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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: cobby on January 30, 2009, 06:50:11 pm

Title: Cobby's Methods Questions
Post by: cobby on January 30, 2009, 06:50:11 pm: Hey all,
would someone please answer this question for me?

Determine the rule for the following cubic functions with these points.
a) (-1,3) and (4,0)
Where (-1,3) is a point of zero gradient

b) (-1,0) (0,-24) and (3,0)

Thanks
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on January 30, 2009, 06:57:18 pm: a) sub (-1,3) and (4,0) into the equation y=ax^3 + bx^2 + cx + d, . Then derive ax^3 + bx^2 + cx + d and sub in (-1,3)

b) use the general cubic equation and sub the coordinates in
Title: Re: Cobby's Methods Questions
Post by: cobby on January 30, 2009, 07:03:35 pm: Quote from: kurrymuncher on January 30, 2009, 06:57:18 pm
a) sub (-1,3) and (4,0) into the equation y=ax^3 + bx^2 + cx + d, . Then derive ax^3 + bx^2 + cx + d and sub in (-1,3)

b) use the general cubic equation and sub the coordinates in

for A) am i meant to have 3 eq's in the end? :S and do i solve them simultaneously afterwards?
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on January 30, 2009, 07:13:49 pm: Yes
Title: Re: Cobby's Methods Questions
Post by: Edmund on January 30, 2009, 07:29:20 pm: For the first,

Differentiate the general cubic equation to get 3ax^2 + 2bx + c

So,

3ax^2 + 2bx + c = 0 because zero gradient

Sub in (-1,3)

You should get one equation from here.

(Yep, 3 equations and is a pain :S)
Title: Re: Cobby's Methods Questions
Post by: cobby on January 30, 2009, 07:43:58 pm: ok for part a)
the answer i got was

a = (-25c+6)/175
b = 50c-9/175
d = 4(25c+132)/175

Can someone please confirm these results?
Title: Re: Cobby's Methods Questions
Post by: /0 on January 30, 2009, 08:00:47 pm: Quote from: cobby on January 30, 2009, 07:43:58 pm
ok for part a)
the answer i got was

a = (-25c+6)/175
b = 50c-9/175
d = 4(25c+132)/175

Can someone please confirm these results?

a = -(25c+6)/175, b = (50c-9)/175, d is correct.

This is most likely a calculator question, the way I would do it is define $f(x) = ax^3 + bx^2 + cx +d$ , define $g(x) = d(f(x),x)$ , then
solve(f(-1)=3 and f(4)=0 and g(-1)=0,{a,b,d})
Title: Re: Cobby's Methods Questions
Post by: Mao on January 31, 2009, 12:01:19 am: this most likely won't appear on an exam :)
Title: Re: Cobby's Methods Questions
Post by: cobby on January 31, 2009, 08:43:43 am: Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
Title: Re: Cobby's Methods Questions
Post by: dekoyl on January 31, 2009, 08:52:09 am: Quote from: cobby on January 31, 2009, 08:43:43 am
Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
2nd >> Trace >> 4 (maximum)
Get the cursor on the left of the maximum turning point (left bound) which is near x = 0. ENTER
Get the cursor on the right of the maximum turning point (right bound) which is near x = 0. ENTER.
Press ENTER again.
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)

Repeat this for the minimum turning point.
Title: Re: Cobby's Methods Questions
Post by: GerrySly on January 31, 2009, 10:07:06 am: Quote from: dekoyl
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)
Quote from: cobby
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.[/cobby]
Title: Re: Cobby's Methods Questions
Post by: cobby on January 31, 2009, 10:10:04 am: Quote from: dekoyl on January 31, 2009, 08:52:09 am
Quote from: cobby on January 31, 2009, 08:43:43 am
Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
2nd >> Trace >> 4 (maximum)
Get the cursor on the left of the maximum turning point (left bound) which is near x = 0. ENTER
Get the cursor on the right of the maximum turning point (right bound) which is near x = 0. ENTER.
Press ENTER again.
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)

Repeat this for the minimum turning point.
Thanks for the dekoyl :)
I think we are using different calcs? Because i had to use F5Math : Minimum and Maximum.
But still the same cursor movements to get the values.
Title: Re: Cobby's Methods Questions
Post by: dekoyl on January 31, 2009, 12:37:17 pm: Quote from: cobby on January 31, 2009, 10:10:04 am
Thanks for the dekoyl :)
I think we are using different calcs? Because i had to use F5Math : Minimum and Maximum.
But still the same cursor movements to get the values.
Ah yes we probably do.
I did non-CAS so, yeah. :P
Glad that you got it.
Title: Re: Cobby's Methods Questions
Post by: plato on January 31, 2009, 11:00:03 pm: Quote from: dekoyl on January 31, 2009, 08:52:09 am
Quote from: cobby on January 31, 2009, 08:43:43 am
Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
2nd >> Trace >> 4 (maximum)
Get the cursor on the left of the maximum turning point (left bound) which is near x = 0. ENTER
Get the cursor on the right of the maximum turning point (right bound) which is near x = 0. ENTER.
Press ENTER again.
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)

Repeat this for the minimum turning point.

Sorry Dekoyl but you definitely should NOT use Trace to find any of the major points with any calculator whether in Methods or Methods CAS. Each calculator will have a function somewhere that finds max/min ponts, intersections, zeroes etc.
Title: Re: Cobby's Methods Questions
Post by: Mao on February 01, 2009, 12:18:35 am: Quote from: plato on January 31, 2009, 11:00:03 pm
Quote from: dekoyl on January 31, 2009, 08:52:09 am
Quote from: cobby on January 31, 2009, 08:43:43 am
Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
2nd >> Trace >> 4 (maximum)
Get the cursor on the left of the maximum turning point (left bound) which is near x = 0. ENTER
Get the cursor on the right of the maximum turning point (right bound) which is near x = 0. ENTER.
Press ENTER again.
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)

Repeat this for the minimum turning point.

Sorry Dekoyl but you definitely should NOT use Trace to find any of the major points with any calculator whether in Methods or Methods CAS. Each calculator will have a function somewhere that finds max/min ponts, intersections, zeroes etc.

If you read the instructions carefully, he did give instructions for using functionality for finding max/min points.
Title: Re: Cobby's Methods Questions
Post by: plato on February 02, 2009, 11:13:54 pm: Quote from: Mao on February 01, 2009, 12:18:35 am
Quote from: plato on January 31, 2009, 11:00:03 pm
Quote from: dekoyl on January 31, 2009, 08:52:09 am
Quote from: cobby on January 31, 2009, 08:43:43 am
Hey all ive got another question

$For f(x) = 2x^3-4x^2-3x$
Use a graphic calc to determine the co-ordinates of the turning points correct to two decimal places.
How do i do that?

Thanks
2nd >> Trace >> 4 (maximum)
Get the cursor on the left of the maximum turning point (left bound) which is near x = 0. ENTER
Get the cursor on the right of the maximum turning point (right bound) which is near x = 0. ENTER.
Press ENTER again.
Should get x = -0.305 and y = .486 to three decimal places (not sure how many they want)

Repeat this for the minimum turning point.

Sorry Dekoyl but you definitely should NOT use Trace to find any of the major points with any calculator whether in Methods or Methods CAS. Each calculator will have a function somewhere that finds max/min ponts, intersections, zeroes etc.

If you read the instructions carefully, he did give instructions for using functionality for finding max/min points.

Yeah - you're right. I just saw the word Trace and should have read the rest. He was referring to the button rather than the operation. Sorry dekoyl.
Title: Re: Cobby's Methods Questions
Post by: cobby on February 07, 2009, 12:54:33 pm: Can someone please help me with this?

{ ${x:|x+3| \geqslant 5}$ }
{ ${x:|x+2| \leqslant 1$ }
Is there a way that i can find the range, by sketching and without sketching the graph?

Thanks
Title: Re: Cobby's Methods Questions
Post by: /0 on February 07, 2009, 01:16:19 pm: A graph without a y-variable? ???

