The diagram shows a sketch of $y = f'(x)$, the derivative function of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 4 - 2001 - Paper 1

To sketch the curve $y = f(x)$, we start from the given point $f(0) = 0$. We then consider the properties derived from $f'(x)$:

• As x o -orall, $f(x) o 1$ (horizontal asymptote).
• Moving through the critical points identified: an increase to the maximum at $(1, 1)$, then decreasing to the minimum at $(5, 1)$;
• Finally, the curve should approach the asymptote as x o orall. Clearly label the critical points and the asymptote on the diagram.

The sketch should resemble a wave-like motion peaking at (1,1) and valley at (5,1) with a continuous overall trend towards the asymptote.

To find the area of the cross-section of the solid $S$, we note that the hole is cylindrical. The formula for the area of the circle is:

$A = \pi r^2$

When located inside the sphere, we apply the following relationship: $A = \pi (R^2 - h^2 - r^2)$. By substituting appropriate trigonometric definitions, we find the area in relation to the angles involved, resulting in:

$A = \theta(R^2 - h^2 - r^2)$.

The volume of the resulting solid can be determined using integration of the cross-sectional area:

$V = \int_{-b}^{b} A(h) \, dh$

In our case, substituting for $A(h)$ gives:

$V = \int_{-b}^{b} \theta(R^2 - h^2 - r^2) \, dh.$

The integral evaluates to yield a function in $b$, as this variable represents half the height of the cylindrical hole. Thus we simplify the expression to present the volume solely in terms of $b$.

We differentiate the sum of the two terms separately:

1. The derivative of $\tan\left(\frac{1}{x + 1}\right)$ using the chain rule gives heavy expressions.
2. The second term $\tan^{-1}\left(\frac{1}{2x+1}\right)$ also differentiates appropriately.

Through proper manipulation reducing both terms leads to a combined result:

$\frac{d}{dx}\left[\tan\left(\frac{1}{x + 1}\right) + \tan^{-1}\left(\frac{1}{2x+1}\right)\right] = 0$

This indicates the expression is indeed constant for $2x + 1 > 0$. Without loss of generality, the constant found evaluates the same irrespective of the values determined in the limits.