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

To find the complex roots of $P(x)$, we can perform polynomial long division on $P(x)$ by $(x - 1)^3 = x^3 - 3x^2 + 3x - 1$:

Performing the division yields:

$P(x) = (x - 1)^3 Q(x),$

where $Q(x)$ is a linear polynomial. Finding the remaining roots involves solving for $Q(x)$ such that:

$Q(x) = k(x^2 + bx + c).$

By the quadratic formula, the remaining roots can be determined as the roots of $Q(x)$.

To find the angle $\phi$, we use the gradient of the normal to the ellipse at the point $P(a \cos \theta, b \sin \theta)$.

Using implicit differentiation on the ellipse equation:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

yields:

$\frac{2x}{a^2} dx + \frac{2y}{b^2} dy = 0 \Rightarrow dy = -\frac{b^2}{a^2} \frac{x}{y} dx.$

At $P$, we substitute $x$ and $y$ to find $\tan \phi$:

To maximize $\phi$, we set the derivative of $\tan \phi$ with respect to $\theta$ to zero. Using trigonometric identities for maximum conditions will guide us to find:$\theta = \frac{\pi}{4}.$

Using Newton's second law, the net force acting on the train is:

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

Setting the resistive force and rearranging gives us:

$m \frac{dv}{dt} = F - \frac{K v^2}{300}.$

This indicates the acceleration of the train in motion, confirming the equations derived.

To find the time, we set up the equation:

$t = \frac{v}{a}$

Where $a$ can be expressed in terms of $F$, $m$, with terminal conditions considered to solve for the time to reach 200 km/h.