VCE Methods Question Thread!

[RuiAce]

There is a first-principles way if you prefer. Given that it's a M/C question I'd have just plugged into a calculator, but if you wanna do it manually you can consider the following.
\[ \text{Rearrange the given equation into}\\ a(4x+y) + b(y+4) = 0 \]
\[ \text{Due to the fact that this must hold }\textbf{for all }a,b\in \mathbb{R}\\ \text{we may }\textbf{equate the coefficients}\text{ on }a\text{ and }b.\]
\[ \text{The coefficient of }b\text{ on the LHS is }y+4.\\ \text{The coefficient of }b\text{ on the RHS is 0.}\\ \text{Hence upon equating, }y+4 = 0\implies \boxed{y=-4}.\]
\[ \text{The coefficient of }a\text{ on the LHS is }4x+y.\\ \text{The coefficient of }a\text{ on the RHS is 0.}\\ \text{Hence upon equating, }4x+y = 0\text{, so subbing }y=-4\text{ gives }\boxed{x=1}.\]
\[ \text{Therefore the point is }(1,-4).\]

[persistent_insomniac]

Hi this may be a very simple question but can someone please explain how they got from the first line to the second?

[MB_]

Hi this may be a very simple question but can someone please explain how they got from the first line to the second?

\begin{align*}x^2 \sqrt{1-\frac{9}{x^2}}  &=x^2 \sqrt{\frac{x^2}{x^2}-\frac{9}{x^2}} \\ &=x^2 \sqrt{\frac{1}{x^2}(x^2-9)} \\ &=x^2 \sqrt{\frac{1}{x^2}}\sqrt{x^2-9} \\ &=x \sqrt{x^2-9}\end{align*}

[persistent_insomniac]

\begin{align*}x^2 \sqrt{1-\frac{9}{x^2}}  &=x^2 \sqrt{\frac{x^2}{x^2}-\frac{9}{x^2}} \\ &=x^2 \sqrt{\frac{1}{x^2}(x^2-9)} \\ &=x^2 \sqrt{\frac{1}{x^2}}\sqrt{x^2-9} \\ &=x \sqrt{x^2-9}\end{align*}

Can you plz explain how you got from the 3rd line to the 4th

[S_R_K]

Can you plz explain how you got from the 3rd line to the 4th

What is the positive square root of x^2?

[MB_]

Can you plz explain how you got from the 3rd line to the 4th

\begin{align*} x^2 \sqrt{\frac{1}{x^2}}\sqrt{x^2-9}&=x^2 \frac{\sqrt{1}}{\sqrt{x^2}}\sqrt{x^2-9}\\ &=x^2 \frac{1}{x}\sqrt{x^2-9}  \\ &=x \sqrt{x^2-9}\end{align*}

[AlphaZero]

Just going to quickly add:

Although the absolute value function is not required in this subject, please note that \[\sqrt{x^2}=\begin{cases}x, & x\geq 0\\
-x, & x<0\end{cases}.\] So, the final result is only valid for  \(x>3(>0)\).

[Arthurmorgan]

"Determine whether the following functional equations hold true for the standard functions with the parameter 'a'.

f(x+y)=f(xy) for f(x)=2^{ax}

Struggling to grasp what Is meant by the parameter, never seen it before in the textbook.

[AlphaZero]

"Determine whether the following functional equations hold true for the standard functions with the parameter 'a'.

f(x+y)=f(xy) for f(x)=2^{ax}

Struggling to grasp what Is meant by the parameter, never seen it before in the textbook.

The word "parameter" just means a quantity that influences the output. Here \(a\) is just a real constant (\(a\in\mathbb{R}\)).

[lzxnl]

What is the positive square root of x^2?

Not x. sqrt((-1)^2) = 1, not -1.

[S_R_K]

Not x. sqrt((-1)^2) = 1, not -1.

Yes of course it should be |x|, not x. But I took it from the question (where the solution was provided) that x was assumed to be positive.

[VanessaS]

Could someone please help me with question 7a?

[AlphaZero]

Could someone please help me with question 7a?

Hint:  consider the angles in the isosceles triangle \(\triangle OXZ\).

[Donut]

How do I solve this question?

The answer is 9.2, but if the portions are reversed shouldn't it be 18 years because there should be a ratio of 5:1 for currawongs to magpies, not a 1:1 ratio.

[S_R_K]

How do I solve this question?

The answer is 9.2, but if the portions are reversed shouldn't it be 18 years because there should be a ratio of 5:1 for currawongs to magpies, not a 1:1 ratio.

You are correct, the proportions are reversed at the time t for which C * (1.12)^t = 25C * (0.94)^t (where C is the initial size of the currawong population).