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2. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1
Question 2
2. Use a SEPARATE writing booklet. (a) By using the substitution $t = \tan \frac{\theta}{2}$, show that \(\frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{... show full transcript
Worked Solution & Example Answer:2. Use a SEPARATE writing booklet - HSC - SSCE Mathematics Extension 1 - Question 2 - 2007 - Paper 1
Step 1
By using the substitution $t = \tan \frac{\theta}{2}$, show that \(\frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2}\).
Answer
To show that (\frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2}), we utilize the substitution (t = \tan \frac{\theta}{2}). Using the half-angle formulas, we have:
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(\sin \theta = \frac{2t}{1+t^2}) and (\cos \theta = \frac{1-t^2}{1+t^2}).
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Thus, (1 - \cos \theta = 1 - \frac{1-t^2}{1+t^2} = \frac{2t^2}{1+t^2}).
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Now substituting these results, we get:
[ \frac{1 - \cos \theta}{\sin \theta} = \frac{\frac{2t^2}{1+t^2}}{\frac{2t}{1+t^2}} = \frac{t}{1} = t. ]
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Therefore, this shows that (\frac{1 - \cos \theta}{\sin \theta} = \tan \frac{\theta}{2} = t).
Step 2
Let \(f(x) = 2\cos^{-1} x\). (i) Sketch the graph of \(y = f(x)\), indicating clearly the coordinates of the endpoints of the graph.
Answer
The function (f(x) = 2\cos^{-1} x) is defined for (-1 \leq x \leq 1), as (\cos^{-1} x) is only valid within this interval.
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The endpoints of the graph occur at:
- When (x = -1): (f(-1) = 2\cos^{-1}(-1) = 2 \cdot \pi = 2\pi).
- When (x = 1): (f(1) = 2\cos^{-1}(1) = 0).
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The graph will start from (( -1, 2\pi)) and end at ((1, 0)), and is generally a decreasing curve.
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Sketching this function will show a smooth curve connecting these two points.
Step 3
State the range of \(f(x)\).
Answer
The range of the function (f(x) = 2\cos^{-1} x) is ([0, 2\pi]). This is because:
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At (x = 1), the value of (f(x) = 0) and
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At (x = -1), the value of (f(x) = 2\pi).
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Since the function is continuous and decreasing on the interval ([-1, 1]), it takes all values between 0 and (2\pi).
Step 4
The polynomial \(P(x) = x^{2} + ax + b\) has a zero at \(x = 2\). When \(P(x)\) is divided by \(x + 1\), the remainder is 18. Find the values of a and b.
Answer
To find the values of (a) and (b):
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Since (P(2) = 0): (P(2) = 2^2 + 2a + b = 0) → (4 + 2a + b = 0) → (2a + b = -4) (i)
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By the Remainder Theorem, (P(-1) = 18): (P(-1) = (-1)^2 + a(-1) + b = 18) → (1 - a + b = 18) → (-a + b = 17) (ii)
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Now we solve the system:
- From (i): (b = -4 - 2a)
- Substituting into (ii): (-a + (-4 - 2a) = 17) → (-3a - 4 = 17) → (-3a = 21) → (a = -7, b = 10).
Step 5
A skydiver jumps from a hot air balloon which is 2000 metres above the ground. The velocity, \(v\) metres per second, at which she is falling \(t\) seconds after jumping is given by \(v = 50(1 - e^{-2t})\). (i) Find her acceleration ten seconds after she jumps. Give your answer correct to one decimal place.
Answer
To find the acceleration, we need to differentiate the velocity function:
- Given: (v = 50(1 - e^{-2t})).
- Differentiating (v) with respect to (t): (a = \frac{dv}{dt} = 50 \cdot 2e^{-2t} = 100e^{-2t}).
- Evaluating at (t = 10): (a = 100e^{-20}).
- Approximating: (e^{-20} \approx 1.38 \times 10^{-9}), so (a \approx 0.0000138) (very small), thus the acceleration is around (0.0) m/s² to one decimal place.
Step 6
Find the distance that she has fallen in the first ten seconds. Give your answer correct to the nearest metre.
Answer
The distance fallen can be found by integrating the velocity function:
- Distance fallen (s) from (t = 0) to (t = 10): (s = \int_{0}^{10} v dt = \int_{0}^{10} 50(1 - e^{-2t}) dt).
- Evaluating the integral:
- (\int 50 dt = 50t |_{0}^{10} = 500)
- (\int 50e^{-2t} dt = -25e^{-2t}|_{0}^{10} = -25(e^{-20} - 1) \approx -25(-1) = 25)
- Hence: (s = 500 + 25 = 525), which is the total distance fallen.
- The answer should be rounded to the nearest metre:
- The skydriver falls approximately 525 metres.