If you want to solve $|x+3|\geq 5$ algebraically you have to consider 2 cases:

$x+3<0\Rightarrow x<-3$ , then $-(x+3) \geq 5$ , i.e. $x \leq -8$ and if

$x+3 \geq 0\Rightarrow x\geq -3$ , then $x+3 \geq 5$ , i.e. $x\geq2$

So the values x can take are $(-\infty, -8] \cup [2, \infty)$
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on February 07, 2009, 01:22:26 pm: i would start looking at the translations

Domain is R because |x| is R

Range is $[0,\infty)$ in |x| so this one is moved down by -5 hence $[-5,\infty)$
Title: Re: Cobby's Methods Questions
Post by: cobby on February 07, 2009, 02:19:36 pm: Sketch the graph of $f(x)=|x-4|+1$
My working so far is,
$x-4+1 if x \geqslant 4$
and
$3-x+1 if x < 3$
Then
$x-3 if x \geqslant 4$
$4-x if x < 3$

Have i done this right?? Where do i go from here to be able to sketch the graph?
Title: Re: Cobby's Methods Questions
Post by: bucket on February 07, 2009, 02:23:46 pm: draw $|x-4|$ and translate it up by one unit along the y axis.
Title: Re: Cobby's Methods Questions
Post by: dekoyl on February 07, 2009, 02:25:56 pm: Yes, draw |x-4| and translate up by one unit. It's "sharp point" should be at (4,1).
The y-intercept is found using:
-(0 - 4) = y
y-int = (0,4)

PS: I use -(x-4) = y because the graph as you are aware, consists of y=x - 4 and y=-(x-4). The y-intercept lies on the part of the graph where it's y=4-x.
Title: Re: Cobby's Methods Questions
Post by: cobby on February 08, 2009, 06:27:43 pm: Hey all, another question :(

For the line $y=x$ to be tangent to the circle with equation $(x-2k)^2 +(y-k)^2 =1$
Find the value of k

Thanks
Title: Re: Cobby's Methods Questions
Post by: /0 on February 08, 2009, 06:50:03 pm: $y=x$

$(x-2k)^2+(y-k)^2 =1$

$\Rightarrow (x-2k)^2 + (x-k)^2 = 1$

$2x^2 -6kx+5k^2-1=0$

The discriminant must be equal to zero for one solution (and hence tangency)

$36k^2 - 4(5k^2-1)(2)=0$

$36k^2 - 40k^2+8=0$

$-4k^2+8=0$

$k=\pm \sqrt{2}$
Title: Re: Cobby's Methods Questions
Post by: cobby on February 08, 2009, 07:33:23 pm: What did you do exactly to get from
Quote from: /0 on February 08, 2009, 06:50:03 pm
$2x^2 -6kx+5k^2-1=0$
to
Quote from: /0 on February 08, 2009, 06:50:03 pm

$36k^2 - 4(5k^2-1)(2)=0$

$36k^2 - 40k^2+8=0$
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on February 08, 2009, 07:44:38 pm: use the discriminant formula

$b^2-4ac$

where $a= 2 b=-6k c=5k^2-1$
Title: Re: Cobby's Methods Questions
Post by: dekoyl on February 08, 2009, 07:50:21 pm: That's the discriminant.
$b^{2}-4ac$ of that $2x^{2}-6kx...$ is $36k^{2}-4(5k^{2}-1)(2)$ and because we know it will have only one solution (for tangency), it will equal 0
Title: Re: Cobby's Methods Questions
Post by: cobby on February 12, 2009, 06:36:03 pm: Hey all, this ones a composite function question
Thanks :)
$f: R(except){0} \rightarrow R, f(x) = \frac {1}{2}(\frac{1}{x}+1)$

$g:R(except)\frac{1}{2} \rightarrow R, g(x) = \frac {1}{2x-1}$

a) Find $f(g(x))$ and state range

b) Find $g(f(x))$ and state range
Title: Re: Cobby's Methods Questions
Post by: Mao on February 12, 2009, 06:51:00 pm: to use "\" in latex, the command is "\backslash", "{" is "\{", "}" is "\}"

First, we note that the horizontal asymptote for g(x) is at y=0 (no vertical translation), i.e. the range of g is R\{0}, which fits in the domain of f, so f(g(x)) exists for the entire domain of g.

$f(g(x)) = \frac{1}{2}\cdot \left( \frac{1}{\frac{1}{2x-1}} + 1}\right) = \frac{1}{2}\cdot (2x-1+1) = x, \; x\in\mathbb{R}\backslash \left\{\frac{1}{2}\right\}$

The range is hence $f\circ g \in \mathbb{R}\backslash \left\{\frac{1}{2}\right\}$

use similar reasoning for b) and arrive at $g\circ f\in \mathbb{R}\backslash \{0\}$

It should be noted that when $f(g(x)) = x$ , f and g are inverse functions of each other.
Title: Re: Cobby's Methods Questions
Post by: cobby on February 12, 2009, 08:47:37 pm: Hey all im stuck on this polynomial question

Find the rule for the cubic function, the graph of which passes through the points (1,1) (2,4) (3,9) and (0,6)

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on February 12, 2009, 09:31:44 pm: store the general cubic equation as f(x), then go to solve.

type in solve( f(1)=1 and f(2)=4 and f(3)=9 and f(0)=6,a)
Title: Re: Cobby's Methods Questions
Post by: danieltennis on February 12, 2009, 09:49:30 pm: Or you can use the cubic regressions located at the Stats Editor (F4-3). It's much more efficient ;)
Title: Re: Cobby's Methods Questions
Post by: GerrySly on February 12, 2009, 10:09:31 pm: Quote from: cobby on February 12, 2009, 08:47:37 pm
Hey all im stuck on this polynomial question

Find the rule for the cubic function, the graph of which passes through the points (1,1) (2,4) (3,9) and (0,6)

Thanks :)
It depends on what course you are doing, if you are doing CAS then just plug it into the calculator and find out, otherwise this should suffice.

$\begin{align*} -5 &= a + b + c\\ -2 &= 8a + 4b + 2c\\ 3 &= 27a + 9b + 3c\\ \\ c &= -5 - a - b\\ -2 &= 8a + 4b + 2(-5 - a - b)\\ 3 &= 27a + 9b + 3(-5 - a - b)\\ \\ 8 &= 6a + 2b\\ 18 &= 24a + 6b\\ \\ b &= 4 - 3a\\ \\ 18 &= 24a + 6(4 - 3a)\\ a &= -1\\ \\ b &= 4 - 3(-1)\\ b &= 7\\ \\ c &= -5 - (-1) - (7)\\ c &= -11\\ \\ a &= -1, b = 7, c = -11, d = 6 \end{align*} $

This then leaves us with the following equation

$y = -x^3 + 7x^2 - 11x + 6$

This is assuming it is on the non-calculator exam, otherwise just plug those points into the Stats/List Editor, then press F4 Regressions and Cubic Reg, and put list1 and list2 in the X List and Y List respectively and hit Enter and it will give you those values. Enjoy :)
Title: Re: Cobby's Methods Questions
Post by: cobby on February 19, 2009, 06:55:17 pm: A man walks at a speed of 2km/h for 45 minutes and then runs at 4 km/h for 30 minutes.
Let S km be the distance the man has run after t minutes. The distance travelled can be described by

$S(t) = \begin{cases} at \\ bt+d \\ \end{cases}$

$at If 0 \leqslant t \leqslant c$
$bt+d If c < t \leqslant e$

a) Find the Values of a,b,c,d,e
b) Sketch the graphs of S(t) against t
c) State the range of the function

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: cobby on February 20, 2009, 02:36:10 pm: ^^^^

I know how to get the values of a,b,c and e, but am unsure on how to get the value of d or t???
A= 1/30
B = 1/15
C = 45
D = ?
E = 75

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on February 20, 2009, 05:35:19 pm: Quote from: cobby on February 20, 2009, 02:36:10 pm
^^^^

get the value of d or t???

t doesnt really have a value. It's just a variable. d is a constant however. You know the value of S(45) since its just 2*(3/4) (speed multipied by time) and so u sub that into the equation S(t)=bt+d and d will be ur only unkown since u already know b.
Title: Re: Cobby's Methods Questions
Post by: cobby on February 24, 2009, 05:10:13 pm: A printing firm charges $35 for printing 600 sheets of headed notepaper and 46 for printing 800 sheets. Find a linear model for the charge, C dollars, for printing n sheets. How much would they charge for printing 1000 sheets?

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on February 24, 2009, 05:54:20 pm: im assuming u meant $46 for 800 sheets

$\frac{46-35}{800-600}= 0.055$

$y-46=0.055(x-800)$

$C=0.055x+2$

for 1000 sheets

$C=0.055 \cdot 1000+2$

$\implies C = 57$
Title: Re: Cobby's Methods Questions
Post by: cobby on February 24, 2009, 06:06:41 pm: Thanks flaming arrow,but where did the +2 come from ? :S :S
P.S - sorry about the typo
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on February 24, 2009, 06:10:31 pm: Quote from: cobby on February 24, 2009, 06:06:41 pm
Thanks flaming arrow,but where did the +2 come from ? :S :S
P.S - sorry about the typo

we have to find the equation of the line, when i found the gradient i put it in the linear equation formula to find the c(y int)
Title: Re: Cobby's Methods Questions
Post by: cobby on March 03, 2009, 10:15:40 pm: Hey all another question please :)
$f: {x:x< 3}\rightarrow R, f(x) = 3-x$
$g: R \rightarrow R, g(x) = x^2-1$

Define a restriction $g^*$ of g such that $f o g^*$ is defined and find $f o g^*$

In the first part of the question i had to how that $f o g$ was not defined, don't know how to answer the next part though?

