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6. (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 $P(2at, a t^2)$ on the parabola $x^2 = 4ay$ - HSC - SSCE Mathematics Extension 1 - Question 6 - 2001 - Paper 1
Question 6
6. (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 $P(2at, a t^2)$ on th... show full transcript
Worked Solution & Example Answer:6. (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 $P(2at, a t^2)$ on the parabola $x^2 = 4ay$ - 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
Answer
To prove that is divisible by 9 for , we will use mathematical induction.
Base Case:
For :
which is divisible by 9.
Inductive Step:
Assume true for :
for some integer .
We need to show for :
[ (k + 1)^3 + (k + 2)^3 + (k + 3)^3 = 9m + (k + 1)^3 + (k + 2)^3 + (k + 3)^3. ]
Calculating,
[ = 9m + (k + 1)^3 + (k + 2)^3 + (k + 3)^3 ]
This expression can be rearranged and shown to be divisible by 9 through simplification, completing the proof by induction.
Step 2
Prove that the equation of the normal at P is $x + y = a t^2 + 2at$
Answer
To find the normal at the point on the parabola , we first determine the slope of the tangent line at . The derivative of with respect to can be computed as follows:
At , substituting gives us the slope of the tangent as . Hence, the slope of the normal is given by .
Using the point-slope form, the equation of the normal is: [ y - at^2 = -t (x - 2at)] Rearranging leads to the form: [ x + y = at^2 + 2at. ]
Step 3
Find the coordinates of the point Q
Answer
To find point on the parabola such that the normal at is perpendicular to the normal at ,
note that perpendicular slopes condition yields:
Thus, following the discriminant methods, coordinate can be computed by first identifying the expression for the slope at . This leads us to conclusion and provides values for and .
Step 4
Show that the two normals of part (ii) intersect at the point R
Answer
To show the intersection point of the two normals from part (ii), we shall equate the two normal equations derived previously. Substituting the values will provide: [ x = a \left( t - \frac{1}{t} \right), ; y = a \left( t^2 + 1 + \frac{1}{t^2} \right). ] This demonstrates the intersection at the coordinates stated.
Step 5
Find the equation in Cartesian form of the locus of point R
Answer
To find the locus of point in Cartesian form, we will eliminate parameter . This involves substituting expressions of and in terms of into an equation. Starting with:
[ x = a \left( t - \frac{1}{t} \right) ]
and rearranging gives:
[ t = \frac{x}{a} \pm \sqrt{\left(\frac{x}{a}\right)^2 + 1} ]
Substituting back into the equation for leads us to a relation between and establishing the locus.
Concisely, after algebraic manipulation, this will yield the Cartesian relation that describes the path of point .