Author Topic: VCE Methods Question Thread! (Read 5701541 times) Tweet Share

TimewaveZero · « **Reply #3900 on:** February 04, 2014, 04:42:27 pm »

I'm kind of stuck on this:

If a factor of P(x) = -7 = ax +5x^2 + 15x^3 bx^4
is (x^2 - 1)

find the values of a and b

Do you have to do long division or something? I'm lost

b^3 · « **Reply #3901 on:** February 04, 2014, 04:56:37 pm »

Hint: If $x^{2}-1$ is a factor of $P$ , then you know that $(x-1)$ and $(x+1)$ are also factors of $P$ as $x^{2}-1=(x-1)(x+1)$ .

additional info

Then as those two are factors, you then know by the factor theorem that $P(1)=0$ and $P(-1)=0$ . So you can now form two simultaneous equations and solve then for $a$ and $b$ .

working

$\begin{alignedat}{1}P\left(x\right) & =ax+5x^{2}+15x^{3}+bx^{4} \\ P\left(1\right) & =0 \\ 0 & =a\left(1\right)+5\left(1\right)+15\left(1\right)+b\left(1\right) \\ 0 & =a+b+20 \\ a+b & =-20\qquad[1] \\ P\left(-1\right) & =0 \\ 0 & =a\left(-1\right)+5\left(1\right)+15\left(-1\right)+b\left(1\right) \\ -10-a+b & =0 \\ b-a & =10 \\ b & =a+10\qquad[2] \\ \text{Substitute }[2]\text{ into }[1] \\ a+a+10 & =-20 \\ 2a & =-30 \\ \implies a & =-15 \\ \text{Substitute into [2]} \\ b & =-15+10 \\ \implies b & =-5 \\ \therefore a=-15,\: b=-5 \end{alignedat}$

Hope that helps

EDIT: Looking at it a second time I'm not sure why they've put the -7 in there, which is kinda confusing. Even if they were saying that the -7 were the remainder, then our original expression wouldn't be a factor of $P$ . Anyways see if what I've said gives you the correct answer and if it is just bad wording/setup of the question or not.

EDIT2: Added a nested spoiler with working.

TimewaveZero · « **Reply #3902 on:** February 04, 2014, 05:25:56 pm »

Quote from: b^3 on February 04, 2014, 04:56:37 pm

Hint: If $x^{2}-1$ is a factor of $P$ , then you know that $(x-1)$ and $(x+1)$ are also factors of $P$ as $x^{2}-1=(x-1)(x+1)$ .

additional info
Then as those two are factors, you then know by the factor theorem that $P(1)=0$ and $P(-1)=0$ . So you can now form two simultaneous equations and solve then for $a$ and $b$ .
working
$\begin{alignedat}{1}P\left(x\right) & =ax+5x^{2}+15x^{3}+bx^{4} \\ P\left(1\right) & =0 \\ 0 & =a\left(1\right)+5\left(1\right)+15\left(1\right)+b\left(1\right) \\ 0 & =a+b+20 \\ a+b & =-20\qquad[1] \\ P\left(-1\right) & =0 \\ 0 & =a\left(-1\right)+5\left(1\right)+15\left(-1\right)+b\left(1\right) \\ -10-a+b & =0 \\ b-a & =10 \\ b & =a+10\qquad[2] \\ \text{Substitute }[2]\text{ into }[1] \\ a+a+10 & =-20 \\ 2a & =-30 \\ \implies a & =-15 \\ \text{Substitute into [2]} \\ b & =-15+10 \\ \implies b & =-5 \\ \therefore a=-15,\: b=-5 \end{alignedat}$

Hope that helps

EDIT: Looking at it a second time I'm not sure why they've put the -7 in there, which is kinda confusing. Even if they were saying that the -7 were the remainder, then our original expression wouldn't be a factor of $P$ . Anyways see if what I've said gives you the correct answer and if it is just bad wording/setup of the question or not.

EDIT2: Added a nested spoiler with working.

Thankyou!!

I'm not sure why that -7 is there either $:-\$

Cort · « **Reply #3903 on:** February 04, 2014, 07:20:53 pm »

Besides knowing the rule that "for F o G to happen, ran g(x) must be a subset of dom f(x)" -- is there a reason why this must happen? My current (and ignorant) assumption is because x gives the output of y(range), plugging in x -> y, which means y-> new x? Forgive me if I'm wrong. Additionally:
Why does ran g(x) need to be smaller than the domain f(x) to occur?
Why on earth that dom F o G = domain of g(x)?

How on earth do you find the restrictions of G in this case then? The example is from the picture below.

PS. Is there a good maths textbook which gives a good conceptual understanding of maths, rather than just Essentials? I like the questions they give, the understanding.. not so much. I understand that the questions I posed is large, so if you can answer any, I'll be swell.