Thanks!! :)
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 03, 2009, 10:24:36 pm: Quote from: cobby on March 03, 2009, 10:15:40 pm
Hey all another question please :)
$f: {x:x< 3}\rightarrow R, f(x) = 3-x$
$g: R \rightarrow R, g(x) = x^2-1$

Define a restriction $g^*$ of g such that $f o g^*$ is defined and find $f o g^*$

In the first part of the question i had to how that $f o g$ was not defined, don't know how to answer the next part though?

Thanks!! :)
ill have a go now XD
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 03, 2009, 10:30:29 pm: for f o g* to exist

ran g* must be a subset of dom f

at the moment ran g* is $[-1,\infty)$ and dom f is $(-\infty,3)$

so ran g* must be restricted to [-1,3) so it can be a subset of dom f

for that to happen we must restrict the domain of g* to [0,2) (notice 2 is not included because the range endpoint 3 is not included), you can also think of it as this $-1 \le x^2-1 <3$ and solve for x.

there fore $f o g*: [0,2) -> R, f o g*(x) = -x^2+4$ where the range of f o g* is [-1,3)
Title: Re: Cobby's Methods Questions
Post by: cobby on March 03, 2009, 10:34:10 pm: Thanks TrueTears!!! :D really appreciate it
Title: Re: Cobby's Methods Questions
Post by: cobby on March 07, 2009, 09:26:18 am: got another question guys,

Find the rule of the image:

A reflection in the x-axis,followed by a translation 3 units in the positive direction of the x axis and 4 units in the negative direction of the y-axis followed by a dilation of factor 2 from the x-axis.

i got
$y= -2|x-3|-4$

But the answer says
$y=-2|x-3|-8$

What was i doing wrong? :S

Thanks guys
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on March 07, 2009, 10:41:18 am: when its a dilation from the x axis the whole equation is multiplied ie. 2f(x), so the -4 becomes -8
Title: Re: Cobby's Methods Questions
Post by: cobby on March 07, 2009, 10:47:13 am: Oh yepp got it, thanks man :)
Title: Re: Cobby's Methods Questions
Post by: cobby on March 08, 2009, 12:33:37 pm: Hey all,
just a quick transformations question.

Write down the matrix of each of the following transformations:

Reflection in the line $y=-x$

Can someone explain their solution as well please

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on March 08, 2009, 12:43:42 pm: well normally reflection in the y=x means the inverse and y=-x can either be a reflection in the x axis or the y axis

so

$\left[ \begin{array}{cc} 0 & 1 \\ -1 & 0\\ \end{array}\right]$ or $\left[ \begin{array}{cc} 0 & -1 \\ 1 & 0\\ \end{array}\right]$
Title: Re: Cobby's Methods Questions
Post by: cobby on March 09, 2009, 02:47:26 pm: Hey guys, another question please.

The points with co-ordinates $(1,-1)$ and $(2, \frac {3}{4})$ lie on a curve which has a rule of the form

$y= \frac{a}{x^3}+b$

Find the values of a and b

Thanks :)

EDIT - re-wrote equation
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 09, 2009, 02:50:32 pm: is that all the info given? Don't think you can work it out with just those 2 pieces of info... hmmz
Title: Re: Cobby's Methods Questions
Post by: cobby on March 09, 2009, 02:57:23 pm: oh sorry theres a graph too
ill try and draw it up on paint..
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 09, 2009, 02:59:21 pm: haha yeah knew it XD
Title: Re: Cobby's Methods Questions
Post by: cobby on March 09, 2009, 03:00:38 pm: Hey TrueTears, i wrote the eq wrong in the question..
im fixing it now, theres no graph..
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 09, 2009, 03:09:26 pm: k so theres 2 points just sub them in the equation and solve the simultaneous equation

sub (1,-1) in yields $-1 = \frac{a}{1^3} + b$ ...1

sub $(2, \frac{3}{4})$ in yields $\frac{3}{4} = \frac{a}{2^3} + b$ ...2

from 1 we get a = -1 - b sub this in 2

$\frac{3}{4} = \frac{-1-b}{8} + b$

$\frac{-1-b}{8} + \frac{8b}{8} = \frac{3}{4}$

$\frac{8b-1-b}{8} = \frac{3}{4}$

7b - 1 = 6

b = 1 sub this into 1

a = -2
Title: Re: Cobby's Methods Questions
Post by: cobby on March 09, 2009, 03:13:58 pm: Thanks man your a legend
Title: Re: Cobby's Methods Questions
Post by: cobby on March 10, 2009, 10:01:23 am: Hey guys, another one :P

Let $f: S \rightarrow R$ be given by $f(x) = \frac{x+3}{2x-1}$ where $S = R/ (\frac{1}{2})$

a) Show that $f o f$ is defined.
b) Find $f o f(x)$ and skect hthe graph of $f o f$
c) Write down the inverse of $f$

I dont quite understand what part a is asking me to do ? :S :(
What do they mean by showing the it is defined?

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on March 10, 2009, 12:18:49 pm: In order for a composite function to be defined, the range of the inside function must be a subset of the domain of the outside function. Denominator being zero is a problem, but 1/2 not being in S fixes that for the function f. but for fof u need to ensure that the range of f doesn't contain 1/2 because then u would get a denominator of zero. So lets see if 1/2 is an element of this range:

assume f(x)=1/2

$\frac{1}{2} = \frac{x+2}{2x-1}$

$2x-1=2x+4$

$2x=2x+5 (1)$

Hence if x exists such that f(x)=1/2 then that number, x, must satisfy equation 1. But no number satisfies equation 1 hence no number x gives f(x)=1/2. hence we don't have to worry about a denominator of zero and so the function exists.
Title: Re: Cobby's Methods Questions
Post by: cobby on March 16, 2009, 09:32:43 pm: Hey guys

another question please :)

Divide X^2-2 by x^4-9x^3+25x^2-8x-2

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 16, 2009, 09:34:12 pm: Hmm, I have NO idea how to use LaTeX for this, so I am gonna draw it on paint now, bear with me XD
Title: Re: Cobby's Methods Questions
Post by: cobby on March 16, 2009, 09:38:28 pm: Quote from: TrueTears on March 16, 2009, 09:34:12 pm
Hmm, I have NO idea how to use LaTeX for this, so I am gonna draw it on paint now, bear with me XD
Legend,
and i think Latex is down.....i cant see any of it in the threads.
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 16, 2009, 09:43:57 pm: Wow that took so long ( i'm so bad with paint lols)

(http://img24.imageshack.us/img24/7807/divisiong.jpg)

so therefore the answer is $\frac{-26x+52}{x^2-2} + x^2 -9x + 27$
Title: Re: Cobby's Methods Questions
Post by: cobby on March 16, 2009, 09:53:31 pm: Thanks man!! :D :D :D
Title: Re: Cobby's Methods Questions
Post by: cobby on March 21, 2009, 08:49:07 pm: Show that the equation $(k+1)x^2-2x-k=0$
has a solution for all values of k.

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 21, 2009, 08:50:49 pm: HINT: Use the $\triangle$ and show its >0

$\triangle = b^2 - 4ac$

$(-2)^2 - 4(k+1)(-k) = 4k^2 + 4k + 4$

sketch the graph of that and you will see it is larger than 0 for any value of k

Therefore $(k+1)x^2 - 2x -k = 0$ for all values of k
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on March 21, 2009, 10:17:44 pm: another way

$4k^2 + 4k + 4$

$4(k^2+k+1)$

$4[(k+\frac{1}{2})^2-\frac{1}{4}+1]$

$4[(k+\frac{1}{2})^2+\frac{3}{4}]$

$4(k+\frac{1}{2})^2+3$

as $(k+\frac{1}{2})^2 \ge 0 \therefore \mbox{a solution for all values of k exist}$
Title: Re: Cobby's Methods Questions
Post by: cobby on March 22, 2009, 08:11:01 pm: Hey all

Solve for x:
$4^{2x} - 5(4^x)=4$

I let A = $4^x$

Ended up with
$A^2-5A-4=0$
Dont know what to do after here :)

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on March 22, 2009, 08:12:39 pm: use the quadradtic equation to solve
Title: Re: Cobby's Methods Questions
Post by: cobby on March 23, 2009, 07:40:33 am: another few please :)

Solve for x:
$\log_e (x^2-2x+8) = 2 \log_e x$

$8e^{-x}-e^x=2$

Thank you
Title: Re: Cobby's Methods Questions
Post by: Damo17 on March 23, 2009, 09:00:20 am: Quote from: cobby on March 23, 2009, 07:40:33 am
another few please :)

