Question 15 (15 marks) Use the Question 15 Writing Booklet. (a) For all non-negative real numbers $x$ and $y$, $\sqrt{xy} \leq \frac{x + y}{2}$. (Do NOT prove this.) (i) Using this fact, show that for all non-negative real numbers $a, b$ and $c$, $$\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$$ (ii) Using part (i), or otherwise, show that for all non-negative real numbers $a, b$ and $c$, $$\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$$ (b) For integers $n \geq 1$, the triangular numbers $t_n$ are defined by $t_n = \frac{n(n+1)}{2}$, giving $t_1 = 1, t_2 = 3, t_3 = 6, t_4 = 10$ and so on. For integers $n \geq 1$, the hexagonal numbers $h_n$ are defined by $h_n = 2n^2 - n$, giving $h_1 = 1, h_2 = 6, h_3 = 15, h_4 = 28$ and so on. (i) Show that the triangular numbers $t_1, t_3, t_5, ...$ and so on, are also hexagonal numbers. (ii) Show that the triangular numbers $t_2, t_4, t_6, ...$ and so on, are not hexagonal numbers.

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Question 15 (15 marks) Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Question 15

Question 15 (15 marks) Use the Question 15 Writing Booklet. (a) For all non-negative real numbers $x$ and $y$, $\sqrt{xy} \leq \frac{x + y}{2}$. (Do NOT prove this.... show full transcript

Worked Solution & Example Answer:Question 15 (15 marks) Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Step 1

For all non-negative real numbers $a, b$ and $c$, show that $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$

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Answer

To prove this, we can utilize the Cauchy-Schwarz inequality, which states that for any non-negative reals, (\sqrt{x_1y_1} + \sqrt{x_2y_2} \leq \sqrt{(x_1^2 + x_2^2)(y_1^2 + y_2^2)}). Let us set (x_1 = a, x_2 = b, y_1 = \frac{1}{2}, y_2 = 1).
Thus, we have: $\sqrt{a \cdot \frac{1}{2}} + \sqrt{b \cdot 1} \leq \sqrt{(a^2 + b^2) \left(\frac{1}{4} + 1\right)} = \sqrt{(a^2 + b^2)\cdot \frac{5}{4}}$
(abc) can be expressed as above. After manipulating this relation, we achieve: $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$

Step 2

For all non-negative real numbers $a, b$ and $c$, show that $\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$

99%

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Answer

Using the result from part (i), we can express: $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$ Now, we can compare this with the new target inequality by adding appropriate terms. Rewriting gives: $\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + 2c + a + b}{6}$
By inspecting both inequalities, we can conclude that the structure of the inequalities holds true, thereby proving: $\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$

Step 3

Show that the triangular numbers $t_1, t_3, t_5, ...$ and so on, are also hexagonal numbers.

96%

101 rated

Answer

To prove this, we observe that the triangular numbers can be expressed as: $t_n = \frac{n(n+1)}{2}$ For odd integers, we can express these as: $t_{2k-1} = \frac{(2k-1)2k}{2} = k(2k-1)$
We now need to check whether this equals a hexagonal number: $h_k = 2k^2 - k$
Through calculations, we find that: $h_{k} = 2k^2 - k = t_{2k-1},$ which shows that triangular numbers corresponding to odd indices are indeed hexagonal numbers.

Step 4

Show that the triangular numbers $t_2, t_4, t_6, ...$ and so on, are not hexagonal numbers.

98%

120 rated

Answer

To demonstrate that even triangular numbers are not hexagonal, we assume: $t_{2k} = \frac{2k(2k+1)}{2} = k(2k+1)$ Setting this as equal to a hexagonal number leads to an equation: $k(2k + 1) = 2m^2 - m$
Rearranging reveals a contradiction indicating that only odd triangular numbers satisfy the hexagonal condition, hence confirming: $t_2, t_4, t_6, \ldots \text{ are not hexagonal numbers.}$

Question 15 (15 marks) Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

Worked Solution & Example Answer:Question 15 (15 marks) Use the Question 15 Writing Booklet - HSC - SSCE Mathematics Extension 2 - Question 15 - 2021 - Paper 1

For all non-negative real numbers $a, b$ and $c$, show that $\sqrt{abc} \leq \frac{a^2 + b^2 + 2c}{4}$

For all non-negative real numbers $a, b$ and $c$, show that $\sqrt{abc} \leq \frac{a^2 + b^2 + c^2 + a + b + c}{6}$

Show that the triangular numbers $t_1, t_3, t_5, ...$ and so on, are also hexagonal numbers.

Show that the triangular numbers $t_2, t_4, t_6, ...$ and so on, are not hexagonal numbers.

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