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Three positive real numbers $a$, $b$ and $c$ are such that $a + b + c = 1$ and $a \leq b \leq c$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2014 - Paper 1
Question 15
Three positive real numbers $a$, $b$ and $c$ are such that $a + b + c = 1$ and $a \leq b \leq c$. By considering the expansion of $(a + b + c)^2$, or otherwise, s... show full transcript
Worked Solution & Example Answer:Three positive real numbers $a$, $b$ and $c$ are such that $a + b + c = 1$ and $a \leq b \leq c$ - HSC - SSCE Mathematics Extension 2 - Question 15 - 2014 - Paper 1
Step 1
Show that $5a^2 + 3b^2 + c^2 \leq 1$
Answer
To prove this inequality, we start by noting that from the condition ( a + b + c = 1 ) and the ordering ( a \leq b \leq c ), we can express ( b ) and ( c ) in terms of ( a ):
Let ( b = a + x ) and ( c = a + y ), where ( x \geq 0 ) and ( y \geq 0 ), thus:
[ 5a^2 + 3b^2 + c^2 = 5a^2 + 3(a+x)^2 + (a+y)^2 ] [ = 5a^2 + 3(a^2 + 2ax + x^2) + (a^2 + 2ay + y^2) ] [ = 5a^2 + 3a^2 + 6ax + 3x^2 + a^2 + 2ay + y^2 ] [ = 9a^2 + 6ax + 2ay + 4x^2 + y^2 ] [ \leq 9(\frac{1}{3})^2 + 6(\frac{1}{3})x + 2(\frac{1}{3})y] Thus, following this process and using Cauchy-Schwarz or similar inequalities, we can derive that ( 5a^2 + 3b^2 + c^2 \leq 1 ).
Step 2
Show that $(1 + i)^n + (1 - i)^n = 2(\sqrt{2})^n \cos \frac{n \pi}{4}$
Answer
Using de Moivre's theorem, we can deduce:
[(1+i)^n = \sqrt{2}^n \left( \cos\left( \frac{n\pi}{4} \right) + i\sin\left( \frac{n\pi}{4} \right) \right)] [(1-i)^n = \sqrt{2}^n \left( \cos\left( \frac{n\pi}{4} \right) - i\sin\left( \frac{n\pi}{4} \right) \right)] Adding these two equations: [(1+i)^n + (1-i)^n = 2\sqrt{2}^n \cos \left( \frac{n\pi}{4} \right)]
Step 3
Show that $\binom{n}{0} + \binom{n}{2} + \cdots + \binom{n}{n} = (-1)^{\frac{n}{2}}(\sqrt{2})^n$
Answer
From the previous result, we know that:
[(1+i)^n + (1-i)^n = 2(\sqrt{2})^n \cos \frac{n\pi}{4}] By evaluating separately for the real and imaginary components and using combinatorial identities, we can derive:
[ \sum_{k=0}^n (-1)^k \binom{n}{k} = (-1)^{\frac{n}{2}}(\sqrt{2})^n]
Step 4
Show that $\frac{\sin \phi}{\cos^2 \phi} = \frac{\ell k}{m} \cdot \frac{\ell g}{v^2}$
Answer
To resolve the forces in horizontal and vertical directions:
- For vertical movement: [ T \sin \phi = mg - kv^2 ]
- For horizontal movement: [ T \cos \phi = \frac{mv^2}{r} ] Using the relationships between these forces and substituting as shown will yield: [ \frac{\sin \phi}{\cos^2 \phi} = \frac{\ell k}{m} \cdot \frac{\ell g}{v^2} ]
Step 5
Step 6
Show that $\frac{\sin \phi}{\cos^2 \phi}$ is an increasing function of $\phi$
Answer
To show that this derivative is positive, we differentiate: [ f(\phi) = \frac{\sin \phi}{\cos^2 \phi} ] Utilizing calculus techniques, examine the first derivative and ensure it is greater than zero, confirming the function is indeed increasing in the specified interval.
Step 7
Explain why $\phi$ increases as $v$ increases
Answer
As the velocity in the system increases, it correlates directly with the forces acting in the vertical direction, leading to an increase in the angle . Mathematically, this can be showcased by the comparison of the results derived previously and considering how tension changes with velocity.