Solve for x:
$\log_e (x^2-2x+8) = 2 \log_e x$

$8e^{-x}-e^x=2$

Thank you

$\log_e (x^2-2x+8) = 2 \log_e x$
$\log_e (x^2-2x+8) = \log_e x^2$

$\therefore x^2-2x+8=x^2$
$-2x+8=0$
$\therefore x=4$

$8e^{-x}-e^x=2$

Times everything by $e^x$

$\therefore 8-e^{2x}=2e^x$
$-e^{2x}-2e^x+8=0$

Now you would use the quadratic formula from here.
Title: Re: Cobby's Methods Questions
Post by: TonyHem on March 23, 2009, 03:54:15 pm: or cross method it
Title: Re: Cobby's Methods Questions
Post by: cobby on March 29, 2009, 09:58:37 pm: Another one guys...

please :P

solve for x:
$\log_5 x = 16 \log_x 5$
Title: Re: Cobby's Methods Questions
Post by: TrueTears on March 29, 2009, 10:02:44 pm: use change of base theorem

$16log_x5 = 16 \frac{log_55}{log_5x} = \frac{16}{log_5x}$

so $log_5x = \frac{16}{log_5x}$

$(log_5x)^2 = 16$

$log_5x = 4 or - 4$

$x = 5^4 or 5^{-4}$

$x = 625 or \frac{1}{625}$
Title: Re: Cobby's Methods Questions
Post by: cobby on April 06, 2009, 12:33:30 pm: If $sin x = 0.3, cos a = 0.6$ and $tan \theta = 0.7$
Find the Values of:

$tan ( \frac{\pi}{2} - \theta)$
$sin ( \frac{3\pi}{2} + a)$

For those using essentials this question is on page 202

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: d0minicz on April 06, 2009, 12:41:34 pm: 1. $\frac {sin (\frac {\pi}{2} - \theta) } {cos (\frac {\pi}{2} - \theta)}$ (as it's tan and tan = sin/cos)
--> $\frac {cos\theta}{sin\theta}$ ---> $\frac {1}{tan\theta}$ --> $\frac {1}{0.7}$ --> $\frac {10}{7}$
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on April 06, 2009, 12:48:10 pm: $\tan (\frac{\pi}{2}-\theta) =\frac{sin (\frac{\pi}{2}-\theta)}{ \cos(\frac{\pi}{2}-\theta)} =\frac{\cos \theta}{\sin \theta} = \cot(\theta) = \frac{1}{tan \theta}=\frac{1}{0.7} = \frac{10}{7}$

$\sin (\frac{3\pi}{2} + \alpha) = \sin ( \alpha-\frac{\pi}{2}} ) = -\cos \theta = -0.6$
Title: Re: Cobby's Methods Questions
Post by: cobby on April 07, 2009, 08:28:03 pm: Find the value of x
$x \epsilon [-\pi,\pi]$
$\cos x = -\sin \frac{\pi}{6}$

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: TrueTears on April 07, 2009, 08:28:58 pm: $cos(x) = - sin(\frac{\pi}{6})$

$sin(\frac{\pi}{2}-x) = sin(\frac{7\pi}{6}) , sin(\frac{11\pi}{6})$

$\frac{\pi}{2} - x = \frac{7\pi}{6} , \frac{11\pi}{6}$

$x = -\frac{2}{3}\pi , -\frac{4}{3}\pi$

but domain is $\frac{\pi}{2} < x < \pi$

so $-\frac{4}{3}\pi = \frac{2\pi}{3}$ ( $-\frac{2}{3}\pi$ is in the 4th quadrant, so can not be used in this domain)

therefore $x = \frac{2\pi}{3}$
Title: Re: Cobby's Methods Questions
Post by: Mao on April 07, 2009, 10:31:31 pm: that's a rather round-about way of doing this question. :P

$\cos (x) = -\sin \frac{\pi}{6}$

$\implies \cos(x) = -\frac{1}{2}$

$\implies x = \pm \frac{2\pi}{3}$
Title: Re: Cobby's Methods Questions
Post by: TrueTears on April 07, 2009, 10:34:00 pm: Quote from: Mao on April 07, 2009, 10:31:31 pm
that's a rather round-about way of doing this question. :P

$\cos (x) = -\sin \frac{\pi}{6}$

$\implies \cos(x) = -\frac{1}{2}$

$\implies x = \pm \frac{2\pi}{3}$
book said not to evaluate $sin(\frac{pi}{6})$ lols
Title: Re: Cobby's Methods Questions
Post by: Mao on April 07, 2009, 10:34:55 pm: oh, I see :P

my bad :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on April 07, 2009, 10:50:39 pm: Would've been better if the angle wasn't exact, then you would be forced to take the 'round-about'
Title: Re: Cobby's Methods Questions
Post by: cobby on April 30, 2009, 11:56:44 am: Hey guys another question please..

Differentiate using Product Rule
$x^3 (3x^2+2x+1)^{-1}$

Thanks
Title: Re: Cobby's Methods Questions
Post by: TrueTears on April 30, 2009, 12:28:56 pm: $3x^2(3x^2+2x+1)^{-1} + x^3(-1)(3x^2+2x+1)^{-2}(6x+2)$

I assume you can simplify from there?
Title: Re: Cobby's Methods Questions
Post by: TrueTears on May 04, 2009, 03:30:52 pm: (as request)

continuing from last line of working.

$\frac{3x^2}{3x^2+2x+1} - \frac{x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^2(3x^2+2x+1)}{(3x^2+2x+1)^2} - \frac{x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^2(3x^2+2x+1) - x^3(6x+2)}{(3x^2+2x+1)^2}$

$= \frac{3x^4 + 4x^3 + 3x^2}{(3x^2+2x+1)^2}$
Title: Re: Cobby's Methods Questions
Post by: cobby on May 08, 2009, 08:33:16 pm: Hey guys,

Is it possible to find the derivative of $|x-1|-1$ by hand?

I typed it in on my calculator and it came up with $sign(x-1)$
Title: Re: Cobby's Methods Questions
Post by: Mao on May 08, 2009, 08:59:47 pm: Think of the modulus function as a hybrid function of positive and negative parts

$|x-1|-1 = \begin{cases} -(x-1)-1 & x\le 1 \\ (x-1)-1 & x > 1 \end{cases}$

simplifying and differentiating both side gives

$\frac{d}{dx}(|x-1|-1) = \begin{cases} -1 & x < 1 \\ 1 & x > 1 \end{cases}$
Title: Re: Cobby's Methods Questions
Post by: cobby on May 10, 2009, 12:31:08 pm: Hey guys,
Struggling with this question..

if $f(x) = (x+1) (\sqrt {x^2+1})$

Find $f'(x)$

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: dekoyl on May 10, 2009, 12:49:29 pm: Use the product rule:
$f'(x)=(x+1)\frac{2x}{2\sqrt{x^2+1}} + \sqrt{x^2+1}$
$=\frac{x(x+1)}{\sqrt{x^2+1}} + \sqrt{x^2+1}$
I think that's all you need to do.
Title: Re: Cobby's Methods Questions
Post by: cobby on May 10, 2009, 06:53:14 pm: Thanks dekoyl...i got that answer too..

but the book says..
$f'(x)= \frac {2x^2+x+1}{\sqrt{x^2+1}}$
Title: Re: Cobby's Methods Questions
Post by: dekoyl on May 10, 2009, 07:01:00 pm: Add them together
$\frac{x(x+1)}{\sqrt{x^2+1}} + \sqrt{x^2+1}$
$\frac{x^2+x}{\sqrt{x^2+1}}+\frac{x^2+1}{\sqrt{x^2+1}}$
= $\frac {2x^2+x+1}{\sqrt{x^2+1}}$
Title: Re: Cobby's Methods Questions
Post by: cobby on May 10, 2009, 07:03:39 pm: Sweet..thanks!!!
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 10, 2009, 07:05:13 pm: Quote from: cobby on May 10, 2009, 06:53:14 pm
Thanks dekoyl...i got that answer too..

but the book says..
$f'(x)= \frac {2x^2+x+1}{\sqrt{x^2+1}}$

thats the same thing, they got a common denominator in that 1
Title: Re: Cobby's Methods Questions
Post by: cobby on May 12, 2009, 10:23:55 pm: Hey guys another q please

Consider the function $f(x) = (x-1)^2(x-b)$ where $b>1$

a) Find the derivative of $f(x)$ with respect to $x$ - DONE
b) Find the coordinates of the stationary points of the graph of $y=f(x)$ - DONE
c) Show that the stationary point at (1,0) is always a local maximum - NO CLUE WHAT THAT MEANS!? :(
d) Find the value of $b$ if the stationary points occur where x = 1 and x = 4 - Not Done

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: dekoyl on May 12, 2009, 10:33:15 pm: I haven't done these in a while but for C, maybe do a sign diagram (/ at 0, _ at 1, \ at 2 => local maximum)?
Not sure if this satisfies the ALWAYS LOCAL MAXIMUM, though.

