2. (a) Let $f(x) = ext{sin}^{-1}(x + 5).$ (i) State the domain and range of the function $f(x)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2006 - Paper 1

To find the gradient of the graph, we first need to differentiate $f(x)$. The derivative of $f(x) = ext{sin}^{-1}(x + 5)$ is given by:
$f'(x) = \frac{1}{\sqrt{1 - (x + 5)^2}}.$
Plugging in $x = -5$:
$f'(-5) = \frac{1}{\sqrt{1 - (-5 + 5)^2}} = \frac{1}{\sqrt{1 - 0}} = 1.$
Thus, the gradient at the point where $x = -5$ is 1.

To sketch the graph of $y = f(x) = ext{sin}^{-1}(x + 5)$, it is essential to consider the transformations involved. Starting from the parent function $y = ext{sin}^{-1}(x)$, we shift left by 5 units.
The graph will be defined between the points (-6, -π/2) and (-4, π/2). The overall shape of the graph resembles the parent inverse sine function, asymmetric about the horizontal axis.

Applying the binomial theorem gives us:
$(1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k$
Differentiating both sides w.r.t. $x$ yields:
$n(1+x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} x^{k-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + ... + n\binom{n}{n}x^{n-1}.$

Substituting $x = r$ in the derived equality from part (i) gives:
$n(1+r)^{n-1} = \sum_{k=0}^{n} \binom{n}{k} r^k.$
Upon multiplying both sides by $n$, we find:
$n3^{n-1} = \sum_{k=0}^{n} \binom{n}{k} r^k,$
this allows us to deduce the required identity.

The coordinates of point $U$ can be found by determining where the chord $PR$ intersects with the axis of the parabola. From the equation of the chord $PR$, set $y = 0$:
$0 = \frac{1}{2}(p + r)x - apr \implies x = \frac{2apr}{p + r}.$
Thus, the coordinates of $U$ are given by:
$U = \left( \frac{2apr}{p + r}, 0 \right).$

The tangent at point $P$ is given by the equation $y = px - aq^2$.
The tangent at point $Q$ can be determined similarly. Setting both equations equal to find the intersection gives:
Solving these simultaneous equations results in finding the coordinates of point $T$:
$T = (a(p + q), aq).$

To show that line segment $TU$ is perpendicular to the axis of the parabola, let's find the slope of line $TU$. The slope can be found using the coordinates of points $T(a(p + q), aq)$ and $U(\frac{2apr}{p + r}, 0)$.
$\text{slope of } TU = \frac{aq - 0}{a(p + q) - \frac{2apr}{p + r}}.$
If this slope equals 0, then it is indeed perpendicular because the axis of the parabola is vertical (undefined slope).