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Prove that \(\sqrt{23}\) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1
Question 12
Prove that \(\sqrt{23}\) is irrational. Prove that for all real numbers \(x\) and \(y\), where \(x^2 + y^2 \neq 0\), \[ \frac{(x+y)^2}{x^2+y^2} \leq 2. \] An objec... show full transcript
Worked Solution & Example Answer:Prove that \(\sqrt{23}\) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1
Step 1
Prove that \(\sqrt{23}\) is irrational.
Answer
To prove that for all real numbers (x) and (y), where (x^2 + y^2 \neq 0):
[ \frac{(x+y)^2}{x^2+y^2} \leq 2, ]
we start with the observation that:
[ (x+y)^2 = x^2 + 2xy + y^2. ]
Thus, the left-hand side can be expressed as:
[ \frac{x^2 + 2xy + y^2}{x^2 + y^2}. ]
Now, we can separate it:
[ \frac{x^2}{x^2+y^2} + \frac{y^2}{x^2+y^2} + \frac{2xy}{x^2+y^2}. ]
Because (\frac{x^2}{x^2+y^2} + \frac{y^2}{x^2+y^2} = 1), we focus on the term (\frac{2xy}{x^2+y^2}). Considering (x^2 + y^2 \geq 2xy) from the AM-GM inequality, we deduce that (\frac{2xy}{x^2+y^2} \leq 1). Thus, we conclude that:
[ \frac{(x+y)^2}{x^2+y^2} \leq 2. ]
Step 2
Show that the resultant force on the object is \(\vec{F} = -m(g\sin \theta)\hat{j}\).
Answer
To find the resultant force (\vec{F}), we resolve the weight of the object into two components: one parallel to the inclined plane and one perpendicular to it. The weight of the object can be expressed as:
[ \vec{W} = mg ]
where (g) is the acceleration due to gravity.
- The component of weight parallel to the incline: [ W_{\parallel} = mg \sin \theta ]
- The component of weight perpendicular to the incline: [ W_{\perpendicular} = mg \cos \theta ]
The normal force (\vec{R}) balances the perpendicular component of weight, thus:
[ \vec{R} = mg \cos \theta ]
The net force acting on the object along the incline (assuming frictionless motion) is:
[ \vec{F} = -W_{\parallel} = -mg \sin \theta \hat{j}. ]
Step 3
Given that the object is initially at rest, find its velocity \(y(t)\) in terms of \(\theta\, R\, \text{and} \; i\,.\)
Answer
When the object is initially at rest, we analyze its motion along the inclined plane using Newton's second law. The net force acting on it along the incline is:
[ F = -mg \sin \theta, ]
Setting this equal to (ma) (where (a) is the acceleration), we have:
[ ma = -mg \sin \theta \Rightarrow a = -g \sin \theta. ]
Integrating this acceleration with respect to time gives us the velocity:
[ v(t) = -g \sin \theta \cdot t + C. ]
Given that at (t = 0), the object is at rest, we have (C = 0). Hence, we can express the velocity as:
[ y(t) = -g \sin \theta , t. ]
Step 4
Find the cube roots of \(2 - 2i\). Give your answer in exponential form.
Answer
Let (z = 2 - 2i). To find the cube roots, we first express it in polar form. The modulus is:
[ |z| = \sqrt{(2^2) + (-2^2)} = \sqrt{8} = 2\sqrt{2}. ]
The argument is:
[ \theta = \tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}. ]
Thus, we can write:
[ z = 2\sqrt{2} \left(\cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right)\right). ]
The cube roots are found using the formula:
[ z_k = r^{1/3} \left(\cos\left(\frac{\theta + 2k\pi}{3}\right) + i \sin\left(\frac{\theta + 2k\pi}{3}\right)\right). ]
where (k = 0, 1, 2). Substituting the values:
For (k = 0): [ z_0 = (2\sqrt{2})^{1/3} \left(\cos\left(-\frac{\pi}{12}\right) + i \sin\left(-\frac{\pi}{12}\right)\right). ]
For (k = 1): [ z_1 = (2\sqrt{2})^{1/3} \left(\cos\left(\frac{5\pi}{12}\right) + i \sin\left(\frac{5\pi}{12}\right)\right). ]
For (k = 2): [ z_2 = (2\sqrt{2})^{1/3} \left(\cos\left(\frac{13\pi}{12}\right) + i \sin\left(\frac{13\pi}{12}\right)\right). ]
Step 5
Step 6
Find the remaining zeros of the polynomial \(P(z)\).
Answer
Given that one zero is (2 + i) and another zero is (2 - i), we can denote the polynomial as:
[ P(z) = (z - (2 + i))(z - (2 - i))(Az^2 + Bz + C). ]
Expanding the factors ((z - (2 + i))(z - (2 - i))) yields:
[ (z - 2)^2 + 1 = z^2 - 4z + 5. ]
Dividing (P(z)) by ((z^2 - 4z + 5)) will give us the remaining quadratic part, and setting the remaining polynomial to zero:
[ Az^2 + Bz + C = 0 ]
will allow us to find the remaining zeros using the quadratic formula:
[ z = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}. ]