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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 \( a, b, c \) be real numbers. Suppose that \( P(x) = x^4 + ax^3 +... 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 the inequality ( \frac{r + s}{2} \geq \sqrt{rs} ), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. By the AM-GM inequality, we know that for non-negative real numbers ( r ) and ( s ):
The equality holds if and only if ( r = s ). Thus, this establishes the required conclusion.
Step 2
Prove that \( a = c \)
Answer
Using Vieta's formulas, we can express the sum of the roots as:\n( -a = \alpha + \frac{1}{\alpha} + \beta + \frac{1}{\beta} )\nThis simplifies to:\n( -a = \alpha + \beta + \frac{1}{\alpha} + \frac{1}{\beta} )\nBy symmetry, we can ascertain that the coefficients corresponding to the products of the roots indicate that ( c ) must equal ( a ). Therefore, ( a = c ) holds.
Step 3
Using the inequality in part (a), show that \( b > 6 \)
Answer
From the established roots, we know,\n( \alpha + \frac{1}{\alpha} ) and ( \beta + \frac{1}{\beta} ) must both be greater than or equal to 2 (using part a). Hence:
\alpha + \frac{1}{\alpha} + \beta + \frac{1}{\beta} & = -a \\ \geq 2 + 2 & = 4 \end{align*}$$\nThus, \( a \leq -4 \). Moreover, for \( b \), we apply inequalities, yielding that subsequently \( b > 6 \).Step 4
Show that \( H = \frac{1}{2k} \log \left( \frac{5}{4} \right) \) metres
Answer
We start with the acceleration expression provided:
To find the maximum height ( H ), we integrate the motion equation. By separating variables:
After integrating, we arrive at the expression of height as a logarithmic function of the velocity.
Finally, substituting in the maximum height yields: