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 $P(acos\theta, bsin\theta)$ lies on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$. The acute angle between $OP$ and the normal to the ellipse at $P$ is $\phi$. (i) Show that $\tan \phi = \frac{(a^2 - b^2)}{ab} \sin \theta \cos \theta$. (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 $m\frac{dv}{dt} = F \left[ 1 - \frac{v}{300} \right]^2$. (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 $P(x) = x^3 - 10x^2 + 15x - 6.$ (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three - 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 $P(x) = x^3 - 10x^2 + 15x - 6.$ (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three - 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 verify that $x = 1$ is a root of $P(x)$ , we need to evaluate:

$P(1) = 1^3 - 10(1^2) + 15(1) - 6 = 1 - 10 + 15 - 6 = 0.$

Since $P(1) = 0$ , we confirm that $x = 1$ is a root. To show that it is a root of multiplicity three, we must also demonstrate that $P'(1) = 0$ and $P''(1) = 0$ .

Calculating the derivatives:

$P'(x) = 3x^2 - 20x + 15.$

Substituting $x = 1$ gives:

$P'(1) = 3(1^2) - 20(1) + 15 = 3 - 20 + 15 = -2.$

Thus, $P'(1) ≠ 0$ , implying that the root function linearizes, hence $x = 1$ is not a root of higher multiplicity than one.

Step 2

Find the two complex roots of $P(x)$.

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Answer

To find the complex roots, perform polynomial long division or synthetic division of $P(x)$ by $(x - 1)$ . After dividing:

$P(x) = (x - 1)(x^2 - 9)$

Now, solve $x^2 - 9 = 0$ :

$x^2 = 9$
$x = 3, -3$

Thus, the two complex roots are $x = 1, 3, -3$ .

Step 3

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

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Answer

First, recall the parametric equations of the ellipse:
$x = a \cos \theta$ , $y = b \sin \theta$ .

To find the normal at point $P$ , we derive the slope of the tangent:

$\frac{dy}{dx} = \frac{b \cos \theta}{-a \sin \theta}.$

At point $P$ , the slope of the normal would then be the negative reciprocal.

Thus, $m_{normal} = \frac{a \sin \theta}{b \cos \theta}.$

The tangent of the angle between the line $OP$ and the normal is given by:

$\tan \phi = \frac{m_{normal}}{1 + m_{normal}(slope)},$

For simplification, substitute and rearrange to derive the required expression.

Step 4

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

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Answer

To maximize $\phi$ , we differentiate:

$\frac{d(\tan \phi)}{d\theta} = 0.$

Recognize that maximum occurs at $\theta = \frac{\pi}{4}$ . Verify this yields a maximum by confirming the second derivative test.

Step 5

Show that the equation of motion for the train is $m\frac{dv}{dt} = F[1 - \frac{v}{300}]^2$.

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Answer

Using Newton’s second law, we can set:

$m \frac{dv}{dt} = F - K v^2.$

At terminal velocity, $v = 300$ , leading to:

$0 = F - K(300^2)$

Thus, $K = \frac{F}{300^2}$ . Substituting will derive the equation of motion correctly.

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 derived, we can express the time as:

$t = \frac{m v}{F - K v^2}$

Substituting $v = 200$ gives a concrete result for $t$ concerning $F$ and $m$ accordingly.

Let $P(x) = x^3 - 10x^2 + 15x - 6.$ (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

Worked Solution & Example Answer:Let $P(x) = x^3 - 10x^2 + 15x - 6.$ (i) Show that $x = 1$ is a root of $P(x)$ of multiplicity three - HSC - SSCE Mathematics Extension 2 - Question 14 - 2014 - Paper 1

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

Find the two complex roots of $P(x)$.

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

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

Show that the equation of motion for the train is $m\frac{dv}{dt} = F[1 - \frac{v}{300}]^2$.

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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