The shaded region bounded by $y=3-x^2$, $y=x+x^2$ and $x=-1$ is rotated about the line $x=-1$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2002 - Paper 1

The volume $V$ of the solid of revolution can be calculated using cylindrical shells. The formula for volume using shells is:

$V = 2\pi \int_{a}^{b} (radius)(height) dx$

Here, the radius of a shell is $|x + 1|$ (distance from the line $x = -1$) and the height is the difference between the two curves:

$height = (3 - x^2) - (x + x^2) = 3 - 2x^2 - x$

Thus, the limits of integration will be from $x = -1$ to $x = 1$. Therefore, we can express the volume as:

$V = 2\pi \int_{-1}^{1} (x + 1)(3 - 2x^2 - x) \, dx$

To evaluate the integral:

$V = 2\pi \int_{-1}^{1} (x + 1)(3 - 2x^2 - x) \, dx$

First, expand the expression inside the integral:

$(x + 1)(3 - 2x^2 - x) = 3x + 3 - 2x^3 - x^2 - 2x^2 - 2x = 3x + 3 - 2x^3 - 3x^2$

So we have:

$V = 2\pi \int_{-1}^{1} (3 - 2x^3 - 3x^2 + 3x) \, dx$

Calculating the integral:

$V = 2\pi \left[3x - \frac{3}{2}x^2 - \frac{1}{2}x^4\right]_{-1}^{1}$

Evaluating the limits gives:

$V = 2\pi \left[(3 - \frac{3}{2} - \frac{1}{2}) - (-3 - \frac{3}{2} - \frac{1}{2}) \right]$

This simplifies to:

$V = 2\pi \left[3 - 1 - (-3 - 1)\right] = 8\pi$

Thus, the volume of the solid of revolution is $8\pi$.

Since A, B, C and D are concyclic, we can utilize the properties of cyclic quadrilaterals. By inscribing triangle DSR within circle ABCD, we note that angles subtended by the same arc are equal. Therefore:

$\angle DSR = \angle DAR$

This concludes the proof.

Triangles R, S, D and T also share similar properties. Utilizing the cyclic nature:

$\angle DST + \angle DCT = \pi$

Rearranging yields:

$\angle DST = \pi - \angle DCT$

From the previous results, we established that ∠DSR = ∠DAR and ∠DST = π - ∠DCT. Thus, if angles ∠DST and ∠DCT are equal, points R, S, and T must be collinear according to the angle relationship of being supplementary. Hence, we conclude R, S, and T are indeed collinear.

To exceed 400 with three drawn digits, the first digit must be 4, 5, 6, 7, 8, or 9. The total ways to choose three digits from nine is:

$\binom{9}{3} = 84$

Now, if we fix the first digit (4, 5, 6, 7, 8, or 9), the remaining two can be chosen from smaller digits. Thus:

The count is:

• 4 as first: Choose any 2 from 1, 2, 3 -> 3 choices
• 5 as first: Choose any 2 from 1, 2, 3, 4 -> 4 choices
• 6 as first: Choose any 2 from 1, 2, 3, 4, 5 -> 5 choices
• etc.

Summing these gives: $3 + 4 + 5 + 6 + 7 + 8 = 33$

Thus, probability = $\frac{33}{84} = \frac{11}{28}$.

A set of three distinct digits can be arranged in descending order in only one way per any combination.

Thus, the probability that a set of three drawn numbers is in descending order is:

$P = \frac{1}{3!} = \frac{1}{6}$

Since we can select any three from the nine digits, the total combinations remain the same: $\binom{9}{3}$. Thus, probability remains:

$P = \frac{1}{6}$