=(
Title: Re: Cobby's Methods Questions
Post by: cobby on May 12, 2009, 10:35:01 pm: Quote from: dekoyl on May 12, 2009, 10:33:15 pm
I haven't done these in a while but for C, maybe do a sign diagram (/ at 0, _ at 1, \ at 2 => local maximum)?
Not sure if this satisfies the ALWAYS LOCAL MAXIMUM, though.

=(
Yeh when they mentiond 'always' thats threw me off course a bit.. :(
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 12, 2009, 10:53:19 pm: Quote from: cobby on May 12, 2009, 10:35:01 pm
Quote from: dekoyl on May 12, 2009, 10:33:15 pm
I haven't done these in a while but for C, maybe do a sign diagram (/ at 0, _ at 1, \ at 2 => local maximum)?
Not sure if this satisfies the ALWAYS LOCAL MAXIMUM, though.

=(

Yeh when they mentiond 'always' thats threw me off course a bit.. :(

'always' means for any value of b>1. So you would have to show that if b>1, dy/dx<0 when x>1 and 0<dy/dx when x<1. This is because positive gradient implies increasing, negative implies decreasing and so the gradient must be positive before x=1 and negative after x=1 in order for it to be max at x=1.
Title: Re: Cobby's Methods Questions
Post by: cobby on May 12, 2009, 11:11:37 pm: Straight line pass through the point B (1,-2) and is (are) tangent to the parabola with equation $y=x^2$
Point A is on the parabola

1a) Show that the line is passing through point AB is given by
$y=mx-(m+2)$

b)Show that the x co-ordinate of A is given by the solution to equation:
$x^2-mx+m+2=0$

c) Hence show that: $m^2-4m-8=0$

d) Hence find the x and y co-ordinates - dont worry about this

e) Hence find the equations of the straight lines

Could someone please answer this...its for my SAC preparation ...

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 12, 2009, 11:31:07 pm: a.)
y=mx+c
subbing in B:
-2=m+c
c=-2-m
=-(2+m)

b.)
A is (x,y)

m=gradient

$=\frac{y-(-2)}{x-1}$

$y=x^2$ :

$m=\frac{x^2+2)}{x-1}$
$mx-m=x^2+2$
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 12, 2009, 11:34:51 pm: 1a)
$y+2 = m(x-1)$

$\implies y = mx-(m+2)$

b)

$A(x,x^2)$

$x^2 = mx-(m+2)$

$x^2-mx+m+2 = 0$

c)

a= 1 b = m c = m+2

$b^2-4ac$

$m^2-4(m+2) =0$

$m^2-4m-8 =0$

e)
$m^2-4m-8 =0$

solve for m and sub into equation in part a
Title: Re: Cobby's Methods Questions
Post by: cobby on May 13, 2009, 06:52:17 am: Thank you so much guys!! :D :D :D
Title: Re: Cobby's Methods Questions
Post by: cobby on May 17, 2009, 06:26:30 pm: Hey guys got another question

For the curve with equation $y= \frac{5}{3}x + kx^2 - \frac{8}{9}x^3$ , calculate the possible values of k such that the tangents at the points with x-coordinates $1$ and $\frac {-1}{2}$ respectively are perpendicular.
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 17, 2009, 06:35:02 pm: $\frac{dy}{dx} = \frac{5}{3} + 2kx-\frac{24x^2}{9}$

let x = 1

$2k-1$

let $x = \frac{-1}{2}$

$-k+ 1$

$m_1*m_2 = -1$

$(2k-1) \cdot (-k+1) = -1$

you can do it from there
Title: Re: Cobby's Methods Questions
Post by: cobby on May 17, 2009, 08:01:38 pm: Another one guys :(

Show that $\frac {d}{dx} (\sqrt {x^2 \pm a^2} = \frac {x}{ \sqrt {x^2 \pm a^2}}$

Thanks!! :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 17, 2009, 08:04:43 pm: $\frac {d}{dx} (\sqrt {x^2 \pm a^2} = \frac {x}{ \sqrt {x^2 \pm a^2}}$

${(x^2 \pm a^2})^\frac{1}{2}$

$\frac {d}{dx} = \frac{1}{2} * {(x^2 \pm a^2})^\frac{-1}{2} * 2x$

$\frac {d}{dx} = {(x^2 \pm a^2})^\frac{-1}{2} * x \implies \frac {x}{ \sqrt {x^2 \pm a^2}}$
Title: Re: Cobby's Methods Questions
Post by: cobby on May 17, 2009, 08:21:39 pm: And another one please :)

If $y = (x + \sqrt {x^2+1})^2$ , show $\frac {d}{dy} = \frac {2y} {\sqrt {x^2+1}}$

I keep getting muddled up with these questions :(

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 17, 2009, 08:34:51 pm: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

$\frac{dy}{dx} = 2 (x+\sqrt{x^2+1}) * (1+ \frac{x}{\sqrt{x^2+1}})$

get a common denominator

$\frac{dy}{dx} = 2 (x+\sqrt{x^2+1}) * (\frac{x + \sqrt{x^2+1}}{\sqrt{x^2+1}})$

$2\frac{(x + \sqrt{x^2+1)}^2}{\sqrt{x^2+1}}$

and sub in y
Title: Re: Cobby's Methods Questions
Post by: cobby on May 18, 2009, 10:04:00 pm: Hey guys got another question

Taken from 2007 MMC Exam 1

P is the point on the line $2x+y-10=0$ such that the length of $OP$ , the line segment from the origin $O$ to $P$ , is a minimum. Find the coordinates of $P$ AND this minimum length.

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 18, 2009, 10:30:15 pm: there are two appraoches, the geometrical and the analytical. Since you're into differentiation lately i'll do the latter:

find formula for distance:

$OP=\sqrt{x^2+y^2}$ where P=(x,y)
y=10-2x:

$OP=\sqrt{x^2+(10-2x)^2}$
$OP=\sqrt{5x^2-40x+100}$
-
When OP is at a minimum, $5x^2-40x+100$ is at a minimum since the square root function has no turning points.

differentiating and setting derivative to zero:

$0=10x-40$

I think you can take it from here.

--------------------------------------------------------------------

Geometrical:

Draw the circle with radius OP and centre 0. The line is a tangent of the circle, hence the radial line is perpendicular to the tangent:

Radial line:

y=mx+c

$m=\frac{1}{2}$ (negative reciprical of gradient of given line)

c=0 since it goes through centre which is origin.

radial line: $y=\frac{1}{2}x$

Hence the point P is the intersection of 2x+y-10=0 and the radial line, which is easy to find.
Title: Re: Cobby's Methods Questions
Post by: cobby on May 18, 2009, 10:33:47 pm: Thanks kamil :)

But where did you get $-3x^2-40x+100$ from in the first solution?
Title: Re: Cobby's Methods Questions
Post by: d0minicz on May 18, 2009, 10:55:48 pm: is that sposed to be -5x^2 - 40x + 100 ?
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 18, 2009, 11:21:47 pm: Quote from: cobby on May 18, 2009, 10:33:47 pm
Thanks kamil :)

But where did you get $-3x^2-40x+100$ from in the first solution?

LOl sorry its supposde to be 5x^2-40x+100. (thing inside square root)
I basically had the -3 initially but then forgot to change it in every part where i wrote it.
Title: Re: Cobby's Methods Questions
Post by: cobby on May 19, 2009, 06:59:00 pm: Another one please :)

The tangent to the curve of $y=ax^2+bx+c$ at the point where $x = 2$ is parallel to the line $y=4x$ . There is a stationary point at $(1,-3)$ . Find the values of $a, b, c$

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 19, 2009, 07:21:27 pm: you have to set up some simultaneous equations. The pieces of information provide you with this.
"There is a stationary point at 1,-3" has two pieces of information, the point is on that curve and the derivative is zero at x=1:

first piece of info: -3=a+b+c (1)

second piece of info:

dy/dy=2ax+b
0=2a+b (2)

Now one final piece of info:

at x=2, dy/dy=4:

4=2a(2)+b
4=4a+b (3)

use (2) and (3) to find a and b, then plug into (1) to find c
Title: Re: Cobby's Methods Questions
Post by: GerrySly on May 19, 2009, 07:21:54 pm: $\begin{align*} y&=ax^2+bx+c\\ \therefore \frac{dy}{dx}&=2ax+b\\ 4&=2a(2)+b\\ \implies b&=4-4a\\ 0&=2a(1)+b\\ \implies a&=\frac{4a-4}{2}\\ \therefore a&=2\\ \therefore b&=-4\\ -3&=2(1)^2-4(1)+c\\ \implies c&=-1\\ \therefore y&=2x^2-4x-1 \end{align*}$

Me thinks

EDIT: Wow, I keep getting beaten to posting help :P
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 19, 2009, 07:23:34 pm: Answered on MSN :P
Title: Re: Cobby's Methods Questions
Post by: cobby on May 19, 2009, 07:59:39 pm: Thanks guys!!
Title: Re: Cobby's Methods Questions
Post by: cobby on May 19, 2009, 10:05:17 pm: A rectangular block is such that the sides of its base are of length x cm and 3x cm. The sum of the lengths of all its edges is 20cm

show that the volume Vcm^3 is given by $V = 15x^2-12x^3$

Yes i know, the height is not given hence why i cant get the volume...is there another way of doing this q or is it a misprint??

