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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: samiira on March 25, 2010, 11:12:16 pm

Title: Multiple choice Questions spesh help URGENT!!
Post by: samiira on March 25, 2010, 11:12:16 pm: http://i299.photobucket.com/albums/mm307/miss-laila/q2.jpg
http://i299.photobucket.com/albums/mm307/miss-laila/q1.jpg
Title: Re: Multiple choice Questions spesh help URGENT!!
Post by: Martoman on March 26, 2010, 11:56:04 pm: With the exception of the last question, these are a tad interesting.

1) Set up parametric equations.

So you know that, by dividing by 9,
$(\frac{x}{3})^2 +(\frac{y}{3})^2 = 1$

Knowing a tiny bit about parametric equations we can say that

$\frac{x}{3} = cos(t)$

$\frac{y}{3} = sin(t)$

Your questions asks parallel to the y, so every y point is multipled by a factor of $\frac{1}{3}$ .

Then $\frac{y}{3} = \frac{1}{3} \times sin(t)$

$\Rightarrow y = sin(t)$

So now going back from parametric to cartesian.

$\frac{x^2}{9} + y^2 = 1$
= $x^2 + 9y^2 = 9$

B

2) Interesting specimen with multiple solutions to it. This is by no means the most elegant.

Notice that the only thing that is of interest is the numerator of the derivative.

$-2(8x+p) = 0$
for stat points by deriving.

$p = -8x, so \; x = -\frac{p}{8}$

Subbing into the original equation.

$\frac{-2}{\frac{p^2}{16}-q^2}$

This is the y point of the stat(s) points.

This will be undefined where the denominator = 0

$p^2 = 16q^2$
$\therefore p = \pm 4q$

Hence the stat points are undefined at $p = \pm 4q$

Which means option B cannot be right. This is because if we take the positive case of q, ( $4q)$ provided q is positive, then p will eventually hit 4q. This cannot happen for a stat point to exist.

Similar reasoning applies to -4q and option C.

Now option A doesn't *hit* any of these problems, and conveniently avoids -4q, so it is right.

3) The domain options are clearly wrong as they don't have fractions. This requires fractions.

4. This is relatively easy. $cos^{-1} (\frac{-\sqrt{3}}{2})$

It is defined over $x \epsilon [o, \pi]$

Cos is negative in the second quadrant. The reference angle is $\frac{\pi}{6}$

But we are in the second quadrant, so $\pi - \frac{\pi}{6}$

$\therefore$ C
Title: Re: Multiple choice Questions spesh help URGENT!!
Post by: samiira on March 29, 2010, 03:45:51 pm: Thank You soo much.. not even my teacher cud do sum of those.. :)