Thanks,
Cort.

b^3 · « **Reply #3904 on:** February 04, 2014, 08:22:47 pm »

You're on the right track with your reasoning. Think of it this way, with $f(g(x))$ you're putting an input $x$ into the $g$ , to give an output $y$ , which is then taken as the input to $f$ . We can only input things into $f$ which give an output of $f$ , that is whatever we put into $f$ should be within the maximal domain of $f$ , otherwise it'll come out as undefined (for example if we had $f(x)=\frac{1}{x-1}$ and we tried to put an output of $g$ , $g(x)=1$ into $f$ , it wouldn't work as $f$ would come out as undefined). Now what are we putting into $f$ ? The output of $g$ , so this means that the range of $g$ has to 'fit into
the domain of $f$ , so that anything we put into $g$ will give an acceptable output that can then be put into $f$ .

Now as to why the domain of $f(g(x))$ is equal to the domain of $g(x)$ . What you're originally doing is inputting into $g$ , this initial input will be your domain of the whole function, but remember it has to be accepted by $g$ and also has to give an output that is accepted by $f$ . If we know that we've restricted the function (if necessary) so that it will be accepted by $f$ , then these values we are putting into the composite function are just the domain of $g$ .

Since we want $f(g(x))$ , we need to restrict $g$ such that the range of $g$ is equal to or a subset of the domain of $f$ . Since we are restricting $g$ , we can't change $f$ so we can first look at the domain of $f$ , which is $x\leq3$ . So now we want to restrict $g$ so that it's range is $g(x)\leq 3$ . If we do a quick sketch we get the graph below.

Now if we solve for when we have $g(x)=3$ we have,
$\begin{alignedat}{1}x^{2}-1 & =3 \\ x^{2} & =4 \\ x & =\pm2 \end{alignedat}$
Now as we can see from the graph, $g$ will be less than or equal to $3$ for $-2\leq x\leq 2$ . So that becomes our restricted domain of $g$ .
$\begin{alignedat}{1}g*:\left[-2,2\right]\rightarrow \mathbb{R}, & \: g\left(x\right)=x^{2}-1\end{alignedat}$
Now now matter what you put into $g$ given the new domain, you'll always get an output that can then be 'accepted' into $f$ .

Hope that helps

Cort · « **Reply #3905 on:** February 05, 2014, 09:13:39 am »

Quote from: b^3 on February 04, 2014, 08:22:47 pm

You're on the right track with your reasoning. Think of it this way, with $f(g(x))$ you're putting an input $x$ into the $g$ , to give an output $y$ , which is then taken as the input to $f$ . We can only input things into $f$ which give an output of $f$ , that is whatever we put into $f$ should be within the maximal domain of $f$ , otherwise it'll come out as undefined (for example if we had $f(x)=\frac{1}{x-1}$ and we tried to put an output of $g$ , $g(x)=1$ into $f$ , it wouldn't work as $f$ would come out as undefined). Now what are we putting into $f$ ? The output of $g$ , so this means that the range of $g$ has to 'fit into
the domain of $f$ , so that anything we put into $g$ will give an acceptable output that can then be put into $f$ .

Now as to why the domain of $f(g(x))$ is equal to the domain of $g(x)$ . What you're originally doing is inputting into $g$ , this initial input will be your domain of the whole function, but remember it has to be accepted by $g$ and also has to give an output that is accepted by $f$ . If we know that we've restricted the function (if necessary) so that it will be accepted by $f$ , then these values we are putting into the composite function are just the domain of $g$ .

Since we want $f(g(x))$ , we need to restrict $g$ such that the range of $g$ is equal to or a subset of the domain of $f$ . Since we are restricting $g$ , we can't change $f$ so we can first look at the domain of $f$ , which is $x\leq3$ . So now we want to restrict $g$ so that it's range is $g(x)\leq 3$ . If we do a quick sketch we get the graph below.

I'm sorry, but that logic always applies right? Since g(x) is larger than f(x), we'll always have to restrict it to the domain of the f(x)?

b^3 · « **Reply #3906 on:** February 05, 2014, 12:32:14 pm »

Quote from: Cort on February 05, 2014, 09:13:39 am

Since g(x) is larger than f(x)

Remember, it's not that $g(x)$ is larger than $f(x)$ , it's that the range of $g(x)$ doesn't fit in or is equal to (i.e. equal to or subset of) the domain of $f(x)$ . But yes, if you have the case where the original two functions don't work, and then are asked to restrict it then it'll be the second function in $f(g(x))$ , as you can't make the domain of the other function bigger, and making it smaller won't achieve anything.

Sometimes it'll fit into the other function nicely and you won't have to restrict it, normally only restrict it if you're asked since you're effectively making a new function $g*(x)$ .

smile+energy · « **Reply #3907 on:** February 05, 2014, 04:07:18 pm »

I need some help. Can anyone show me how to solve this question(step by step):
Kx-y-z=8
3x+ky+2z=2
X+3y+z=-6
For what values of k, is there
a) a unique solution?
b) no solution?
c)an infinite number of solutions?