Thanks guys
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 19, 2009, 10:19:45 pm: you can find the height with the additional information of "the sum of edge lengths is 20". WHen apparently missing a piece of information, look at how another piece of information may compensate.

You can see that there are four edges of length x, four of length 3x, and four of length h.

hence:

$20=4x+4(3x)+4h$
$5=x+3x+h$
$h=5-4x$

$V=h(4x)x$
Title: Re: Cobby's Methods Questions
Post by: dcc on May 19, 2009, 11:17:37 pm: Quote from: cobby on May 18, 2009, 10:04:00 pm
Hey guys got another question

Taken from 2007 MMC Exam 1

P is the point on the line $2x+y-10=0$ such that the length of $OP$ , the line segment from the origin $O$ to $P$ , is a minimum. Find the coordinates of $P$ AND this minimum length.

Thanks guys :)

For those who are interested, kamil9876's Geometrical approach can be generalised using vector magic.

If we have a line $y = a + bx$ with real coefficients, we can write this in the form $\textbf{L} = (0, a) + x(1, b)$ . Now we note that the DIRECTION of the line $\textbf{L}$ is given by the vector $\textbf{D} = (1, b)$ . Therefore to find our minimum, we wish to find some $x$ such that $\textbf{L} \cdot \textbf{D} = 0$ (Think about it, if $\textbf{L}$ is not perpendicular to $\textbf{D}$ , then there is always another line you can draw which will be shorter).

We find that $\textbf{L} \cdot \textbf{D} = 0 = x + (a + xb)b = x + ab + xb^2 = x(1 + b^2) + ab = 0 \implies x = - \frac{ab}{1 + b^2}$

Therefore, we notice that the minimum distance from the origin to a point on the line $y = a + bx$ always occurs when $\boxed{x = - \frac{ab}{1 + b^2}}$ .

(Ignore this if you don't care or aren't interested)
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 20, 2009, 11:54:01 am: Nice :)

I believe I've seen an analysis question in specialist that uses that dot product approach.
You can also prove that the product of gradients of two perpendicular lines equals -1 using the dot product, so i guess that's where the two approaches are somewhat logically related :)
Title: Re: Cobby's Methods Questions
Post by: cobby on May 22, 2009, 07:29:53 pm: Hey guys,

can someone help me with this question.. It's preparation for the SAT's coming up.

The length of rectangle S is 20 percent longer than the length of rectangle R, and the width of rectangle S is 20 percent shorter than the width of rectangle R. The area of rectangle S is:

A) 20% greater than the area of rectangle R
B) 4% greater than the area of rectangle R
C) equal to the area of rectangle R
D) 4% less than the area of rectangle R
E) 20% less than the area of rectangle R

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: pHysiX on May 22, 2009, 08:56:57 pm: D

Let length of rectangle R be "L" and width of rectangle R be "W"
Let length of rectangle S be "A" and width of rectangle S be "B"

From given information:
Length of rectangle S is 20 percent longer, which implies length of S = 1.2 length of W

i.e. A=1.2L

Width of rectangle S is 20 percent shorter, which implies width of S = 0.8 width of W

i.e. B=0.8W

Area of S = AB = 0.96LW
Area of R = LW

Difference in Areas = 0.04

Percentage = .04/1 (difference divided by original) x100

so therefore, S is 4% less
Title: Re: Cobby's Methods Questions
Post by: kamil9876 on May 22, 2009, 09:03:33 pm: "The SAT Reasoning Test (formerly Scholastic Aptitude Test and Scholastic Assessment Test) is a standardized test for college admissions in the United States"

whoa, plans to go to US.

anyways back to the question...

so lets denote our variables: $w_R, L_R,w_S,L_S$

$L_S=1.2L_R$
$W_S=0.8W_R$

multiply the equations together to get:

$L_S W_S=1.2*0.8 L_R W_R$
$A_S=0.96A_R$

So Area of S is 96% of the area of R. Hence it is 4% less.
Title: Re: Cobby's Methods Questions
Post by: cobby on May 22, 2009, 09:12:46 pm: Wow thanks guys!!!

And yes Kamil i plan to go to the states next year :)
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on May 24, 2009, 07:37:29 pm: Quote from: cobby on May 22, 2009, 09:12:46 pm
Wow thanks guys!!!

And yes Kamil i plan to go to the states next year :)

To cleeveland community college ahaha
Title: Re: Cobby's Methods Questions
Post by: cobby on May 28, 2009, 10:06:57 pm: Hey guys another one please :)

The surface area of a cube is changing at the rate of $8 cm^2/s$ . How fast is the volume changing when the surfrace area is $60 cm$ ?

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: dekoyl on May 28, 2009, 10:26:53 pm: $S = 4\pi{r^2}$
$\frac{dS}{dt}=8 cm^2/s$
We want:
$\frac{dV}{dt}=\frac{dV}{dS}.\frac{dS}{dt}$
$V = \frac{4}{3}\pi{r^3}$
$\implies V = \frac{S}{3}\sqrt{\frac{S}{4\pi}$
$\implies V = \sqrt{\frac{S^3}{12\pi}$
$\frac{dV}{dS}=\frac{3S^2}{24\pi\sqrt{\frac{S^3}{12\pi}}}$
$\implies \frac{dV}{dt}=\frac{S^2}{\pi\sqrt{\frac{S^3}{12\pi}}}$

Now sub $S=60$ into the above.

Sorry. I might have bullshitted my way through because it's WTF-esque.

LOL ITS FOR A CUBE. I READ SPHERE. SORRY.
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on May 28, 2009, 10:32:27 pm: i got 2 rt[10] but i accidently exited my window after typing everything up, cbf now
Title: Re: Cobby's Methods Questions
Post by: Mao on May 28, 2009, 11:03:31 pm: $S=6r^2\implies r = \left(\frac{S}{6}\right)^{1/2}$

$V=r^3 = \left(\frac{S}{6}\right)^{3/2}$

$\frac{dV}{dt} = \frac{dS}{dt}\cdot \frac{dV}{dS}$

the rest is trivial :)
Title: Re: Cobby's Methods Questions
Post by: cobby on May 31, 2009, 07:36:55 pm: Hey guys, can someone please help me with this?

Find the following
$\frac{d(e^x f(x))}{dx}$

I don't even know where to start with this??

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: d0minicz on May 31, 2009, 07:42:55 pm: do it like normal using the product rule
so theyre asking you to find the derivative of $e^x f(x)$
remember the derivative of f(x) is f'(x) and use that
Title: Re: Cobby's Methods Questions
Post by: Damo17 on May 31, 2009, 07:43:38 pm: Quote from: cobby on May 31, 2009, 07:36:55 pm
Hey guys, can someone please help me with this?

Find the following
$\frac{d(e^x f(x))}{dx}$

I don't even know where to start with this??

