For real numbers $a, b \geq 0$ prove that $ \frac{a+b}{2} \geq \sqrt{ab} $. A particle is moving in a straight line with acceleration $a = 12 - 6t$. The particle starts from rest at the origin. What is the position of the particle when it reaches its maximum velocity? A particle with mass 1 kg is moving along the x-axis. Initially, the particle is at the origin and has speed in m s$^{-1}$. The particle experiences a resistive force of magnitude $v + 3t^{2}$ newtons, where $v$ is m s$^{-1}$ is the speed of the particle after $t$ seconds. The particle is never at rest. Show that $ \frac{dv}{dx} = -(1 + 3y). $ Hence, or otherwise, find $x$ as a function of $v$. Using partial fractions, evaluate $ \int_{2}^{n} \frac{4 + x}{(1-x)(4+x)} \, dx, $ giving your answer in the form $\frac{1}{2} \ln f(n)$, where $f(n)$ is a function of $n$. Given the complex number $z = e^{i\theta}$, show that $w = \frac{z^{2}-1}{z^{2}+1}$ is purely imaginary.

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For real numbers $a, b \geq 0$ prove that $ \frac{a+b}{2} \geq \sqrt{ab} $ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2022 - Paper 1

Question 12

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For real numbers $a, b \geq 0$ prove that $ \frac{a+b}{2} \geq \sqrt{ab} $. A particle is moving in a straight line with acceleration $a = 12 - 6t$. The particle ... show full transcript

Worked Solution & Example Answer:For real numbers $a, b \geq 0$ prove that $ \frac{a+b}{2} \geq \sqrt{ab} $ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2022 - Paper 1

Step 1

For real numbers $a, b \geq 0$ prove that $ \frac{a+b}{2} \geq \sqrt{ab} $

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Answer

To prove this inequality, we will use the method of squaring both sides. We start from:

[ \left( \frac{a+b}{2} \right)^2 \geq ab ]

Expanding the left-hand side gives us:

[ \frac{a^2 + 2ab + b^2}{4} \geq ab ]

Multiplying both sides by 4 leads to:

[ a^2 + 2ab + b^2 \geq 4ab ]

Rearranging terms yields:

[ a^2 - 2ab + b^2 \geq 0 ]

This can be factored as:

[ (a - b)^2 \geq 0 ]

Since the square of any real number is non-negative, the inequality holds true, hence:

[ \frac{a+b}{2} \geq \sqrt{ab} ]

Step 2

A particle is moving in a straight line with acceleration $a = 12 - 6t$. The particle starts from rest at the origin. What is the position of the particle when it reaches its maximum velocity?

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Answer

Given the acceleration is:

[ a = \frac{dv}{dt} = 12 - 6t ]

Integrating the acceleration to find the velocity:

[ v = \int (12 - 6t) dt = 12t - 3t^2 + C ]

Since the particle starts from rest at $t=0$ , we have:

[ v(0) = 0 \Rightarrow C = 0\ ]

Thus:

[ v = 12t - 3t^2 ]

To find the time at which the maximum velocity occurs, we set the derivative of velocity to zero:

[ \frac{dv}{dt} = 0 \Rightarrow 12 - 6t = 0 \Rightarrow t = 2 ]

Now substituting $t=2$ back into the position function:

[ x = \int v , dt = \int (12t - 3t^2) dt = 6t^2 - t^3 + C ]

Evaluating this from $0$ to $t=2$ gives:

[ x = 6(2^2) - (2^3) = 24 - 8 = 16\ ]

Thus, the position of the particle is 16 units to the right of the origin.

Step 3

Show that $ \frac{dv}{dx} = -(1 + 3y) $

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Answer

Applying Newton's second law, we have:

[ F = ma = \frac{dv}{dt}]

The resistive force from the problem gives:

[ F = -(v + 3t^2)]

Setting the two equal gives:

[ \frac{dv}{dt} = -(1 + 3y) ]

Step 4

Hence, or otherwise, find $x$ as a function of $v$.

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Answer

To find $x$ as a function of $v$ , we separate variables from:

[ \frac{dv}{dx} = -(1 + 3y) \implies dx = -\frac{1}{(1 + 3y)}dv ]

Integrating both sides:

[ \int dx = -\int \frac{1}{(1 + 3y)} dv ]

Results in:

[ x = -\frac{1}{3} \ln(1 + 3y) + C ]

Step 5

Using partial fractions, evaluate $ \int_{2}^{n} \frac{4 + x}{(1-x)(4+x)} \, dx, $ giving your answer in the form $\frac{1}{2} \ln f(n)$.

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Answer

To evaluate the integral:

[ \int \frac{4 + x}{(1-x)(4+x)} , dx ]

We decompose into partial fractions:

[ \frac{A}{1-x} + \frac{B}{4+x}\n]

Finding $A$ and $B$ leads to:

[ A = 1, B = 1\ ]

Thus:

[ \int \left( \frac{1}{1-x} + \frac{1}{4+x} \right) dx ]

Integrating gives:

[ -\ln|1-x| + \ln|4+x| + C\ ]

Evaluate from $2$ to $n$ provides the final answer in required form.

Step 6

Given the complex number $z = e^{i\theta}$, show that $w = \frac{z^{2}-1}{z^{2}+1}$ is purely imaginary.

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Answer

From the definition: $z = e^{i\theta} = \cos\theta + i \sin\theta$ , we compute:

[ z^{2} = e^{2i\theta} = \cos(2\theta) + i \sin(2\theta)\ ]

Then substituting into w gives:

[ w = \frac{\cos(2\theta) + i \sin(2\theta) - 1}{\cos(2\theta) + i \sin(2\theta) + 1}\ ]

Simplifying this expression shows that the real part cancels out, confirming that $w$ is purely imaginary.

For real numbers $a, b \geq 0$ prove that \( \frac{a+b}{2} \geq \sqrt{ab} \) - HSC - SSCE Mathematics Extension 2 - Question 12 - 2022 - Paper 1

Worked Solution & Example Answer:For real numbers $a, b \geq 0$ prove that \( \frac{a+b}{2} \geq \sqrt{ab} \) - HSC - SSCE Mathematics Extension 2 - Question 12 - 2022 - Paper 1

For real numbers $a, b \geq 0$ prove that \( \frac{a+b}{2} \geq \sqrt{ab} \)

A particle is moving in a straight line with acceleration $a = 12 - 6t$. The particle starts from rest at the origin. What is the position of the particle when it reaches its maximum velocity?

Show that \( \frac{dv}{dx} = -(1 + 3y) \)

Hence, or otherwise, find $x$ as a function of $v$.

Using partial fractions, evaluate \( \int_{2}^{n} \frac{4 + x}{(1-x)(4+x)} \, dx, \) giving your answer in the form $\frac{1}{2} \ln f(n)$.

Given the complex number $z = e^{i\theta}$, show that $w = \frac{z^{2}-1}{z^{2}+1}$ is purely imaginary.

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