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 1 m s$^{-1}$. The particle experiences a resistsive force of magnitude $v + 3t^2$ newtons, where $v$ m s$^{-1}$ is the speed of the particle after $t$ seconds. The particle is never at rest. Show that $\frac{dv}{dt} = -(1 + 3t)$. 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 st... 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 that $\frac{a+b}{2} \geq \sqrt{ab}$ , we can use the Cauchy-Schwarz inequality or the method of squares:

Start by rearranging the inequality: $\left(\frac{a + b}{2}\right)^2 \geq ab.$
Expanding the left-hand side: $\frac{(a+b)^2}{4} \geq ab,$ which simplifies to $a^2 + 2ab + b^2 \geq 4ab.$
Rearranging gives: $a^2 - 2ab + b^2 \geq 0.$
This inequality is always true since it can be factored as $(a - b)^2 \geq 0$ . Thus, we conclude that $\frac{a+b}{2} \geq \sqrt{ab}$ .

Step 2

What is the position of the particle when it reaches its maximum velocity?

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Answer

Given the acceleration $a(t) = 12 - 6t$ , we can find the velocity by integrating: $v(t) = \int a(t) \ dt = \int (12 - 6t) \ dt = 12t - 3t^2 + C.$

Initially, at $t = 0$ , the particle starts from rest, meaning $v(0) = 0$ . Thus, $C = 0$ . Therefore, $v(t) = 12t - 3t^2.$

To find when the velocity is at its maximum, take the derivative and set it to zero: $\frac{dv}{dt} = 12 - 6t = 0 \implies t = 2.$

Now, we can find the position $x(t)$ by integrating the velocity: $x(t) = \int v(t) \ dt = \int (12t - 3t^2) \ dt = 6t^2 - t^3 + C.$

At $t = 0$ , $x(0) = 0$ implies $C = 0$ . Thus, $x(t) = 6t^2 - t^3.$

Substituting $t = 2$ , we have: $x(2) = 6(2^2) - (2^3) = 24 - 8 = 16.$

Therefore, the position of the particle at maximum velocity is 16 units to the right of the origin.

Step 3

Show that $\frac{dv}{dt} = -(1 + 3t)$.

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Answer

The force acting on the particle can be described as: $F = ma = m\frac{dv}{dt} = - (v + 3t^2).$

Since $m = 1$ kg, we have: $\frac{dv}{dt} = - (v + 3t^2).$

Substituting the given equation gives: $\frac{dv}{dt} = -(1 + 3t),$ as needed.

Step 4

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

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From the previous equation we have: $\frac{dv}{dt} = -(1 + 3t) \implies \frac{dv}{1 + 3t} = - dt.$

Integrating both sides: $\int \frac{1}{1 + 3t} \ dv = - \int dt.$

This gives: $\frac{1}{3} \ln |1 + 3t| = -t + C.$

Solving for $t$ gives: $t = \frac{1}{3}{\ln |1 + 3t|} + C.$

We also have $dx = v dt$ to relate $x$ to $v$ , allowing us to further find $x$ as a function of $v$ .

Step 5

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

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Answer

We can use partial fractions to rewrite: $\frac{4 + x}{(1 - x)(4 + x)} = \frac{A}{1 - x} + \frac{B}{4 + x}.$

Multiplying by $(1 - x)(4 + x)$ gives: $4 + x = A(4 + x) + B(1 - x).$

Setting up equations by substituting convenient values will lead us to the values of A and B. After obtaining A and B, integrate each term from 2 to n and compute the definite integral to express the result in the desired logarithmic 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

Starting with: $z = e^{i\theta} = \cos \theta + i \sin \theta.$

Thus: $z^2 = \cos(2\theta) + i \sin(2\theta).$

Plugging this into $w$ gives: $w = \frac{(\cos(2\theta) + i \sin(2\theta)) - 1}{(\cos(2\theta) + i \sin(2\theta)) + 1}.$

This can be simplified further to show that the real part cancels out, leaving us with a purely imaginary number, hence proving the assertion.

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 $\frac{a+b}{2} \geq \sqrt{ab}$.

What is the position of the particle when it reaches its maximum velocity?

Show that $\frac{dv}{dt} = -(1 + 3t)$.

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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