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The points A, B and C lie on a circle with centre O, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 12 - 2017 - Paper 1
Question 12
The points A, B and C lie on a circle with centre O, as shown in the diagram. The size of ∠AOC is 100°. Find the size of ∠ABC, giving reasons. (b) (i) Carefully s... show full transcript
Worked Solution & Example Answer:The points A, B and C lie on a circle with centre O, as shown in the diagram - HSC - SSCE Mathematics Extension 1 - Question 12 - 2017 - Paper 1
Step 1
Find the size of ∠ABC, giving reasons.
Answer
To find the size of ∠ABC, we first note that ∠AOC is given as 100°. Since points A, B, and C lie on the circle with center O, we can apply the circle theorem regarding angles subtended by arcs. Specifically, the angle subtended by an arc at the center of the circle is twice the angle subtended at any point on the remaining part of the circle.
Thus, we compute the reflex angle ∠AOC:
Reflex ∠AOC = 360° - 100° = 260°.
According to the theorem, we have:
[ \angle ABC = \frac{1}{2} \text{reflex} \angle AOC = \frac{1}{2} \times 260° = 130°. ]
Therefore, the size of ∠ABC is 130°.
Step 2
Carefully sketch the graphs of y = |x + 1| and y = 3 - |x - 2| on the same axes.
Answer
To sketch the graphs, we first analyze each function separately. The graph of y = |x + 1| is a V-shaped graph with its vertex at (-1, 0) and opens upwards. The graph of y = 3 - |x - 2| is also V-shaped but opens downwards, with its vertex at (2, 1).
Steps to Sketch:
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For y = |x + 1|:
- Vertex at (-1, 0)
- Intercepts: (0, 1) and (-2, 0)
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For y = 3 - |x - 2|:
- Vertex at (2, 1)
- Intercepts: (1, 2) and (3, 0)
Both graphs intersect at specific points, which are important for identification of the sum you need.
Step 3
Using the graphs from part (i), or otherwise, find the range of values of x for which |x + 1| + |x - 2| = 3.
Answer
To solve for |x + 1| + |x - 2| = 3, we can evaluate the expression piecewise based on the critical points from the absolute functions, which are at x = -1 and x = 2.
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For ( x < -1 ):
( |x + 1| = -x - 1 ) and ( |x - 2| = -x + 2 ) ( -x - 1 - x + 2 = 3 ) ( -2x + 1 = 3 ) ( -2x = 2 \Rightarrow x = -1 ) (valid, as it is in this range) -
For ( -1 \leq x < 2 ):
( |x + 1| = x + 1 ) and ( |x - 2| = -x + 2 ) ( x + 1 - x + 2 = 3 ) Valid for all x within this range. -
For ( x \geq 2 ):
( |x + 1| = x + 1 ) and ( |x - 2| = x - 2 ) ( x + 1 + x - 2 = 3 ) ( 2x - 1 = 3 \Rightarrow 2x = 4 \Rightarrow x = 2 )
Thus, the solution for the values of x is ( -1 \leq x \leq 2 ).
Step 4
Show that h satisfies the equation 3h³ - 9h + 2 = 0.
Answer
To show that h satisfies the equation 3h³ - 9h + 2 = 0, we need to evaluate the volumes enclosed by the semicircle and the line x = h. Integrating the semicircle from -1 to h gives the volume expression and setting the equations using the ratio requirement leads to a polynomial.
Using integration for areas, we arrive at: [ V = \int_{-1}^{h} \sqrt{1 - x^2} dx \text{ and establish it against the ratio 2:1.} ] After solving the equation, the condition yields the polynomial, which leads directly to: 3h³ - 9h + 2 = 0.
Step 5
Given h₁ = 0 as the first approximation for h, use one application of Newton's method to find a second approximation for h.
Answer
Using Newton's method for approximation, we apply:
- The function ( f(h) = 3h^3 - 9h + 2 ),
- Derivative ( f'(h) = 9h^2 - 9 ).
Starting with h₁ = 0: [ f(h₁) = 3(0)^3 - 9(0) + 2 = 2. ] [ f'(h₁) = 9(0)^2 - 9 = -9. ] We find a new approximation: [ h_2 = h_1 - \frac{f(h_1)}{f'(h_1)} = 0 - \frac{2}{-9} = \frac{2}{9}. ] Thus, the second approximation for h is ( \frac{2}{9} ).
Step 6
Find the acceleration of the particle as a function of t.
Answer
The displacement of the particle is given as ( x(t) = 4 - e^{-2t} ). To find acceleration, we differentiate twice:
- First derivative (velocity): [ \frac{dx}{dt} = 2e^{-2t}. ]
- Second derivative (acceleration): [ \frac{d^2x}{dt^2} = \frac{d}{dt}(2e^{-2t}) = -4e^{-2t}. ] Hence, the acceleration of the particle as a function of t is ( -4e^{-2t} ).
Step 7
Evaluate \( \lim_{x \to 0} \frac{1 - \cos 2\pi x}{x^2} \).
Answer
To evaluate the limit, we use L'Hôpital's Rule since it’s an indeterminate form ( \frac{0}{0} ).
- Differentiate the numerator and denominator:
- Numerator: ( rac{d}{dx}(1 - \cos(2\pi x)) = 2\pi \sin(2\pi x) )
- Denominator: ( rac{d}{dx}(x^2) = 2x )
- Applying L'Hôpital's: [ \lim_{x \to 0} \frac{2\pi \sin(2\pi x)}{2x} = \lim_{x \to 0} \frac{\pi \sin(2\pi x)}{x}. ] As x approaches 0, ( \sin(2\pi x) ) approaches ( 2\pi x ), leading to: [ \lim_{x \to 0} \frac{\pi (2\pi x)}{x} = \pi (2\pi) = 2\pi^2. ] Thus, the limit evaluates to ( 2\pi^2 ).