(a) Prove that $ \sqrt{23} $ is irrational. (b) Prove that for all real numbers $ x $ and $ y $, where $ x^{2} + y^{2} \neq 0 $, \[ \frac{(x+y)^{2}}{x^{2} + y^{2}} \leq 2. \] (c) An object with mass $ m $ kilograms slides down a smooth inclined plane with velocity $ y(t) $, where $ t $ is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle $ \theta $ with the horizontal, as shown in the diagram. The normal reaction force is $ R $ and has magnitude $ g $. There are no other forces act on the object. The vectors $ \hat{i} $ and $ \hat{j} $ are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram. (i) Show that the resultant force on the object is $ \mathbf{F} = - (mg \sin \theta) \hat{j}. $ (ii) Given that the object is initially at rest, find its velocity $ y(t) $ in terms of $ \theta, \; R \; \text{and} \; g $. (d) Find the cube roots of $ 2 - 2i $. Give your answer in exponential form. (e) The complex number $ 2 + i $ is a zero of the polynomial $ P(z) = z^{4} - z^{3} + cz^{2} + dz - 30 $ where $ c $ and $ d $ are real numbers. (i) Explain why $ 2 - i $ is also a zero of the polynomial $ P(z) $. (ii) Find the remaining zeros of the polynomial $ P(z) $.

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(a) Prove that $ \sqrt{23} $ is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Question 12

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(a) Prove that $ \sqrt{23} $ is irrational. (b) Prove that for all real numbers $ x $ and $ y $, where $ x^{2} + y^{2} \neq 0 $, \[ \frac{(x+y)^{2}}{x^{2}... show full transcript

Worked Solution & Example Answer:(a) Prove that $ \sqrt{23} $ is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Step 1

Prove that $ \sqrt{23} $ is irrational.

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Answer

Assume for contradiction that ( \sqrt{23} ) is rational. Then we can express it as ( \sqrt{23} = \frac{p}{q} ) with ( p ) and ( q ) as coprime integers (no common factors). Squaring both sides gives:

23 = \frac{p^2}{q^2}

This leads to:
( p^2 = 23q^2 ).
Since 23 is prime, it implies that ( p^2 ) is divisible by 23, and thus ( p ) must also be divisible by 23. Let ( p = 23k ) for some integer ( k ), substituting back yields:

23k^2 = q^2

indicating that ( q^2 ) is also divisible by 23, meaning ( q ) is divisible by 23 as well. This contradicts the assumption that ( p ) and ( q ) are coprime. Therefore, ( \sqrt{23} ) is irrational.

Step 2

Prove that $ \frac{(x+y)^{2}}{x^{2} + y^{2}} \leq 2. $

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Answer

To prove this inequality, apply the Cauchy-Schwarz inequality:

(x+y)^{2} \leq (1^{2} + 1^{2})(x^{2} + y^{2}) = 2(x^{2} + y^{2}).

Rearranging gives:

\frac{(x+y)^{2}}{x^{2} + y^{2}} \leq 2.

Thus, the result is proved.

Step 3

Show that the resultant force on the object is $ \mathbf{F} = - (mg \sin \theta) \hat{j}. $

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To find the resultant force, we need to resolve the weight ( mg ) into components parallel and perpendicular to the inclined plane:

The component along the slope:

$F_{parallel} = mg \sin \theta\n$

The component perpendicular to the slope gives: $F_{perpendicular} = mg \cos \theta.$

Since the only force acting perpendicular is the normal force ( R ), we have: ( \mathbf{F} = - (mg \sin \theta) \hat{j}. )

Step 4

Given that the object is initially at rest, find its velocity $ y(t) $ in terms of $ \theta, \; R \; \text{and} \; g $.

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Starting from Newton's second law, the net force acting on the object equals mass times acceleration:

Integrating the acceleration equation: $rac{d^2 y}{dt^2} = g \sin \theta \implies \frac{dy}{dt} = g \sin \theta t + C.$ The constant ( C = 0 ) (initially at rest), and integrating again: $y(t) = \frac{1}{2} g \sin \theta t^{2}.$

Step 5

Find the cube roots of $ 2 - 2i $.

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Answer

First, express the complex number in polar form:

$r = \sqrt{2^2 + (-2)^2} = \sqrt{8} = 2\sqrt{2},$

and the argument is: $\theta = \tan^{-1}\left(\frac{-2}{2}\right) = -\frac{\pi}{4}.$

So we write:\n $2 - 2i = 2\sqrt{2} \left( \cos\left(-\frac{\pi}{4}\right) + i \sin\left(-\frac{\pi}{4}\right) \right).$

The cube roots are: $\sqrt[3]{2\sqrt{2}} \left( \cos \left(\frac{\theta + 2k\pi}{3}\right) + i \sin \left(\frac{\theta + 2k\pi}{3}\right) \right), \quad k = 0, 1, 2.$

Step 6

Explain why $ 2 - i $ is also a zero of the polynomial $ P(z) $.

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Answer

Since the coefficients of ( P(z) ) are real, complex roots must occur in conjugate pairs. Since ( 2 + i ) is a zero, its conjugate ( 2 - i ) must also be a zero.

Step 7

Find the remaining zeros of the polynomial $ P(z) $.

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Given the known zeros ( 2 + i ) and ( 2 - i ), we can express:

$P(z) = (z - (2 + i))(z - (2 - i))(z - \alpha)(z - \beta)$

where ( \alpha ) and ( \beta ) are the remaining roots. Now we can expand the product of the two known factors and set the polynomial equal to zero to find ( \alpha ) and ( \beta ). Expanding gives: $(z^2 - 4z + 5)(z^2 + az + b) \text{ for some real } a, b.$

(a) Prove that \( \sqrt{23} \) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Worked Solution & Example Answer:(a) Prove that \( \sqrt{23} \) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Prove that \( \sqrt{23} \) is irrational.

Prove that \( \frac{(x+y)^{2}}{x^{2} + y^{2}} \leq 2. \)

Show that the resultant force on the object is \( \mathbf{F} = - (mg \sin \theta) \hat{j}. \)

Given that the object is initially at rest, find its velocity \( y(t) \) in terms of \( \theta, \; R \; \text{and} \; g \).

Find the cube roots of \( 2 - 2i \).

Explain why \( 2 - i \) is also a zero of the polynomial \( P(z) \).

Find the remaining zeros of the polynomial \( P(z) \).

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