Author Topic: VCE Methods Question Thread! (Read 5429567 times) Tweet Share

hongs- · « **Reply #1245 on:** November 04, 2012, 08:50:51 pm »

Can someone help me on this question please?

A normal to the graph of y=√ x has equation y=-4x+a , where a is a real constant. Find the value of a

And for question 7b on VCAA 2006 exam , why is the range [0,20] ?
I would've assumed it was [16,20] if i sub the values of in the domain into the equation...

http://www.vcaa.vic.edu.au/Documents/exams/mathematics/2006mmcas1-w.pdf

b^3 · « **Reply #1246 on:** November 04, 2012, 08:56:53 pm »

1.
We know that the normal is perpendicular to the tangent, $m_{normal}\times m_{tangent}=-1$ . From the equation we can see that the gradient of the normal is $m_{normal}=-4$ .

So that makes the gradient of the tangent to be $m_{tangent}=\frac{1}{4}$ .

Now we need to find when the curve has that gradient. So finding the derivative $\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$
$\begin{alignedat}{1}\frac{1}{2\sqrt{x}} & =\frac{1}{4} \\ 2 & =\sqrt{x} \\ x & =4 \end{alignedat}$

Now we know that the normal and the function cross at $x=4$ , so we can find $a$ .
$\begin{alignedat}{1}\sqrt{4} & =-4(4)+a \\ a & =2+16 \\ a & =18 \end{alignedat}$

2. When you draw $|f(x)|$ out for that domain, you will still have an x-intercept at $x=-3$ and $x=0$ , with the $y$ value being $0$ at those two points, the end points will give the maximum value of 20.

https://www.desmos.com/calculator/z6dlrtttbc

e^1 · « **Reply #1247 on:** November 04, 2012, 11:19:52 pm »

Hi, got another question:

Find the smallest value of $a$ , such that there are no local maximum values of $|f(x) - g(x)|$ in the interval $0 \leq x \leq 6$ ,

where $f(x) = 4 cos(\frac{\pi x}{6})$ &
$g(x) = a(x+3)(x-3)(x-9)$

This comes from Insight Exam 2 2012.

Jenny_2108 · « **Reply #1248 on:** November 05, 2012, 12:33:59 am »

Quote from: xp3r009 on November 04, 2012, 11:19:52 pm

Hi, got another question:

Find the smallest value of $a$ , such that there are no local maximum values of $|f(x) - g(x)|$ in the interval $0 \leq x \leq 6$ ,

where $f(x) = 4 cos(\frac{\pi x}{6})$ &
$g(x) = a(x+3)(x-3)(x-9)$

This comes from Insight Exam 2 2012.

Is the smallest value of a $=\frac{ \pi ^2}{162} ?$

If yes, I will type my working out because it takes a bit of time to type LaTex and I'm not sure if its a right ans

e^1 · « **Reply #1249 on:** November 05, 2012, 01:24:09 am »

Quote from: Ennjy on November 05, 2012, 12:33:59 am

Quote from: xp3r009 on November 04, 2012, 11:19:52 pm
Hi, got another question:

Find the smallest value of $a$ , such that there are no local maximum values of $|f(x) - g(x)|$ in the interval $0 \leq x \leq 6$ ,

where $f(x) = 4 cos(\frac{\pi x}{6})$ &
$g(x) = a(x+3)(x-3)(x-9)$

This comes from Insight Exam 2 2012.

Is the smallest value of a $=\frac{ \pi ^2}{162} ?$

If yes, I will type my working out because it takes a bit of time to type LaTex and I'm not sure if its a right ans

The answer from the solutions is actually something else. Do you want me to show the solutions instead? I don't get how they do it.

Homer · « **Reply #1250 on:** November 05, 2012, 09:53:17 am »

f(x)=x^3-6x^2+9x-4 Find the real values of p for which the equation f(x)=p has exactly one solution.

e^1 · « **Reply #1251 on:** November 05, 2012, 10:43:17 am »

Quote from: Jai on November 05, 2012, 09:53:17 am

f(x)=x^3-6x^2+9x-4 Find the real values of p for which the equation f(x)=p has exactly one solution.

The function looks likes this, and for $f(x) = p$ to have exactly one solution, then it must not be within the range of the turning points, otherwise we'll get two or three solutions.

$\frac{df(x)}{dx} = 3x^2 - 12x + 9$

For maximum and minimum:

$\frac{df(x)}{dx} = 0$
$0 = 3x^2 - 12x + 9$

Then $(x - 3)(x - 1) = 0$

$\therefore x = 3 or x= 1$

Find y-values:
$f(1) = 0$ and $f(3) = -4$
So within the range from -4 to 0, we'll get two or three solutions.

Therefore, with the help of the graph, $p \in (-\infty, -4) \cup (0, \infty)$ , for $f(x) = p$ to have exactly one solution.

Typo fixed. Thanks Jai!

nina_rox · « **Reply #1252 on:** November 05, 2012, 10:51:31 am »

Thank you b^3 and TrueTears, that makes sense now!

