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} = {n rack 1} + 2 {n rack 2} x + ext{...} + {n rack n} x^{n - 1}$. (ii) Hence deduce that n3^{n} = {n rack 0} + {n rack 1} imes (1 + 1) + ext{...} + {n rack n} (2^{n}). (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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Let $f(x) = ext{sin}^{-1}(x + 5)$ - HSC - SSCE Mathematics Extension 1 - Question 2 - 2006 - Paper 1

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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 $... show full transcript

Worked Solution & Example Answer:Let $f(x) = ext{sin}^{-1}(x + 5)$ - 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 must lie within the interval [-1, 1]. Therefore, the inequality $-1 \\leq x + 5 \\leq 1$
leads us to the domain:
$-6 \\leq x \\leq -4$
The range of $f(x)$ given the nature of the arcsin function is the interval:
$[-\frac{\pi}{2}, \frac{\pi}{2}]$ .

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 at $x = -5$ , we differentiate $f(x) = ext{sin}^{-1}(x + 5)$ with respect to $x$ . The derivative is given by the formula: $f'(x) = \frac{1}{\sqrt{1 - (x+5)^{2}}}$
Substituting $x = -5$ , we can deduce: $f'(-5) = \frac{1}{\sqrt{1 - (0)^{2}}} = 1$
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 $y = f(x)$ , we begin with the domain of $-6 \leq x \leq -4$ and determine the corresponding range values. The endpoints yield:

For $x = -6$ , $f(-6) = ext{sin}^{-1}(-1) = -\frac{\pi}{2}$
For $x = -4$ , $f(-4) = ext{sin}^{-1}(1) = \frac{\pi}{2}$
From this, we can draw a curve starting from the point $(-6, -\frac{\pi}{2})$ , smoothly moving to $(-4, \frac{\pi}{2})$ , reflecting the characteristics of the arcsin function.

Step 4

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

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Answer

Using the Binomial Theorem:
$(1+x)^{n} = \sum_{k=0}^{n} {n \choose k} x^{k}$
Differentiating both sides gives:
$n(1+x)^{n-1} = {n \choose 1} + 2{n \choose 2} x + \cdots + n{n \choose n} x^{n-1}$ .

Step 5

Hence deduce that

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By substituting $x=1$ into the previous result, we obtain:
$n3^{n-1} = {n \choose 0} + {n \choose 1}(2) + {n \choose 2}(1^{2}) + \cdots + {n \choose n}(2^{n})$
thus leading to the required expression.

Step 6

Find the coordinates of $U$

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To find the coordinates of $U$ , we first set the equation of the chord $PR$ to meet the $y$ -axis. Since the equation of the line is given as:
$y = \frac{1}{2}(p + r)x - apr$
Setting $x = 0$ , we then find:
$y = -apr = U.$
Therefore, $U$ is at $(0, -apr)$ .

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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From the tangents' equations, we can find the intersection point. The two tangents are given as:
$y = px - aq^{2}$
$y = qx - ap^{2}$
Equating and solving for $x$ , results in the coordinates of $T$ being determined as:
$T = (a(p + q), aq).$

Step 8

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

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To show that line $TU$ is perpendicular to the parabola's axis, we can demonstrate their slopes to be negative reciprocals. The slope of line $TU$ can be determined from $T$ and $U$ coordinates while the slope of the parabola's axis is vertical.
Since the axis slope is undefined and the product of slopes must equal -1, we conclude that $TU$ is indeed perpendicular to the axis of the parabola.

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

Worked Solution & Example Answer:Let $f(x) = ext{sin}^{-1}(x + 5)$ - 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

Hence deduce that

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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