Question 6 (15 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 2 - Question 6 - 2002 - Paper 1

Starting from the vertical force balance, we set up the equation: $T + N = \frac{mg}{\sin \alpha}.$ From this, we can solve for $T$: $T = \frac{mg}{\sin \alpha} - N.$ Next, using the horizontal component of forces, we recognize that $T\cos \alpha = \frac{mv^2}{l}$ where $v = l\omega$. This leads to: $T = \frac{m (l\omega)^2}{l \cos \alpha} + N.$ Combining these two equations allows us to derive an expression for $T - N$.

When $N = 0$, the object is on the verge of losing contact with the cone. Substituting $N = 0$ into our previous equation gives: $T = \frac{mg}{\sin \alpha}.$ Now, using the relationship for tension: $T = \frac{ml^2 \omega^2}{g \sin \alpha}.$ Setting these equal provides: $\frac{ml^2 \omega^2}{g \sin \alpha} = \frac{mg}{\sin \alpha}.$ From this, we can isolate $\omega$ and solve for: $\omega = \sqrt{\frac{g}{l \cos \alpha}}.$

Starting with the integral: $I_1 = \int_{0}^{\frac{\pi}{2}} \tan \theta \, d\theta,$ we can use the substitution $u = \sec \theta$, which transforms the integral limits and yields: $I_1 = \frac{1}{2} \ln(2).$ This makes the verification straightforward.

To derive this relationship, we apply integration by parts on $I_n$ and observe the recurrence relations in terms of previous integrals. By careful analysis, we deduce that: $I_n + I_{n-2} = \frac{1}{n - 1},$ proving the formula.

The relation $I_n < I_{n-2}$ follows from observing the decreasing nature of the integrands. Thus, by applying the properties of integration and limiting behavior, we conclude: $\frac{1}{2(n+1)} < I_n < \frac{1}{2(n-1)}$ giving us the required bounds.

Using the recurrence relation derived previously, we compute: $I_5 = I_3 + I_1 = \frac{1}{4} \ln 2 + \frac{1}{2} \ln 2.$ This leads us to find boundaries, ultimately yielding the result: $\frac{2}{3} < \ln 2 < \frac{3}{4}.$ This concludes our computation.