And what about this question? I tried to substitute the x-intercepts into the equation(s) but I still got the wrong answers.
a)The function F(x)=x^3+x^2+bx-64 has x-intercepts (-2, 0) and (4, 0). Find the values of a and b.
b)The functions f(x)=x^3-2x^2+ax+10 and y=6+(a+b)x-4x^2-x^3 both have (-1, 0) as an x-intercept. Find the values of a and b.
Thanks in advance

Anchy · « **Reply #3908 on:** February 05, 2014, 08:57:30 pm »

x+y+z=9
-x+2y-3z=-15
x+5y+3z=29

Come someone please show the steps in solving this simultaneously without using matrix.

Thanks.

IndefatigableLover · « **Reply #3909 on:** February 05, 2014, 09:22:14 pm »

Quote from: Anchy on February 05, 2014, 08:57:30 pm

x+y+z=9
-x+2y-3z=-15
x+5y+3z=29

Come someone please show the steps in solving this simultaneously without using matrix.

Thanks.

Label each equation as (1), (2), & (3)

x+y+z=9 (1)
-x+2y-3z=-15 (2)
x+5y+3z=29 (3)

Now we see that 'x' is the most easiest to get rid of since we want two equations with two unknowns to solve for (like normal simultaneous equations).

So let's go (1) + (2) which will give us 3y - 2z = - 6 (which we will label as [4])
Next we'll go (1) - (3) which will give us -4y - 2z = -20 (which we will label as [5])

Now we have something that looks like something we know!

So we'll go [4] - [5] which gives us 7y = 14
$\therefore y = 2$

Now let's sub the 'y' value into [4] so then we'll have
3(2) - 2Z = -6
-2z = -12
z = 6

Now we'll sub both values into (1)
x+ (2) + (6) = 9
x = 1

$\therefore x=1, y=2, z=6$

Anchy · « **Reply #3910 on:** February 05, 2014, 09:38:40 pm »

Quote from: IndefatigableLover on February 05, 2014, 09:22:14 pm

Label each equation as (1), (2), & (3)

x+y+z=9 (1)
-x+2y-3z=-15 (2)
x+5y+3z=29 (3)

Now we see that 'x' is the most easiest to get rid of since we want two equations with two unknowns to solve for (like normal simultaneous equations).

So let's go (1) + (2) which will give us 3y - 2z = - 6 (which we will label as [4])
Next we'll go (1) - (3) which will give us -4y - 2z = -20 (which we will label as [5])

Now we have something that looks like something we know!

So we'll go [4] - [5] which gives us 7y = 14
$\therefore y = 2$

Now let's sub the 'y' value into [4] so then we'll have
3(2) - 2Z = -6
-2z = -12
z = 6

Now we'll sub both values into (1)
x+ (2) + (6) = 9
x = 1

$\therefore x=1, y=2, z=6$

Thank you!

<3

swagsxcboi · « **Reply #3911 on:** February 05, 2014, 10:14:19 pm »

if 3x^2 + 4x +a = p(x+b)^2 +3 for all real values of x, find the values of a, b and p

IndefatigableLover · « **Reply #3912 on:** February 05, 2014, 10:29:46 pm »

Quote from: swagsxcboi on February 05, 2014, 10:14:19 pm

if 3x^2 + 4x +a = p(x+b)^2 +3 for all real values of x, find the values of a, b and p

Going to take a stab at this question but would you just complete the square for the LHS and then solve for a on the LHS afterwards since 'b' & 'p' would already be given when you complete the square?

EDIT:

$3x^2+4x+a$

$=3(x^2+\frac{4x}{3}+\frac{4}{9})+a-\frac{4}{3}$

$\therefore = 3(x+\frac{2}{3})^2-\frac{3a-4}{3}$ (just made 'a' have the same denominator as $\frac{4}{3}$

From here we can see that 'P' = 3 and b = 2/3
So let's solve for 'a'!

$\frac{3a-4}{3}=3$

$3a-4=9$

$3a=13$

$\therefore a = \frac{13}{3}$

So a = 13/3, b = 2/3, p = 3

(Hope that's right LOL >.<)

nhmn0301 · « **Reply #3913 on:** February 05, 2014, 10:39:19 pm »

Quote from: swagsxcboi on February 05, 2014, 10:14:19 pm

if 3x^2 + 4x +a = p(x+b)^2 +3 for all real values of x, find the values of a, b and p

Expand the RHS, p(x+b)^2 +3 = px^2 + 2pbx + (pb^2 +3 ) (I intentionally put bracket for the last 2 terms)
When you compare the above eqn with the LHS, you can start equating coefficient:
3 = p
4= 2pb
a = pb^2 +3
I think you could do the rest from here

Hope this helps!

ETTH96 · « **Reply #3914 on:** February 06, 2014, 05:55:32 pm »

Hey guys, can someone help me out with these transformations of a truncus

From y=1/x^2 --> y=8/(2x+1)^2 -1

Is this right?
- dilation factor 8 from x axis
- dilation factor 1/2 from y axis
- translation 1/2 units left
- translation 1 unit down

I'm a bit confused about dilation from the y axis.. Can someone clarify how to find this? Thanks!

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