SSCE HSC Mathematics Extension 2 Proofs involving inequalities: Find the minimum value of $P(x)

Question

Find the minimum value of $P(x) = 2x^3 - 15x^2 + 24x + 16$, for $x \geq 0$.

Hence, or otherwise, show that for $x \geq 0$,
$$(x + 1) \left( x^2 + (x + 4)^2 \right) \geq 25x^2.$$

Hence, or otherwise, show that for $m \geq 0$, 
$$(m + n)^2 + (m + n + 4) \geq \frac{100mm}{m + n + 1}.$$

A small bead P of mass m can freely move along a string. The ends of the string are attached to fixed points S and S', where S' lies vertically above S. The bead undergoes uniform circular motion with radius r and constant angular velocity $ \omega $ in a horizontal plane.

The forces acting on the bead are the gravitational force and the tension forces along the string. The tension forces along PS and PS' have the same magnitude T.

The length of the string is 2a and SS' = 2ae, where 0 < e < 1. The horizontal plane through P meets SS' at Q. The midpoint of SS' is O and $ \beta = \angle ZSPQ $. The parameter $ \theta $ is chosen so that OQ = a \cos \theta.

What information indicates that P lies on an ellipse with foci S and S', and with eccentricity e?

Using the focus-directrix definition of an ellipse, or otherwise, show that $SP = a(1 - e \cos \theta)$.

Show that $\sin \beta = \frac{e + \cos \theta}{1 + e \cos \theta}$.

By considering the forces acting on P in the vertical direction, show that
$$mg = \frac{27(1 - e^2) \cos \theta}{1 - e^2 \cos^2 \theta}.$$

Show that the force acting on P in the horizontal direction is
$$mr \omega^2 = \frac{27(1 - e^2) \sin \theta}{1 - e^2 \cos^2 \theta}.$$

Show that $\tan \beta = \frac{r \omega^2}{g(1 - e^2)}$.

Find the minimum value of $P(x) = 2x^3 - 15x^2 + 24x + 16$, for $x \geq 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2013 - Paper 1

Worked Solution & Example Answer:Find the minimum value of $P(x) = 2x^3 - 15x^2 + 24x + 16$, for $x \geq 0$ - HSC - SSCE Mathematics Extension 2 - Question 16 - 2013 - Paper 1

Show that for $m \geq 0$, $(m + n)^2 + (m + n + 4) \geq \frac{100mm}{m + n + 1}$.

Show that $\tan \beta = \frac{r \omega^2}{g(1 - e^2)}$.

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