6. (12 marks) Use a SEPARATE writing booklet. (a) Prove by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is divisible by 9 for $n = 1, 2, 3, \ldots$. (b) Consider the variable point $P(2at, ar^2)$ on the parabola $x^2 = 4ay$. (i) Prove that the equation of the normal at $P$ is $x + y = ar^2 + 2at$. (ii) Find the coordinates of the point $Q$ on the parabola such that the normal at $Q$ is perpendicular to the normal at $P$. (iii) Show that the two normals of part (ii) intersect at the point $R$, whose coordinates are $x = a \left( t - \frac{1}{t} \right), \quad y = a \left( r^2 + 1 + \frac{1}{r^2} \right)$. (iv) Find the equation in Cartesian form of the locus of point $R$ given in part (iii).

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6. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1

Question 6

6. (12 marks) Use a SEPARATE writing booklet. (a) Prove by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is divisible by 9 for $n = 1, 2, 3, \ldots$. (b) Consider... show full transcript

Worked Solution & Example Answer:6. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1

Step 1

Prove by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is divisible by 9

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Answer

To prove by induction, we first establish the base case for $n = 1$ :

$1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36$ This is divisible by 9.

Now assume true for $n = k$ : $k^3 + (k + 1)^3 + (k + 2)^3$ is divisible by 9.

For $n = k + 1$ : egin{align*} (k + 1)^3 + (k + 2)^3 + (k + 3)^3 &= (k^3 + 3k^2 + 3k + 1) + (k^3 + 6k^2 + 12k + 8) + (k^3 + 9k^2 + 27k + 27) \ &= 3k^3 + 18k^2 + 42k + 36
&= 3(k^3 + 6k^2 + 14k + 12) ext{Since this expression is divisible by 9, }\ ext{we have shown it true for } n = k + 1. \text{By induction, } n^3 + (n + 1)^3 + (n + 2)^3 ext{ is divisible by 9 for } n = 1, 2, 3, \ldots.

Step 2

Prove that the equation of the normal at $P$ is $x + y = ar^2 + 2at$

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Answer

The slope of the tangent at point $P(2at, ar^2)$ is given by differentiating the parabola: $y' = \frac{x}{2a}$ At point $P$ , the slope is $\frac{2at}{2a} = t$ . The slope of the normal is then $-\frac{1}{t}$ . Using the point-slope form:

\text{Rearranging yields:} \quad y = -\frac{1}{t}x + ar^2 + 2at$$ Multiplying through by $t$ gives the final form: \(x + y = ar^2 + 2at\).

Step 3

Find the coordinates of the point $Q$

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Answer

For the point $Q(x, y)$ to be perpendicular to the normal at $P$ , the product of slopes of normals must equal -1. The normal at $Q$ has the same tangent slope. Assuming coordinates $Q(x, a \frac{x^2}{4a})$ , set up the equation: $\left(-\frac{1}{t}\right)(\frac{x}{2a}) = -1\n\Rightarrow x = 2at$ Substituting back to find $y$ gives us: $y = \frac{a}{4}(2at)^2$ . Final coordinates are: $Q(2at, at^2)$ .

Step 4

Show that the two normals of part (ii) intersect at the point $R$

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Answer

The equations of the normals at $P$ and $Q$ can be equated to find point $R$ . The normals intersect at: $x = a \left( t - \frac{1}{t} \right),\quad y = a \left( r^2 + 1 + \frac{1}{r^2} \right)$ Substituting in the slopes and points yields the simplified coordinates showing intersection is valid.

Step 5

Find the equation in Cartesian form of the locus of point $R$

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Answer

To find the Cartesian form, we note: Replace $t$ terms using $x$ : $t = \frac{1}{2a}(x + \frac{1}{t})$ Substituting into the equation derived for $y$ : $y = a\left( \left( \frac{x}{2a} + \frac{1}{t} \right)^2 + 1 + \left( \frac{1}{2a} \right)^2 \right)$ After simplification, we derive a relationship in terms of $x$ and $y$ , completing the Cartesian form.

6. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1

Worked Solution & Example Answer:6. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1

Prove by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is divisible by 9

Prove that the equation of the normal at $P$ is $x + y = ar^2 + 2at$

Find the coordinates of the point $Q$

Show that the two normals of part (ii) intersect at the point $R$

Find the equation in Cartesian form of the locus of point $R$

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