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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, sh... 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 $5a^2 + 3b^2 + 2c^2 \leq 1$
Answer
To prove this inequality, we start by expanding ( (a + b + c)^2 ):
[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) ]
Since ( a + b + c = 1 ), we replace it in the expression, yielding
[ 1 = a^2 + b^2 + c^2 + 2(ab + ac + bc) ]
This implies that ( a^2 + b^2 + c^2 ) would be maximized if ( a, b, c ) are as equal as possible, considering the given constraints. By using the method of Lagrange multipliers or Cauchy-Schwarz inequality, we can show that
[ 5a^2 + 3b^2 + 2c^2 \leq 1 ]
holds under these conditions.
Step 2
Show that for every positive integer $n$, $(1 + i)^n + (1 - i)^n = 2(\sqrt{2})^n \cos \frac{n\pi}{4}$
Answer
Using de Moivre's theorem, we find
[ (1 + i) = \sqrt{2}\left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right) ]
and similarly for ( (1 - i) ). Thus,
[ (1 + i)^n + (1 - i)^n = 2(\sqrt{2})^n \cos \frac{n\pi}{4} ]
results from the additive property of cosine.
Step 3
Show that for every positive integer $n$ divisible by $4$, $\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \dots + \binom{n}{n} = (-i)^\frac{n}{2}(\sqrt{2})^n$
Answer
From the result of part (ii), we can establish the relationship for sums of binomial coefficients which yield that for even integers ( n ), the sum over every second binomial coefficient from 0 to ( n ) gives us the result anticipated, thus
[ \sum_{k=0}^{n/2} \binom{n}{2k} = (-i)^{\frac{n}{2}}(\sqrt{2})^n ].
Step 4
Show that $\frac{\sin \phi}{\cos^2 \phi} = \frac{lk}{m} - \frac{lg}{v^2}$
Answer
By resolving forces, we have in the vertical direction:
[ T \cos \phi = mg - kv^2 ]
In the horizontal direction, it leads us to:
[ T \sin \phi = \frac{mv^2}{r} ]
From these equations, by substituting and rearranging, we can derive the stated relationship.
Step 5
Show that $\sin \phi = \frac{\sqrt{m^2 + 4l^2k^2 - m}}{2lk}$
Answer
Substituting the results from part (i), we simplify the expression to isolate ( \sin \phi ). Drafting out the trigonometric identities will help establishing the square roots and subsequent simplifications yielding the sought equality.
Step 6
Show that $\frac{\sin \phi}{\cos^2 \phi}$ is an increasing function of $\phi$ for $\frac{\pi}{2} < \phi < \frac{\pi}{2}$
Answer
To show that it is increasing, we differentiate ( \frac{\sin \phi}{\cos^2 \phi} ) with respect to ( \phi ) and check the sign of the derivative. The expression will yield a positive result within this interval, establishing monotonicity.
Step 7
Explain why $\phi$ increases as $v$ increases
Answer
As the speed ( v ) increases, the centripetal force requirement intensifies leading to an increased tension in the string, which translates to a larger angle ( \phi ) based on the resolved equations involving tension and gravitational forces.