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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
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 particl... 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 \( \frac{a + b}{2} \geq \sqrt{ab} \).
Answer
To prove this inequality, we can use the method of squaring both sides. Starting with:
[\left(\frac{a + b}{2}\right)^2 \geq ab]
Expanding the left-hand side gives:
[\frac{a^2 + 2ab + b^2}{4} \geq ab]
Multiplying through by 4 leads to:
[a^2 - 2ab + b^2 \geq 0]
This simplifies to:
[(a - b)^2 \geq 0]
Since the square of a real number is always non-negative, the inequality holds.
Step 2
What is the position of the particle when it reaches its maximum velocity?
Answer
The acceleration of the particle is defined as:
[a = 12 - 6t]
Integrating with respect to time gives the velocity:
[v = \int (12 - 6t) dt = 12t - 3t^2 + c]
Since the particle starts from rest, this means , which gives . Thus:
[v = 12t - 3t^2]
To find the time at which the velocity is maximum, we take the first derivative and set it to zero:
[\frac{dv}{dt} = 12 - 6t = 0 \Rightarrow t = 2]
Substituting back into the equation for position gives:
[x = \int v dt = \int (12t - 3t^2) dt = 6t^2 - t^3 + c]
Using the initial condition , we find . Thus:
[x = 6(2)^2 - (2)^3 = 24 - 8 = 16 \text{ units to the right of origin.}]
Step 3
(i) Show that \( \frac{dv}{dx} = -(1 + 3y) \).
Answer
Starting from the force acting on the particle:
[F = ma = m\frac{dv}{dt}]
Given the resistive force:
[F = -(v + 3t^2)]
Setting these equal yields:
[m\frac{dv}{dt} = -(v + 3t^2)]
Substituting , and rearranging gives:
[\frac{dv}{dt} = -(1 + 3y)] where .
Step 4
(ii) Hence, or otherwise, find $x$ as a function of $v$.
Answer
To find as a function of , we can separate variables from:
[\frac{dv}{1 + 3t} = -dx]
Integrating both sides results in:
[ \int \frac{dv}{1 + 3t} = -\int dx]
This gives us:
[\frac{1}{3} \ln |1 + 3t| = -x + c]
Rearranging gives:
[x = -\frac{1}{3} \ln |1 + 3t| + c] where can be determined from initial conditions.
Step 5
Using partial fractions, evaluate \( \int_{2}^{n} \frac{4 + x}{(4 + x^2)} \, dx \).
Answer
To perform the integral, we can decompose:
[\frac{4 + x}{4 + x^2} = \frac{A}{4 + x^2} + \frac{Bx + C}{4 + x^2}]
Finding coefficients through substitution:
Integrating leads to:
[\int_{2}^{n} \frac{4 + x}{(4 + x^2)} , dx = \frac{1}{2} \ln |f(n)| - \frac{1}{2}[\ln |1 + x| + 4] from the limits.
Step 6
Given the complex number $z = e^{i\theta}$, show that $w = \frac{z^2 - 1}{z^2 + 1}$ is purely imaginary.
Answer
Starting with:
[z = e^{i\theta} \Rightarrow z^2 = e^{2i\theta}]
Thus:
[w = \frac{e^{2i\theta} - 1}{e^{2i\theta} + 1}]
This can be expressed in the form:
[w = \frac{(\cos 2\theta + i \sin 2\theta) - 1}{(\cos 2\theta + i \sin 2\theta) + 1}]
Rearranging leads to demonstrating that the real part cancels, emphasizing the purely imaginary property.