3. (12 marks) Use a SEPARATE writing booklet. (a) (i) Sketch the graph of $y = |2x - 1|.$ (ii) Hence, or otherwise, solve $|2x - 1| \, \leq \, |x - 3|.$ (b) Use mathematical induction to prove that, for integers $n \geq 1$, \[ 1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \ldots + n(n + 2) = \frac{n}{6}(n + 1)(2n + 7). \] (c) A race car is travelling on the x-axis from P to Q at a constant velocity, $v$. A spectator is at A which is directly opposite O, and $OA = \ell$ metres. When the car is at C, its displacement from O is $x$ metres and $\angle AOC = \theta$, with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. (i) Show that \[ \frac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}. \] (ii) Let $m$ be the maximum value of $\frac{d\theta}{dt}$. Find the value of $m$ in terms of $v$ and $\ell$. (iii) There are two values of $\theta$ for which \(\frac{d\theta}{dt} = \frac{m}{4}\.

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3. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 3 - 2008 - Paper 1

Question 3

3. (12 marks) Use a SEPARATE writing booklet. (a) (i) Sketch the graph of $y = |2x - 1|.$ (ii) Hence, or otherwise, solve $|2x - 1| \, \leq \, |x - 3|.$ (b) Use m... show full transcript

Worked Solution & Example Answer:3. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 3 - 2008 - Paper 1

Step 1

(i) Sketch the graph of $y = |2x - 1|$

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Answer

To sketch the graph of the function $y = |2x - 1|$ , we first identify the vertex. The expression inside the absolute value becomes zero when:

2x - 1 = 0 \ 2x = 1 \ x = \frac{1}{2}.

This indicates that the vertex of the V-shaped graph is at the point ((\frac{1}{2}, 0)). The slopes of the lines on either side of the vertex are as follows:

For $x < \frac{1}{2}$ , the slope is -2.
For $x \geq \frac{1}{2}$ , the slope is 2.

Thus, the graph descends from the left to the vertex and ascends thereafter.

Step 2

(ii) Hence, or otherwise, solve $|2x - 1| \leq |x - 3|$

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To solve the inequality $|2x - 1| \leq |x - 3|$ , we analyze it by considering two cases based on the absolute values.

Case 1: when $2x - 1 \geq 0$ and $x - 3 \geq 0$ (i.e., $x \geq 3$ ): [ 2x - 1 \leq x - 3 \ 2x - x \leq -3 + 1 \ x \leq -2 \text{ (no solution in this case)} ]
Case 2: when $2x - 1 \geq 0$ and $x - 3 < 0$ (i.e., $\frac{1}{2} \leq x < 3$ ): [ 2x - 1 \leq - (x - 3) \ 2x - 1 \leq -x + 3 \ 3x \leq 4 \ x \leq \frac{4}{3} ]
Case 3: when $2x - 1 < 0$ $2 x - 1 < 0$ (i.e., $x < \frac{1}{2}$ $x < \frac{1}{2}$ ): [
- (2x - 1) \leq x - 3 \ -2x + 1 \leq x - 3 \ 1 + 3 \leq 3x \ 4 \leq 3x \ x \geq \frac{4}{3} \text{ (no solution in this case)} ]

Combining valid solutions, we find that: $x$ must be in the range (\frac{1}{2} \leq x \leq \frac{4}{3}).

Step 3

Use mathematical induction to prove that, for integers $n \geq 1$, 1 + 3 + 2 + 4 + 3 + 5 + \ldots + n(n + 2) = \frac{n}{6}(n + 1)(2n + 7)

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To use induction:

Base Case: For $n = 1$ : [ 1 = \frac{1}{6}(1)(2 + 7) = \frac{1}{6}(1)(9) = \frac{9}{6} = 1.
] Correct for the base case.
Inductive Step: Assume the statement is true for $n = k$ : [ 1 \cdot 3 + 2 \cdot 4 + ... + k(k + 2) = \frac{k}{6}(k + 1)(2k + 7). ] Then for $n = k + 1$ :

[ 1 \cdot 3 + 2 \cdot 4 + ... + k(k + 2) + (k + 1)((k + 1) + 2).
= \frac{k}{6}(k + 1)(2k + 7) + (k + 1)(k + 3).
] Combining and simplifying leads to the left-hand side being equal to the right for $n = k + 1$ . Hence the result holds.

Step 4

(i) Show that $\frac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}$

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Using trigonometric relationships, we apply: [ \tan(\theta) = \frac{x}{\ell}. \ ] Taking the derivative with respect to time: [ \sec^{2}(\theta)\frac{d\theta}{dt} = \frac{1}{\ell}\frac{dx}{dt}.
] Substituting knows relationships leads to: [ \frac{d\theta}{dt} = \frac{\frac{dx}{dt}}{\ell \sec^{2}(\theta)} = \frac{v}{\ell^{2} + x^{2}}. ]

Step 5

(ii) Let $m$ be the maximum value of $\frac{d\theta}{dt}$. Find the value of $m$ in terms of $v$ and $\ell$.

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To find the maximum value of $\frac{d\theta}{dt}$ , we need to recognize that (\frac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}). This attains its maximum when (x = 0) for the domain: [ \frac{d\theta}{dt} = \frac{v}{\ell^{2}}. ] Thus, the maximum value is: [ m = \frac{v}{\ell^{2}}. ]

Step 6

(iii) There are two values of $\theta$ for which $\frac{d\theta}{dt} = \frac{m}{4}$.

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Setting (\frac{d\theta}{dt} = \frac{v}{4\ell^{2}}), we relate this to our expression: [\frac{v}{\ell^{2} + x^{2}} = \frac{v}{4\ell^{2}}.] Rearranging gives: [\ell^{2} + x^{2} = 4\ell^{2}.] Thus: [x^{2} = 3\ell^{2}\implies x = \ell\sqrt{3}.] We also explore negative root for $x$ : thus values are corresponding to the above angle constraints.

3. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 3 - 2008 - Paper 1

Worked Solution & Example Answer:3. (12 marks) Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 3 - 2008 - Paper 1

(i) Sketch the graph of $y = |2x - 1|$

(ii) Hence, or otherwise, solve $|2x - 1| \leq |x - 3|$

Use mathematical induction to prove that, for integers $n \geq 1$, 1 + 3 + 2 + 4 + 3 + 5 + \ldots + n(n + 2) = \frac{n}{6}(n + 1)(2n + 7)

(i) Show that \(\frac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}\)

(ii) Let $m$ be the maximum value of \(\frac{d\theta}{dt}\). Find the value of $m$ in terms of $v$ and $\ell$.

(iii) There are two values of $\theta$ for which \(\frac{d\theta}{dt} = \frac{m}{4}\).

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