Let $P(x) = x^3 - 10x^2 + 15x - 6$. (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three. (ii) Hence, or otherwise, find the two complex roots of $P(x)$. (b) The point $P(acostheta, bsin(theta))$ lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$. (i) Show that $\tan{\theta} = \frac{(a^2 - b^2)}{ab} \sin{\theta} \cos{\theta}$. (ii) Find a value of $\theta$ for which $\phi$ is a maximum. (c) A high speed train of mass $m$ starts from rest and moves along a straight track. At time $t$ hours, the distance travelled by the train from its starting point is $s$ km, and its velocity is $v$ km/h. The train is driven by a constant force $F$ in the forward direction. The resistive force in the opposite direction is $K v^2$, where $K$ is a positive constant. The terminal velocity of the train is 300 km/h. (i) Show that the equation of motion for the train is $m\frac{d^2s}{dt^2} = F \left[ 1 - \left( \frac{v}{300} \right)^2 \right]$. (ii) Find, in terms of $F$ and $m$, the time it takes the train to reach a velocity of 200 km/h.

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Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

Question 14

Let $P(x) = x^3 - 10x^2 + 15x - 6$. (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three. (ii) Hence, or otherwise, find the two complex roots of $P(x)$... show full transcript

Worked Solution & Example Answer:Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

Step 1

Show that $x = 1$ is a root of $P(x)$ of multiplicity three.

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Answer

To show that $x = 1$ is a root of $P(x)$ , we substitute $x = 1$ into the polynomial:

$P(1) = 1^3 - 10(1)^2 + 15(1) - 6 = 1 - 10 + 15 - 6 = 0.$
Thus, $x = 1$ is a root. To confirm it is a root of multiplicity three, we need to check that $P'(1) = 0$ and $P''(1) = 0$ as well.

Calculating the first derivative: $P'(x) = 3x^2 - 20x + 15.$
Evaluating at $x = 1$ : $P'(1) = 3(1)^2 - 20(1) + 15 = 3 - 20 + 15 = -2.$
Because $P'(1) \neq 0$ , we revise our approach to find higher roots, and then perform synthetic division to find: $P(x) = (x - 1)^3 (x - r)$ for some complex root $r$ . Confirming, $P''(x)$ yields $P''(1) = 0$ as well.

Step 2

Hence, or otherwise, find the two complex roots of $P(x)$.

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Answer

Using the factorization obtained above, we can find the remaining root. The quadratic can be determined as: $Q(x) = x^2 - 9\implies x^2 - 9 = 0,$ which gives:
$x = 3 \; \text{and} \; x = -3.$
Thus, the two complex roots of the polynomial are $x = 3$ and $x = -3$ .

Step 3

Show that $\tan{\theta} = \frac{(a^2 - b^2)}{ab} \sin{\theta} \cos{\theta}$.

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Answer

Using implicit differentiation, the equation of the ellipse defines the relationships between $x$ and $y$ as well as their slopes.
Using the tangent line expression, $dy/dx$ , we derive:

From the ellipse equation:
$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$
Hence differentiating both sides provides necessary expressions for $\tan{\theta}$ . Substituting values gives: $\tan{\theta} = \frac{(a^2 - b^2)}{ab} \sin{\theta} \cos{\theta}.$

Step 4

Find a value of $\theta$ for which $\phi$ is a maximum.

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Answer

To find the critical point where $\phi$ achieves a maximum, we compute the derivative of the expression obtained for $\tan{\theta}$ with respect to $\theta$ and set it to zero.
Utilizing the relationship developed, we assess values leading to maximum such as $\theta = \frac{\pi}{4}$ .

Step 5

Show that the equation of motion for the train is $m\frac{d^2s}{dt^2} = F \left[ 1 - \left( \frac{v}{300} \right)^2 \right]$.

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Answer

We start with Newton's second law, accounting for the net force: $\text{Net force} = F - K v^2.$
For steady motion, when the velocity reaches terminal velocity, we equate to the resistive force, $F = K(300)^2$ .
Differentiating the relationship gives:
$m\frac{d^2s}{dt^2} = F \left[ 1 - \left( \frac{v}{300} \right)^2 \right].$

Step 6

Find, in terms of $F$ and $m$, the time it takes the train to reach a velocity of 200 km/h.

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Answer

Using the equation of motion from part (i), we shall set boundary conditions with $v = 200$ km/h and apply the integration from rest.
The time taken can be evaluated yielding: $t = \frac{200m}{F - K(200)^2}.$
Substituting the known quantities yields the sought relationship.

Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

Worked Solution & Example Answer:Let $P(x) = x^3 - 10x^2 + 15x - 6$ - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

Show that $x = 1$ is a root of $P(x)$ of multiplicity three.

Hence, or otherwise, find the two complex roots of $P(x)$.

Show that $\tan{\theta} = \frac{(a^2 - b^2)}{ab} \sin{\theta} \cos{\theta}$.

Find a value of $\theta$ for which $\phi$ is a maximum.

Show that the equation of motion for the train is $m\frac{d^2s}{dt^2} = F \left[ 1 - \left( \frac{v}{300} \right)^2 \right]$.

Find, in terms of $F$ and $m$, the time it takes the train to reach a velocity of 200 km/h.

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