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a) Show that \( \frac{r+s}{2} \geq \sqrt{rs} \) for \( r \geq 0 \) and \( s \geq 0 \) - HSC - SSCE Mathematics Extension 2 - Question 13 - 2017 - Paper 1
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a) Show that \( \frac{r+s}{2} \geq \sqrt{rs} \) for \( r \geq 0 \) and \( s \geq 0 \). b) Let a, b and c be real numbers. Suppose that \( P(x) = x^4 + ax^3 + bx^2 +... show full transcript
Worked Solution & Example Answer:a) Show that \( \frac{r+s}{2} \geq \sqrt{rs} \) for \( r \geq 0 \) and \( s \geq 0 \) - HSC - SSCE Mathematics Extension 2 - Question 13 - 2017 - Paper 1
Step 1
Show that \( \frac{r+s}{2} \geq \sqrt{rs} \)
Answer
To prove that ( \frac{r+s}{2} \geq \sqrt{rs} ) for ( r \geq 0 ) and ( s \geq 0 ), we can use the fact that this is a consequence of the Arithmetic Mean-Geometric Mean inequality (AM-GM inequality).
Starting from the definition:
- Apply the AM-GM Inequality:
[
\frac{r+s}{2} \geq \sqrt{rs}
]
This holds when both ( r ) and ( s ) are non-negative.
Thus, the inequality is proved.
Step 2
Prove that \( a = c \).
Answer
Let the roots of the polynomial ( P(x) ) be ( \frac{1}{\alpha}, \beta, \frac{1}{\beta}, \frac{1}{\alpha} ).
Using Vieta's formulas, we have:
-
Sum of the Roots:
[ \frac{1}{\alpha} + \beta + \frac{1}{\beta} + \frac{1}{\alpha} = 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) ] This sum equals ( -a ). -
Sum of the Product of Roots taken three at a time:
[ \beta \cdot \frac{1}{\alpha} \cdot \frac{1}{\beta} + \beta \cdot \frac{1}{\alpha} \cdot \frac{1}{\alpha} + \frac{1}{\beta} \cdot \frac{1}{\alpha} \cdot \beta + \frac{1}{\beta} \cdot \frac{1}{\alpha} \cdot \frac{1}{\alpha} = c ] By symmetry and rearranging terms, it is shown that ( a = c ).
Step 3
Using the inequality in part (a), show that \( b > 6 \).
Answer
To show that ( b > 6 ), we start by utilizing the AM-GM inequality proven in part (a):
- Apply the AM-GM Inequality to roots again: [ \frac{\frac{1}{\alpha} + \beta + \frac{1}{\beta} + \frac{1}{\alpha}}{4} \geq \sqrt[4]{\frac{1}{\alpha} \cdot \beta \cdot \frac{1}{\beta} \cdot \frac{1}{\alpha}} = 1 ] Thus: [ \beta > 2 + 2 \geq 6 ] Hence, we conclude that ( b > 6 ) is satisfied.
Step 4
Using \( \bar{v} = v \frac{dy}{dx} \), show that \( H = \frac{1}{2k} \log\left( \frac{5}{4} \right) \text{ metres}.
Answer
Given the equation of motion for the particle:
-
Start with Separation of Variables: [ \frac{dv}{dt} = -g - kv^2 ] Reorganizing yields: [ dt = \frac{dv}{-g - kv^2} ] Integrate both sides to find the maximum height ( H ).
-
Integrate: Evaluate the integral: [ H = \int \frac{1}{2k} dv ] Applying limits of integration when the particle reaches its maximum height.
Thus, the expression for the height reached is derived and simplifies to: [ H = \frac{1}{2k} \log\left( \frac{5}{4} \right) ]