Author Topic: VCE Specialist 3/4 Question Thread! (Read 2546617 times) Tweet Share

Jaswinder · « **Reply #1186 on:** January 24, 2013, 11:22:12 am »

After dividing I get, $2 + \frac{3x-9}{(x-3)^2}$

$\Rightarrow 2 + \frac{a}{(x-3)} + \frac{b}{(x-3)^2}$

then a(x-3)+b=3x-9

a=3 b=0

$2x+3ln \left | x-3 \right | +c$

I don't know where I am going wrong? :/

polar · « **Reply #1187 on:** January 24, 2013, 11:34:58 am »

$2 + \frac{3x-9}{x^2-6x+9} = \frac{2x^2 - 12x + 18 +3x - 9}{x^2-6x+9} = \frac{2x^2 - 9x + 9}{x^2-6x+9}$

the numerator has a 9 instead of a 7 at the end

Jaswinder · « **Reply #1188 on:** January 24, 2013, 11:55:19 am »

when i express $2 + \frac{3x-9}{x^2-6x+9}$ in partial fractions

I get $\Rightarrow 2 + \frac{3}{(x-3)} + \frac{0}{(x-3)^2}$

sorry for the questions im really confused :/

polar · « **Reply #1189 on:** January 24, 2013, 12:04:40 pm »

$2 + \frac{3x-9}{x^2-6x+9}$ should be $2 + \frac{3x-11}{x^2-6x+9}$

$\begin{aligned} \frac{3x-11}{(x-3)^2} &= \frac{A}{x-3} + \frac{B}{(x-3)^2} \\ 3x-11 &= A(x-3) + B \\ \text{Let \;} x= 3 \\ B &= 3(3) - 11 \\ &= -2 \\ \text{Let \;} x = 4 \\ A - 2 &= 1 \\ A &= 3 \end{aligned}$

Jaswinder · « **Reply #1190 on:** January 24, 2013, 12:20:43 pm »

I get it thanks polar!

zvezda · « **Reply #1191 on:** January 26, 2013, 04:15:16 pm »

Essentials textbook Ex 1C Q9, not sure why they give the answer as 5sqrt3 plus-minus sqrt39?

Question 9: In triangle ABC, ACB has magnitude 30 degrees, AC=10cm and AB=8cm. Find the distance BC using the cosine rule

b^3 · « **Reply #1192 on:** January 26, 2013, 04:46:39 pm »

Working through the working
$\begin{alignedat}{1}8^{2} & =10^{2}+x^{2}-2\left(10\right)\left(x\right)\cos\left(30^{o}\right) \\ 64 & =100+x^{2}-10\sqrt{3}x \\ x^{2}-10\sqrt{3}x+36 & =0 \\ x & =\frac{10\sqrt{3}\pm\sqrt{\left(10\sqrt{3}\right)^{2}-4\left(1\right)\left(36\right)}}{2\left(1\right)} \\ & =\frac{10\sqrt{3}\pm\sqrt{4\times39}}{2} \\ & =\frac{10\sqrt{3}\pm2\sqrt{39}}{2} \\ & =5\sqrt{3}\pm\sqrt{39} \end{alignedat}$

Now if we look at the two answers we get, we see that they can correspond to the two triangles below (when you get answers like this, normally we check if one is negative to disregard it, but in this case they both work with the information given).

zvezda · « **Reply #1193 on:** January 26, 2013, 11:39:53 pm »

Quote from: b^3 on January 26, 2013, 04:46:39 pm

Working through the working
$\begin{alignedat}{1}8^{2} & =10^{2}+x^{2}-2\left(10\right)\left(x\right)\cos\left(30^{o}\right) \\ 64 & =100+x^{2}-10\sqrt{3}x \\ x^{2}-10\sqrt{3}x+36 & =0 \\ x & =\frac{10\sqrt{3}\pm\sqrt{\left(10\sqrt{3}\right)^{2}-4\left(1\right)\left(36\right)}}{2\left(1\right)} \\ & =\frac{10\sqrt{3}\pm\sqrt{4\times39}}{2} \\ & =\frac{10\sqrt{3}\pm2\sqrt{39}}{2} \\ & =5\sqrt{3}\pm\sqrt{39} \end{alignedat}$

Now if we look at the two answers we get, we see that they can correspond to the two triangles below (when you get answers like this, normally we check if one is negative to disregard it, but in this case they both work with the information given).

Oh my, that was so simple. All I did initially was use the sin rule (i think, dont have the working on me) to find the included angle, and then used to cosine rule to find BC. That must mean this was an ambiguous case with the sine rule if im not mistaken?

Jaswinder · « **Reply #1194 on:** January 27, 2013, 10:50:01 am »

$\frac{x^2}{(x+1)(x+2)^2}$ how would i convert this into partial fractions?

brightsky · « **Reply #1195 on:** January 27, 2013, 10:54:52 am »

A/(x+1) + (Bx+c)/(x+2)^2 or A/(x+1) + B/(x+2)^2 + C/(x+2). the two are equivalent.

Jaswinder · « **Reply #1196 on:** January 27, 2013, 11:38:50 am »

I tried that but i couldn't get all of them, only got two values -4/3 ad 1/9

polar · « **Reply #1197 on:** January 27, 2013, 02:23:15 pm »

Jaswinder · « **Reply #1198 on:** January 27, 2013, 03:54:23 pm »

thanks so much polar! but instead of (x+1) there's supposed to be (x-1) in the question. I made an error typing it :/ sowwi. Also i get how to do this but when i use the calculator it has three partial fractions

Jenny_2108 · « **Reply #1199 on:** January 27, 2013, 05:06:02 pm »

Do as Polar, you get

$\begin{aligned} \frac{x^2}{(x-1)(x+2)^2} &= \frac{A}{x-1} + \frac{B}{x+2} + \frac{C}{(x+2)^2} \\ A(x+2)^2 + B(x-1)(x+2) + C(x-1) &= x^2 \\ \text{Let \;} x=-2 \\ A(-2+2)^2 + B(-2-1)(-2+2) + C(-2-1) &= (-2)^2 \\ C &= \frac{-4}{3} \\ \text{Let \;} x=1 \\ A(1+2)^2 + B(1-1)(1+2) + C(1-1) &= 1 \\ A &= \frac{1}{9} \\ \text{Let \;} x=0 \\ A(0+2)^2 + B(0-1)(0+2) + C(0-1) &= 0 \\ 4A - 2B - C &=0 \\ \text{Since \;} A=\frac{1}{9}, C=\frac{-4}{3} \\ 4(\frac{1}{9}) - 2B + \frac{4}{3}&= 0 \\ B &= \frac{8}{9} \\ \; \\ \text{Hence \;} \frac{x^2}{(x+1)(x+2)^2} = \frac{1}{9(x+1)}+\frac{8}{9(x+2)} - \frac{4}{3(x+2)^2}\end{aligned}$

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Author Topic: VCE Specialist 3/4 Question Thread! (Read 2546617 times) Tweet Share

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