2. (a) Let $f(x) = ext{sin}^{-1}(x + 5).$ (i) State the domain and range of the function $f(x)$. (ii) Find the gradient of the graph of $y = f(x)$ at the point where $x = -5$. (iii) Sketch the graph of $y = f(x)$. (b) (i) By applying the binomial theorem to $(1 + x)^n$ and differentiating, show that $$ n(1+x)^{n-1} = inom{n}{0} + inom{n}{1} x + inom{n}{2} x^2 + inom{n}{3} x^3 + ext{...} + inom{n}{n} x^{n-1}. $$ (ii) Hence deduce that $$ n3 n-1 = inom{n}{0} + inom{n}{1} r + inom{n}{2} r^2 + ext{...} + inom{n}{n} r^{n-1} $$ (c) The points $P(2ap, aq^2)$, $Q(2aq, aq^2)$ and $R(2ar, ar^2)$ lie on the parabola $x^2 = 4ay$. The chord $QR$ is perpendicular to the axis of the parabola. The chord $PR$ meets the axis of the parabola at $U$. The equation of the chord $PR$ is $y = rac{1}{2}(p + r)x - apr$. (Do NOT prove this.) The equation of the tangent at $P$ is $y = px - aq^2$. (Do NOT prove this.) (i) Find the coordinates of $U$. (ii) The tangents at $P$ and $Q$ meet at the point $T$. Show that the coordinates of $T$ are $(a(p + q), aq)$. (iii) Show that $TU$ is perpendicular to the axis of the parabola.

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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

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

2. (a) Let $f(x) = ext{sin}^{-1}(x + 5).$ (i) State the domain and range of the function $f(x)$. (ii) Find the gradient of the graph of $y = f(x)$ at the poin... show full transcript

Worked Solution & Example Answer: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

Step 1

State the domain and range of the function f(x).

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Answer

The domain of the function $f(x) = ext{sin}^{-1}(x + 5)$ is determined by the requirement that the argument of the inverse sine must lie within the interval [-1, 1]. Therefore, we solve the inequality:
$-1 \\leq x + 5 \\leq 1$
which simplifies to:
$-6 \\leq x \\leq -4.$
Thus, the domain is:
$x ext{ in } [-6, -4].$
The range of the inverse sine function is always [-π/2, π/2], so the range of $f(x)$ is:
$y ext{ in } [- rac{ ext{π}}{2}, rac{ ext{π}}{2}].$

Step 2

Find the gradient of the graph of y = f(x) at the point where x = -5.

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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.

Step 3

Sketch the graph of y = f(x).

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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.

Step 4

By applying the binomial theorem to (1 + x)^n and differentiating, show that n(1+x)^{n-1} = ...

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Answer

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:
$n(1+x)^{n-1} = \sum_{k=1}^{n} k \binom{n}{k} x^{k-1} = \binom{n}{1} + 2\binom{n}{2}x + 3\binom{n}{3}x^2 + ... + n\binom{n}{n}x^{n-1}.$

Step 5

Hence deduce that n3^{n-1} = ...

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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.

Step 6

Find the coordinates of U.

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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:
$U = \left( \frac{2apr}{p + r}, 0 \right).$

Step 7

The tangents at P and Q meet at the point T. Show that the coordinates of T are (a(p + q), aq).

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Answer

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 $T$ :
$T = (a(p + q), aq).$

Step 8

Show that TU is perpendicular to the axis of the parabola.

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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).

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

Worked Solution & Example Answer: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

State the domain and range of the function f(x).

Find the gradient of the graph of y = f(x) at the point where x = -5.

Sketch the graph of y = f(x).

By applying the binomial theorem to (1 + x)^n and differentiating, show that n(1+x)^{n-1} = ...

Hence deduce that n3^{n-1} = ...

Find the coordinates of U.

The tangents at P and Q meet at the point T. Show that the coordinates of T are (a(p + q), aq).

Show that TU is perpendicular to the axis of the parabola.

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