating constants.

Step 7

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

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Using the derived equation:

$\text{log}_e \bigg( \frac{1}{x} \bigg) = 3t + c$

We can exponentiate both sides:

$\frac{1}{x} = e^{3t + c} \Rightarrow x = \frac{1}{e^{3t + c}}$

Using initial conditions will provide the constant's specific value.

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

Find the value of r.

Find the value of s + t.

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

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

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

Hence, or otherwise, show that \int \frac{1}{x(1+x)} dx = -3t.

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

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