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
Question 13
a) Show that \( \frac{r+s}{2} \geq \sqrt{rs} \) for \( r \geq 0 \) and \( s \geq 0 \).
b) Let \( a, b, c \) be real numbers. Suppose that \( P(x) = x^4 + ax^3 + bx^... 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 \( \frac{r+s}{2} \geq \sqrt{rs} \)
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Answer
To demonstrate ( \frac{r+s}{2} \geq \sqrt{rs} ), we can employ the AM-GM inequality. For any non-negative real numbers ( r ) and ( s ), we have:
2r+s≥rs
This holds true, as equality occurs when ( r = s ) which is the desired inequality.
Step 2
(i) Prove that \( a = c \)
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Answer
Given roots ( \alpha, \frac{1}{\alpha}, \beta, \frac{1}{\beta} ), we know that the sum of the roots ( \alpha + \frac{1}{\alpha} + \beta + \frac{1}{\beta} = -a ).
Using Vieta's formulas, we have:
( \alpha \cdot \frac{1}{\alpha} + \beta \cdot \frac{1}{\beta} = b )
Thus:
( 1 + 1 = b ) implies ( b = 2 )
Since both sums equal ( -a = -c ), it follows that ( a = c ).
Step 3
(ii) Using the inequality in part (a), show that \( b > 6 \)
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Answer
Using the inequality from part (a) with roots relations, we see that:
2β+β1≥β⋅β1=1
It follows that ( \beta + \frac{1}{\beta} \geq 2 ) also holds. Then,
Adding these terms gives:
( b = 2 + 2 ) implies ( b > 6 ).
Step 4
Using \( \frac{v}{dx} \) or otherwise, show that \( H = \frac{1}{2k} \log \left( \frac{5}{4} \right) \) metres.
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Answer
We start with:
( \frac{dv}{dt} = -g - kv^2 )
Integrating and separating variables:
∫g+kv21dv=−∫dt
Using initial conditions, we can find maximum height ( H ) as:
H=2k1log(45)
This derives from energy balance, applying limits during the motion of the particle.
