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1. (a) Write \((1 + \sqrt{5})^3\) in the form \(a + b\sqrt{5}\), where \(a\) and \(b\) are integers - HSC - SSCE Mathematics Extension 1 - Question 1 - 2007 - Paper 1
Question 1
1. (a) Write \((1 + \sqrt{5})^3\) in the form \(a + b\sqrt{5}\), where \(a\) and \(b\) are integers. (b) The interval \(AB\), where \(A\) is \((4, 5)\) and \(B\) is... show full transcript
Worked Solution & Example Answer:1. (a) Write \((1 + \sqrt{5})^3\) in the form \(a + b\sqrt{5}\), where \(a\) and \(b\) are integers - HSC - SSCE Mathematics Extension 1 - Question 1 - 2007 - Paper 1
Step 1
Write \((1 + \sqrt{5})^3\) in the form \(a + b\sqrt{5}\)
Answer
To express ((1 + \sqrt{5})^3) in the desired form, we will expand it using the binomial theorem:
[ (1 + \sqrt{5})^3 = \sum_{k=0}^{3} \binom{3}{k}(1)^{3-k} (\sqrt{5})^k ]
Calculating each term:
- For (k=0): (\binom{3}{0}(1)^3(\sqrt{5})^0 = 1)
- For (k=1): (\binom{3}{1}(1)^2(\sqrt{5})^1 = 3\sqrt{5})
- For (k=2): (\binom{3}{2}(1)^1(\sqrt{5})^2 = 3 \times 5 = 15)
- For (k=3): (\binom{3}{3}(1)^0(\sqrt{5})^3 = \sqrt{125} = 5\sqrt{5})
Combining the results, we have:
[
1 + 3\sqrt{5} + 15 + 5\sqrt{5} = 16 + 8\sqrt{5}
]
Thus, (a = 16) and (b = 8).
Step 2
The interval \(AB\), where \(A\) is \((4, 5)\) and \(B\) is \((19, -5)\), is divided internally in the ratio \(2 : 3\) by the point \(P(x, y)\)
Answer
To find the coordinates of point (P(x, y)) dividing segment (AB) in the ratio (2:3), we use the section formula:
[
x = \frac{m_2x_1 + m_1x_2}{m_1 + m_2}, \quad y = \frac{m_2y_1 + m_1y_2}{m_1 + m_2}
]
Here, (m_1 = 2), (m_2 = 3), (A(4, 5)) and (B(19, -5)).
Calculating (x):
[
x = \frac{3(4) + 2(19)}{2 + 3} = \frac{12 + 38}{5} = \frac{50}{5} = 10
]
Calculating (y):
[
y = \frac{3(5) + 2(-5)}{2 + 3} = \frac{15 - 10}{5} = \frac{5}{5} = 1
]
Thus, the values of (x) and (y) are (10) and (1), respectively.
Step 3
Differentiate \(\tan^{-1}(x^4)\) with respect to \(x\)
Answer
To differentiate (\tan^{-1}(x^4)), we apply the chain rule:
[
\frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}
]
Where (u = x^4), thus (\frac{du}{dx} = 4x^3). Therefore:
[
\frac{d}{dx} \tan^{-1}(x^4) = \frac{1}{1 + (x^4)^2} \cdot 4x^3 = \frac{4x^3}{1 + x^8}
]
Step 4
The graphs of the line \(-2y + 3 = 0\) and the curve \(y = x^3 + 1\) intersect at \((1, 2)\). Find the exact value, in radians, of the acute angle between the line and the tangent to the curve at the point of intersection.
Answer
To find the angle (\theta) between the line and the tangent line to the curve at the point of intersection, we first need to determine the slope of the line and the slope of the tangent to the curve.
-
The line (-2y + 3 = 0) simplifies to (y = \frac{3}{2}), with slope (m_1 = 0).
-
Next, we differentiate the curve (y = x^3 + 1): [\frac{dy}{dx} = 3x^2]
Evaluating at (x = 1):
[\frac{dy}{dx} \Big|_{x=1} = 3(1^2) = 3] Thus, the slope of the tangent line is (m_2 = 3). -
The tangent of the angle (\theta) between the two lines is given by: [ tan(\theta) = \left| \frac{m_2 - m_1}{1 + m_1 m_2} \right| = \left| \frac{3 - 0}{1 + 0} \right| = 3
] -
Finally, we find the angle (\theta): [\theta = \tan^{-1}(3)]
Thus, the exact value of the acute angle in radians is (\theta = \tan^{-1}(3)).
Step 5
Use the substitution \(u = 25 - x^2\) to evaluate \(\int_{3}^{4} \frac{2x}{\sqrt{25 - x^2}} dx\)
Answer
For the integral (\int_{3}^{4} \frac{2x}{\sqrt{25 - x^2}} dx), we will use the substitution (u = 25 - x^2), which gives:
[ du = -2x dx \quad \Rightarrow \quad dx = \frac{du}{-2x} ]
Next, we change the limits when substituting:
- When (x = 3): (u = 25 - 3^2 = 16)
- When (x = 4): (u = 25 - 4^2 = 9)
Now substituting into the integral:
[ \int_{16}^{9} \frac{2x}{\sqrt{u}} \left(\frac{du}{-2x}\right) = -\int_{16}^{9} \frac{1}{\sqrt{u}} du = \int_{9}^{16} \frac{1}{\sqrt{u}} du ]
This evaluates to: [\left[ 2\sqrt{u} \right]_{9}^{16} = 2(4) - 2(3) = 8 - 6 = 2]
Thus, the value of the integral is (2).