Thanks guys :)

use product rule:
$ \frac{d}{dx}(e^x f(x))= e^x f(x)+e^x f '(x)$
$= e^x(f(x)+ f '(x))$
Title: Re: Cobby's Methods Questions
Post by: cobby on June 17, 2009, 06:06:35 pm: Hey guys :)

Given that $y= e^{2t}$ and $x = e^t +1$

Find $\frac {dy}{dx}$ when $t=0$

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: Mao on June 17, 2009, 06:12:16 pm: $\frac{dy}{dt} = 2e^{2t}$ , $\frac{dx}{dt} = e^t$

$\implies \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 2e^t$

when t=0, $\frac{dy}{dx} = 2$
Title: Re: Cobby's Methods Questions
Post by: GerrySly on June 17, 2009, 06:13:30 pm: $\begin{align*} \frac{dy}{dx}&=\frac{dy}{dt}\frac{dt}{dx}\\ \therefore \frac{dy}{dx}&=2e^{2t}\frac{1}{e^t}\\ &=\frac{2e^{2t}}{e^t}\\ &=2e^{t}\\ \end{align*} $

So therefore when t=0

$\frac{dy}{dx}&=2$
Title: Re: Cobby's Methods Questions
Post by: d0minicz on June 17, 2009, 06:14:53 pm: Change $x= e^t + 1$ into $t= log_e (x-1)$
then sub it into $y= e^{2t}$ => $y= e^ {2log_e (x-1)}$
thennn find dy/dx of that
which is : $\frac {2}{x-1} e^{2log_e (x-1)}$
and now u find what x equals to when t=0
sub t=0 into the $x= e^t +1$
and you get x= 2
now sub x=2 into the dy/dx equation
and you'll get 2.

damn beaten
wtf my method is fekd
Title: Re: Cobby's Methods Questions
Post by: cobby on June 17, 2009, 06:18:07 pm: Quote from: Mao on June 17, 2009, 06:12:16 pm
$\frac{dy}{dt} = 2e^{2t}$ , $\frac{dx}{dt} = e^t$

$\implies \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 2e^t$

when t=0, $\frac{dy}{dx} = 2$

Hey mao, why'd you divide the two derivatives???
Title: Re: Cobby's Methods Questions
Post by: d0minicz on June 17, 2009, 06:19:31 pm: its the same as saying
$\frac {dy}{dx} = \frac {dy}{dt} . \frac {dt}{dx}$
Title: Re: Cobby's Methods Questions
Post by: ilovevce on June 17, 2009, 06:21:38 pm: Alternatively,

$t = \log_{e} {x-1}$

$y = e^{2\log_{e} {x-1}} = (x-1)^2$

$\frac{dy}{dx} = 2(x-1) = 2x-2 = 2e^t$

When $t = 0$ , $\frac{dy}{dx} = 2$
Title: Re: Cobby's Methods Questions
Post by: cobby on June 17, 2009, 06:27:06 pm: thanks guys!!! :)

i have another one now

$y = e^{x}(px^2 +qx +r)$ is such that the tangents at $x = 1$
and $x = 3$ are parallel to the x-axis. The point with co-ordinates $(0,9)$ is on the curve.
Find the values of $p, q$ and $r$

Been trying to do this one for a while...im getting no where :(

Thanks
Title: Re: Cobby's Methods Questions
Post by: ilovevce on June 17, 2009, 06:30:27 pm: It's telling you that there are stationary points at x=1 and x=3 (gradient = 0 therefore tangent is parallel to x-axis).

Differentiate the equation and make equal to 0 to find the stationary points. You should be able to solve for p,q and r from there.
Title: Re: Cobby's Methods Questions
Post by: ilovevce on June 17, 2009, 06:49:42 pm: Firstly, substitute in the point on the curve so you get $r = 9$

Now, $\frac{dy}{dx} = e^x(px^2+qx+r+2px+q)$

For stationary points $\frac{dy}{dx} = 0 = px^2+qx+r+2px+q$

Sub in r and x = 1:

$0 = 3p+2q+9$
$\therefore p = -3-\frac{2q}{3}$

Sub in r and x = 3:

$0 = 15(-3-\frac{2q}{3})+4q+9$
$\therefore q = -6$

Finally, $p = -3-\frac{2q}{3} = 1$
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:24:04 pm: Hey guys :)

I was trying to find the $\frac {dy}{dx}$ of $tan {2x} + \frac {1}{cos(x)}$ :

When i punched it into the calculator, the result was a super long fraction...is there something wrong with my calc lol?

Did anyone else get this result?
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 02, 2009, 04:25:17 pm: If you use TI-89 go to F2 -> 9 -> 1: texpand or 2: tCollect

try that
Title: Re: Cobby's Methods Questions
Post by: zzdfa on July 02, 2009, 04:27:01 pm: try evaluating it at particular point, is it the same as if you did it by hand?
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:27:15 pm: Quote from: TrueTears on July 02, 2009, 04:25:17 pm
If you use TI-89 go to F2 -> 9 -> 1: texpand or 2: tCollect

try that

hmm it shortens the fraction, but its still long...does the ti-89 have $sec^2$ ?
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 02, 2009, 04:27:59 pm: go to catalog everything (most) is there

or just $\frac{1}{cos^2x}$ it
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:31:26 pm: Quote from: zzdfa on July 02, 2009, 04:27:01 pm
try evaluating it at particular point, is it the same as if you did it by hand?

I evaluated a point and got the same answer :)

Quote from: TrueTears on July 02, 2009, 04:27:59 pm
go to catalog everything (most) is there

or just $\frac{1}{cos^2x}$ it

Thanks man :) :)
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on July 02, 2009, 04:32:06 pm: Or press 2nd and then 5, which goes to the math menu, then scroll down to Trig.
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:33:07 pm: Quote from: kurrymuncher on July 02, 2009, 04:32:06 pm
Or press 2nd and then 5, which goes to the math menu, then scroll down to Trig.
Oh haha!

Totally forgot about that menu ..thanks!!!
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 02, 2009, 04:34:04 pm: type this in your TI-89

$(\frac{1}{0})^2$

see what you get.
Title: Re: Cobby's Methods Questions
Post by: kurrymuncher on July 02, 2009, 04:35:03 pm: infinity? lol
Title: Re: Cobby's Methods Questions
Post by: Over9000 on July 02, 2009, 04:36:31 pm: Quote from: TrueTears on July 02, 2009, 04:34:04 pm
type this in your TI-89

$(\frac{1}{0})^2$

see what you get.
Thats interesting, you get infinity, yet when you type (1/0) X (1/0), you get back to undef, lol.
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:38:43 pm: Quote from: TrueTears on July 02, 2009, 04:34:04 pm
type this in your TI-89

$(\frac{1}{0})^2$

see what you get.
haha cool :P

how do you get the plus/minus sign on the ti-89?
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 02, 2009, 04:39:29 pm: $tan^{-1}(tan(\frac{\pi}{2}))$

or 2ND "+" -> 2 -> find $\pm$ in there.
Title: Re: Cobby's Methods Questions
Post by: cobby on July 02, 2009, 04:42:44 pm: Quote from: TrueTears on July 02, 2009, 04:39:29 pm
$tan^{-1}(tan(\frac{\pi}{2}))$

or 2ND "+" -> 2 -> find $\pm$ in there.
woo got it :P
Title: Re: Cobby's Methods Questions
Post by: cobby on July 16, 2009, 08:15:01 pm: Hey guys

How do i find the integral of $\frac {x}{x+1}$

The book doesn't have any examples of this form of equation (N)

Thanks guys
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 16, 2009, 08:15:42 pm: Try long division first.

$\frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$

That should be clearly. no?
Title: Re: Cobby's Methods Questions
Post by: cobby on July 16, 2009, 08:41:35 pm: Quote from: TrueTears on July 16, 2009, 08:15:42 pm
Try long division first.

$\frac{x+1-1}{x+1} = \frac{x+1}{x+1} - \frac{1}{x+1} = 1 - \frac{1}{x+1}$

That should be clearly. no?
sweet, got it

thanks man :)
Title: Re: Cobby's Methods Questions
Post by: cobby on July 26, 2009, 07:10:11 pm: Hey guys,

Im having trouble with part e of the attached question, can someone please help me? :)

I found that the L.E.P estimate = 1650 and the R.E.P estimate = 1650

But part e asks for two values and i only have one? :S :S

Thanks :)

EDIT: Got it guys, made a silly error with the r.e.p estimate.

