(a) Let $f(x) = \sin^{-1}(x + 5)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2006 - Paper 1

To find the gradient, we first differentiate $f(x)$:

$f'(x) = \frac{1}{\sqrt{1 - (x + 5)^{2}}}$

Substituting $x = -5$, we get:

$f'(-5) = \frac{1}{\sqrt{1 - 0^{2}}} = 1$

Therefore, the gradient of the graph at the point where $x = -5$ is 1.

The graph of $y = f(x) = \sin^{-1}(x + 5)$ is a reflection of the arc sine function. It will be defined on the interval $[-6, -4]$ and will have a range of $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. The curve will start at the point corresponding to $x = -6$ and will end at $x = -4$, smoothly rising from(-\frac{\pi}{2}) to $\frac{\pi}{2}$.

Applying the binomial theorem, we have:

$(1 + x)^{n} = \sum_{r=0}^{n} \binom{n}{r} x^{r}$

Differentiating with respect to $x$:

$n(1 + x)^{n-1} = \sum_{r=1}^{n} r \binom{n}{r} x^{r-1}$

Therefore, we can express the sum as:

$n(1 + x)^{n-1} = \sum_{r=0}^{n} r \binom{n}{r} x^{r-1}$.

From the differentiation, we deduce:

$n3n3^{n-1} = \sum_{r=0}^{n} \binom{n}{r} r^{2} + \sum_{r=0}^{n} \binom{n}{r} 2^{n - r}$.

To find the coordinates of the point $U$, we know that the chord $PR$ can be expressed in terms of its endpoints, which are points on the parabola. Since $QR$ is perpendicular to the axis of the parabola, it would intersect the y-axis vertically. We need to plug the necessary values for $y$ in terms of $x$ to find the coordinates.

To show this, we will calculate the slopes of the tangents at points $P$ and $Q$. The equations of the tangents are:

$y = px - aq^{2}$
$y = \frac{1}{2}(p + r)x - apr$

Setting these two equations equal will yield the coordinates of point $T$.

To demonstrate that line segment $TU$ is perpendicular to the axis of the parabola, we must investigate the slopes of the segments $TU$ and the axis. If the product of their slopes is $-1$, this will confirm that they are indeed perpendicular.