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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: ice_blockie on April 20, 2008, 09:57:32 am

Title: More specialist questions...
Post by: ice_blockie on April 20, 2008, 09:57:32 am: Sorry to annoy you all on a wonderful Sunday morning when EVERYONE should be outside...but I'm stuck on some spesh and I hope you can help me:

1. The curve $C$ has equation $\frac{2ax^2+2bx+7}{2x+5}$ where $x\not=\frac{-5}{2}$ and $a$ and $b$ are non zero constants. The oblique asymptote cuts the x axis at (-1,0). What is the relationship between $a$ and $b$ ?

Answer is: $7a = 2b$

2.What is the implied domain of $y=Tan(3arcos(2x))$ ?

Answer is $(\frac{\sqrt3}{4},\frac{1}{2})$
Title: Re: More specialist questions...
Post by: Collin Li on April 20, 2008, 10:15:28 am: $\frac{2ax^2+2bx+7}{2x+5} = \dfrac{(2x+5)\left(ax + b- \frac{5}{2}a\right) - 5b + \frac{25}{2}a + 7}{2x+5}$

$= ax + b- \frac{5}{2}a + \dfrac{- 5b + \frac{25}{2}a + 7}{2x+5}$

Therefore, the oblique asymptote is:

$y = ax + b - \frac{5}{2}a$

Since (-1, 0) lies on the oblique asymptote:

$\implies 0 = -a + b -\frac{5}{2}a$

$\implies 7a = 2b$
Title: Re: More specialist questions...
Post by: Collin Li on April 20, 2008, 10:37:26 am: $y = \mbox{Tan}(3\arccos(2x))$

Consider the range of $3\arccos(2x)$ : $\left[0,\,3\pi\right]$

Consider the domain of $\mbox{Tan} x$ : $\left(-\frac{\pi}{2},\,\frac{\pi}{2}\right)$

Now, restrict the domain of $3\arccos(2x)$ such that its new range will be the intersection of $\left[0,\,\pi\right]$ and $\left(-\frac{\pi}{2},\,\frac{\pi}{2}\right)$ , i.e: $\left[0,\,\frac{\pi}{2}\right)$

Thus, $0 \leq 3\arccos(2x) < \frac{\pi}{2}$

Since $3\arccos(2x)$ is a monotonic decreasing function, we can say:

$\cos\frac{\frac{\pi}{2}}{3} < 2x \leq \cos\frac{0}{3}$

$\implies \frac{\sqrt{3}}{2} < 2x \leq 1$

$\therefore \frac{\sqrt{3}}{4} < x \leq \frac{1}{2}$

The domain of $y$ is the domain of $3\arccos(2x)$ , so therefore, $\mbox{dom}\; y = \left(\frac{\sqrt{3}}{4},\,\frac{1}{2}\right]$

(I'm pretty sure the square bracket is right)
Title: Re: More specialist questions...
Post by: ice_blockie on April 20, 2008, 06:18:01 pm: Thanks muchly!