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 $n \geq 1$, $$1 \cdot 3 + 2 \cdot 4 + 3 \cdot 5 + \cdots + n(n + 2) = \frac{n}{6}(n + 1)(2n + 7).$$ (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 $\angle OAC = \theta$, with $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$. (i) Show that $rac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}$. (ii) Let $m$ be the maximum value of $\frac{d\theta}{dt}$. (iii) Find the 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 mathemat... 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

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

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Answer

To sketch the graph of $y = |2x - 1|$ , identify the vertex where the expression inside the absolute value becomes zero:

Set $2x - 1 = 0$ , which gives $x = \frac{1}{2}$ .
The graph is V-shaped, opening upwards.
For $x < \frac{1}{2}$ , $y = 1 - 2x$ ; for $x \geq \frac{1}{2}$ , $y = 2x - 1$ .

Plot points around the vertex to accurately sketch the graph.

Step 2

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

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To solve the inequality $|2x - 1| \leq |x - 3|$ , consider the critical points where each expression changes:

The critical points are $\frac{1}{2}$ and $3$ .
Test intervals:
- For $x < \frac{1}{2}$ , we have $-(2x - 1) \leq -(x - 3)$ .
- For $\frac{1}{2} \leq x < 3$ , we have $2x - 1 \leq -(x - 3)$ .
- For $x \geq 3$ , we have $2x - 1 \leq x - 3$ .
Solve each case and find the solution set by combining results from all intervals.

Step 3

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

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Answer

To prove this by induction:

Base case ( $n = 1$ ):
- Left hand side: $1 \cdot 3 = 3$ .
- Right hand side: $\frac{1}{6}(1+1)(2 \cdot 1 + 7) = 3$ .
- Hence, base case holds true.
Induction step: Assume true for $n=k$ ; prove for $n = k + 1$ .
- Left hand side for $k + 1$ : $1 \cdot 3 + 2 \cdot 4 + \cdots + k(k + 2) + (k + 1)(k + 3)$ .
- Use induction hypothesis: $\frac{k}{6}(k + 1)(2k + 7) + (k + 1)(k + 3)$ .
- After simplification, show it equals $\frac{k + 1}{6}(k + 2)(2k + 9)$ , confirming the induction hypothesis.

Step 4

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

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To find $\frac{d\theta}{dt}$ , use the chain rule:

From geometric considerations, use the relationship: $\tan(\theta) = \frac{x}{\ell}$ .
Differentiate both sides: (rac{d\theta}{dt} = \frac{1}{\ell} \frac{dx}{dt}) with respect to time, where $\frac{dx}{dt} = v$ .
Hence, $\frac{d\theta}{dt} = \frac{v}{\ell^{2} + x^{2}}$ .

Step 5

(c) (ii) Let $m$ be the maximum value of $\frac{d\theta}{dt}$

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To determine the maximum value of $\frac{d\theta}{dt}$ , set conditions:

Consider the expression $\frac{d\theta}{dt} = \frac{v}{\ell^2 + x^2}$ . Since $v$ is constant, maximize by minimizing $\ell^2 + x^2$ .
The minimum occurs when $x = 0$ , leading to a maximum value of $\frac{d\theta}{dt} = \frac{v}{\ell^2}$ .

Step 6

(c) (iii) Find the two values of $\theta$ for which $\frac{d\theta}{dt} = \frac{m}{4}$

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Answer

Using the previous maximum value, we set:

$\frac{v}{\ell^{2}} = \frac{m}{4}$ , leading to $m = \frac{4v}{\ell^2}$ .
Substitute this into $\tan(\theta) = \frac{x}{\ell}$ to express values of $\theta$ in terms of $x$ .
Solve for $\theta$ considering the conditions for $x$ to find two valid angles, likely through trigonometric identities.

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

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

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

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

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

(c) (ii) Let $m$ be the maximum value of $\frac{d\theta}{dt}$

(c) (iii) Find the two values of $\theta$ for which $\frac{d\theta}{dt} = \frac{m}{4}$

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