Homer · « **Reply #1253 on:** November 05, 2012, 10:52:09 am »

you're absolutely right, just a typo but the (-infinity, 4) should be (-infinity, -4). Thanks xp3r009

Jenny_2108 · « **Reply #1254 on:** November 05, 2012, 11:41:15 am »

Quote from: xp3r009 on November 05, 2012, 01:24:09 am

Quote from: Ennjy on November 05, 2012, 12:33:59 am
Quote from: xp3r009 on November 04, 2012, 11:19:52 pm
Hi, got another question:

Find the smallest value of $a$ , such that there are no local maximum values of $|f(x) - g(x)|$ in the interval $0 \leq x \leq 6$ ,

where $f(x) = 4 cos(\frac{\pi x}{6})$ &
$g(x) = a(x+3)(x-3)(x-9)$

This comes from Insight Exam 2 2012.

Is the smallest value of a $=\frac{ \pi ^2}{162} ?$

If yes, I will type my working out because it takes a bit of time to type LaTex and I'm not sure if its a right ans

The answer from the solutions is actually something else. Do you want me to show the solutions instead? I don't get how they do it.

What I did is: I found double derivative and it said "no local maximum", thus d^y/dx^2>=0 then I sub value of x corresponding to find min value of a within the domain x E [0,6]

Yeah, maybe you can post here and if I understand, I can explain

e^1 · « **Reply #1255 on:** November 05, 2012, 11:45:05 am »

Quote from: Ennjy on November 05, 2012, 11:41:15 am

Quote from: xp3r009 on November 05, 2012, 01:24:09 am
Quote from: Ennjy on November 05, 2012, 12:33:59 am
Quote from: xp3r009 on November 04, 2012, 11:19:52 pm
Hi, got another question:

Find the smallest value of $a$ , such that there are no local maximum values of $|f(x) - g(x)|$ in the interval $0 \leq x \leq 6$ ,

where $f(x) = 4 cos(\frac{\pi x}{6})$ &
$g(x) = a(x+3)(x-3)(x-9)$

This comes from Insight Exam 2 2012.

Is the smallest value of a $=\frac{ \pi ^2}{162} ?$

If yes, I will type my working out because it takes a bit of time to type LaTex and I'm not sure if its a right ans

The answer from the solutions is actually something else. Do you want me to show the solutions instead? I don't get how they do it.

What I did is: I found double derivative and it said "no local maximum", thus d^y/dx^2>=0 then I sub value of x corresponding to find min value of a within the domain x E [0,6]

Yeah, maybe you can post here and if I understand, I can explain

It's on the attachment.
Also I appreciate your help

Jenny_2108 · « **Reply #1256 on:** November 05, 2012, 12:10:13 pm »

^ They use CAS and try to plot with different a values to see the change of a values corresponding to the change of |f(x)-g(x)|
Then they limit the graph until there is no local maximum existed, which gives a values from [0.05, 0.06]. For this interval of a, they use CAS to find x-coordinate of the min turning point which is x=3. Sub x=3 into derivative to find a

Dw too much, I don't think VCAA has this kinda ques. Its like you spend time using CAS, plotting and trying with many a values to see the shape of the graph then estimate and narrow the domain of a.

e^1 · « **Reply #1257 on:** November 05, 2012, 12:35:32 pm »

Quote from: Ennjy on November 05, 2012, 12:10:13 pm

^ They use CAS and try to plot with different a values to see the change of a values corresponding to the change of |f(x)-g(x)|
Then they limit the graph until there is no local maximum existed, which gives a values from [0.05, 0.06]. For this interval of a, they use CAS to find x-coordinate of the min turning point which is x=3. Sub x=3 into derivative to find a

Dw too much, I don't think VCAA has this kinda ques. Its like you spend time using CAS, plotting and trying with many a values to see the shape of the graph then estimate and narrow the domain of a.

The solution feels a bit like a trial-and-error method, but thanks for explaining.

breadkay · « **Reply #1258 on:** November 05, 2012, 02:48:43 pm »

Hi,

Could someone please show me how:
(2x / |x|) = (2|x| / x)

Thanks

Lasercookie · « **Reply #1259 on:** November 05, 2012, 02:55:57 pm »

Quote from: breadkay on November 05, 2012, 02:48:43 pm

Hi,

Could someone please show me how:
(2x / |x|) = (2|x| / x)

Thanks

So we know that $|x| = \sqrt{x^2}$ . It's just a bit of rearranging things around + rationalising the denominator.

$\frac{2x}{|x|} = \frac{2x}{\sqrt{x^2}} = \frac{2x}{\sqrt{x^2}} \times \frac{\sqrt{x^2}}{\sqrt{x^2}} = \frac{2x\sqrt{x^2}}{x^2} = \frac{2\sqrt{x^2}}{x} = \frac{2|x|}{x}$

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hongs-

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