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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) Consid... 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 start with the base case:

For $n = 1$ : $1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36,$ which is divisible by 9.

Now, assume it holds for $n = k$ , where: $k^3 + (k + 1)^3 + (k + 2)^3 = 9m,$ for some integer $m$ .

For $n = k + 1$ : $(k + 1)^3 + (k + 2)^3 + (k + 3)^3$ can be expanded to: $(k^3 + 3k^2 + 3k + 1) + (k^3 + 6k^2 + 12k + 8) + (k^3 + 9k^2 + 27k + 27).$

Summing these gives: $3k^3 + 18k^2 + 42k + 36.$

Factoring out the common terms: $3(k^3 + 6k^2 + 14k + 12)$ which shows that it is divisible by 3. Since the inductive hypothesis confirms that the first part is divisible by 9, we conclude that the overall expression is also divisible by 9. Thus, the proposition holds for all positive integers $n$ .

Step 2

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

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Answer

To find the normal at point $P(2at, ar^2)$ , we first determine the slope of the tangent. The derivative of $y$ with respect to $x$ for the parabola $x^2 = 4ay$ is: $\frac{dy}{dx} = \frac{x}{2a}.$

At $P$ , substituting $x = 2at$ , we get: $\frac{dy}{dx} = \frac{2at}{2a} = t.$

The slope of the normal is the negative reciprocal: $-\frac{1}{t}.$

Using point-slope form, the equation of the normal is: $y - ar^2 = -\frac{1}{t}(x - 2at).$

Simplifying, we get: $ty + ar^2t = -x + 2a$ which rearranges to: $x + y = ar^2 + 2at.$ This proves the normal's equation.

Step 3

Find the coordinates of the point Q on the parabola such that the normal at Q is perpendicular to the normal at P

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Answer

To find the coordinates of point $Q$ , denote it as $(2as, as^2)$ on the parabola. For the normals at $P$ and $Q$ to be perpendicular, the product of their slopes must equal -1: $m_1 \cdot m_2 = -1.$

The slope at $Q$ is: $\frac{2as}{2a} = s.$

Thus, $-\frac{1}{t} \cdot s = -1$ (for perpendicularity), leading to: $s = t.$

So, the coordinates of $Q$ can be found by substituting $s = t$ into the equation of the parabola: $Q = (2at, at^2).$

Step 4

Show that the two normals of part (ii) intersect at the point R, whose coordinates are given

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Answer

Finding the normals from points $Q_1$ and $Q_2$ should yield two equations:

Normal at $Q_1$
Normal at $Q_2$

Both use the form $y - y_1 = m(x - x_1).$ By finding the intersection of these two lines:

Set the equations equal and solve for the coordinates of $R$ .
Rearranging those should yield $R$ coordinates: $x = a\left( t - \frac{1}{t} \right), \quad y = a\left( t^2 + 1 + \frac{1}{t^2} \right).$

Step 5

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

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Answer

To find the locus of point $R$ , we replace $t$ with $\frac{x}{2a}$ and transform the $y$ relationship accordingly:

From: $y = a\left(\left(\frac{x}{2a}\right)^2 + 1 + \frac{1}{\left(\frac{x}{2a}\right)^2}\right)$
Rearranging will yield a Cartesian equation in $x$ and $y$ . Collect terms and simplify to present the locus of normal points.

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 on the parabola such that the normal at Q is perpendicular to the normal at P

Show that the two normals of part (ii) intersect at the point R, whose coordinates are given

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

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