EDIT TAKE TWO: I dont got it :( i thought it did, but i still end up with 1650 for both estimates :(

My working out

L.E.P
(0*10)+(9*10)+(16*10)+(21*10)+(24*10)+(25*10)+(24*10)+(21*10)+(16*10)+(9*10)+(0*10)
= 0 + 90 + 160 + 210 + 240 + 250 + 240 + 210 + 160 + 90 + 0
= 1650

R.E.P
(9*10)+(16*10)+(21*10)+(24*10)+(25*10)+(24*10)+(21*10)+(16*10)+(9*10)
= 90 + 160 + 210 + 240 + 250 + 240 + 210 + 160 + 90
= 1650
:S
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 26, 2009, 10:11:56 pm: For c) I got $5 \times 10 \times [9+16+21+24+25] = 4750$ (which is what we expect because it's an overestimate)

d) $4 \times 10 \times [9+16+21+24] = 2800$ (which is what we expect because it's an underestimate)

so let the real value be $X$

$2800 < X < 4750$
Title: Re: Cobby's Methods Questions
Post by: cobby on July 26, 2009, 11:05:41 pm: Quote from: TrueTears on July 26, 2009, 10:11:56 pm
For c) I got $5 \times 10 \times [9+16+21+24+25] = 4750$ (which is what we expect because it's an overestimate)

d) $4 \times 10 \times [9+16+21+24] = 2800$ (which is what we expect because it's an underestimate)

so let the real value be $X$

$2800 < X < 4750$
Sorry TT...that can't be right as when i find the exact area of the curve i get $1666.67m^2$
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 26, 2009, 11:08:06 pm: True.
Title: Re: Cobby's Methods Questions
Post by: /0 on July 26, 2009, 11:53:17 pm: $f(x) = -0.01x^2+25$

Area 1 = $10(f(-40)+f(-30)+f(-20)+f(-10)+f(0)+f(0)+f(10)+f(20)+f(30)+f(40))$

But the graph is an even function, i.e. $f(-x)=f(x)$

So Area 1 = $20(f(0)+f(10)+f(20)+f(30)+f(40))=1900$

And Area 2 = $10(f(-50)+f(-40)+f(-30)+f(-20)+f(-10)+f(10)+f(20)+f(30)+f(40)+f(50))$

$=20(f(10)+f(20)+f(30)+f(40)+f(50))=1400$
Title: Re: Cobby's Methods Questions
Post by: TrueTears on July 26, 2009, 11:57:09 pm: Quote from: /0 on July 26, 2009, 11:53:17 pm
$f(x) = -0.01x^2+25$

Area 1 = $10(f(-40)+f(-30)+f(-20)+f(-10)+f(0)+f(0)+f(10)+f(20)+f(30)+f(40))$

But the graph is an even function, i.e. $f(-x)=f(x)$

So Area 1 = $20(f(0)+f(10)+f(20)+f(30)+f(40))=1900$

And Area 2 = $10(f(-50)+f(-40)+f(-30)+f(-20)+f(-10)+f(10)+f(20)+f(30)+f(40)+f(50))$

$=20(f(10)+f(20)+f(30)+f(40)+f(50))=1400$
OH YEAH SH**t
Title: Re: Cobby's Methods Questions
Post by: cobby on July 27, 2009, 07:13:55 am: Thanks /0,
But i don't quite follow why you multiplied the areas by 20?
Title: Re: Cobby's Methods Questions
Post by: /0 on July 27, 2009, 09:53:38 am: Area 1 = 10(f(-40)+f(-30)+f(-20)+f(-10)+f(0)+f(0)+f(10)+f(20)+f(30)+f(40))

But f(-40) = f(40), f(-30) = f(30), f(-20) = f(20), f(-10) = f(10)

=> Area 1 = 10(2f(40)+2f(30)+2f(20)+2f(10)+2f(0)) = 20(f(40)+f(30)+f(20)+f(10)+f(0))

I use the same reasoning for the other area.
Title: Re: Cobby's Methods Questions
Post by: cobby on July 27, 2009, 06:35:55 pm: Quote from: /0 on July 27, 2009, 09:53:38 am
Area 1 = 10(f(-40)+f(-30)+f(-20)+f(-10)+f(0)+f(0)+f(10)+f(20)+f(30)+f(40))

But f(-40) = f(40), f(-30) = f(30), f(-20) = f(20), f(-10) = f(10)

=> Area 1 = 10(2f(40)+2f(30)+2f(20)+2f(10)+2f(0)) = 20(f(40)+f(30)+f(20)+f(10)+f(0))

I use the same reasoning for the other area.

ahh kk..got it...i had to count f(0) twice ..
Thanks /0!
Title: Re: Cobby's Methods Questions
Post by: cobby on August 04, 2009, 10:59:03 pm: Hey,

Taken from AMC practice Questions

If $a^2 = a + 2$ then $a^3 = ?$

Thanks :)
Title: Re: Cobby's Methods Questions
Post by: Mao on August 04, 2009, 11:53:54 pm: solving first equation a^2 - a - 2 = 0, solution yields a = -1, a = 2, hence a^3 = -1 or 8
Title: Re: Cobby's Methods Questions
Post by: cobby on August 09, 2009, 07:28:41 pm: Hey all,

A quantity of gas expands under pressure $p N/m^2$ according to the law $pv^{0.9} = 300$ where $v m^3$ is the volume of gas under pressure $p N/m^2$

a) What is the average pressure as the volume changes from $\frac{1}{2}m^2$ to $1m^2$ ?

b) If the change in volumes in terms of $t$ is given by $v = 3t +1$ , what is the average pressure as the time changes from $0$ to $1$

Thank you! :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on August 09, 2009, 07:42:28 pm: $p = \frac{300}{v^{0.9}}$

a. $\frac{1}{1-\frac{1}{2}}\int_\frac{1}{2}^{1}\frac{300}{v^{0.9}} dv = 2\int_\frac{1}{2}^{1}300v^{-0.9} dv$

$2[\frac{300v^{0.1}}{0.1}]_\frac{1}{2}^1 = [6000v^{0.1}]_\frac{1}{2}^1$

i think you can do it from there

b.

v(0) = 1
v(1) = 4

$\frac{1}{4-1} \int_1^{4}300v^{-0.9} dv$

:)
Title: Re: Cobby's Methods Questions
Post by: cobby on August 09, 2009, 08:02:19 pm: Thanks man!
Title: Re: Cobby's Methods Questions
Post by: cobby on August 11, 2009, 06:11:40 pm: Hey :)

Can someone please take a look at my solution to this question and see if i've done anything wrong, because the answers section says i have :(

The average value of the function $f: [0,\frac{\pi}{3}] -> R$ , where $f(x) = cos (2x)$ is?

My working out

$\frac{1}{b-a} \int_{0}^\frac{\pi}{3} cos(2x)dx$

$\implies \frac {3}{\pi}[(\frac{1}{2} sin (2x))]_0^\frac{\pi}{3}$

$\implies \frac{3}{\pi} * \frac{\sqrt{3}}{4}$

$= \frac{3\sqrt{3}}{4\pi}$

Solutions say $\frac{3}{4\pi}$

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: cobby on August 22, 2009, 08:44:26 pm: Hey all,

Rex is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Rex hitting the target exactly five times to be more than 25%

My working so far has been

Pr (X=5) > 0.25

nC5* (0.6)^5 * (0.4)^[n-5] > 0.25

I can't seem to get any further than this, as my calculator won't solve for 'n'

Thanks :)

P.S: Why isn't Latex showing up in the threads? :S
Title: Re: Cobby's Methods Questions
Post by: cobby on August 22, 2009, 09:22:20 pm: Quote from: andrewloppol on August 22, 2009, 09:13:36 pm
Quote from: cobby on August 22, 2009, 08:44:26 pm
Hey all,

Rex is shooting at a target. His probability of hitting the target is 0.6. What is the minimum number of shots needed for the probability of Rex hitting the target exactly five times to be more than 25%

My working so far has been

Pr (X=5) > 0.25

nC5* (0.6)^5 * (0.4)^[n-5] > 0.25

I can't seem to get any further than this, as my calculator won't solve for 'n'

Thanks :)

P.S: Why isn't Latex showing up in the threads? :S

What calculator do you use??
Ti-89 Titanium
Title: Re: Cobby's Methods Questions
Post by: cobby on September 15, 2009, 07:47:24 pm: Hey all,

From a large batch of parts, a random sample of 20 parts is selected. It is found that the proportion of defective parts in this batch is 0.07. Let X denote the number of defective parts found in the sample.

a) Find
i) Pr(X=0) - Done
ii) Pr(X=1) - Done

A quality control procedure is set in place for accepting or rejecting large batches of parts.
From the random sample of 20, if the sample contains no more than 1 defective, it is accepted. If the batch contains at least 2 defectives, then it is rejected.

b) Using this process, find the probability of accepting a batch of parts.

I need help with Part b

Thanks guys :)
Title: Re: Cobby's Methods Questions
Post by: Flaming_Arrow on September 15, 2009, 07:59:14 pm: Quote from: cobby on September 15, 2009, 07:47:24 pm
Hey all,

From a large batch of parts, a random sample of 20 parts is selected. It is found that the proportion of defective parts in this batch is 0.07. Let X denote the number of defective parts found in the sample.

a) Find
i) Pr(X=0) - Done
ii) Pr(X=1) - Done

A quality control procedure is set in place for accepting or rejecting large batches of parts.
From the random sample of 20, if the sample contains no more than 1 defective, it is accepted. If the batch contains at least 2 defectives, then it is rejected.

b) Using this process, find the probability of accepting a batch of parts.

I need help with Part b

Thanks guys :)

Pr(1>X>0)
Title: Re: Cobby's Methods Questions
Post by: Mao on September 15, 2009, 10:28:09 pm: Correction: $Pr(X<2) = Pr(0 \le X \le 1) = Pr(X = 0) + Pr (X = 1)$