3 Unit Revising Marathon HSC '10 (1 Viewer)

cutemouse · Jan 13, 2010

The lines y=0, 3x-4y+3=0 and 3x+4y-15=0 are the sides of a triangle.

(i) Find the coordinates of the centre of the circle inscribed in the triangle.

(ii) Hence or otherwise write down the equation of the circle.

xV1P3R · Jan 13, 2010

3x - 4y + 3 = 0 => y = 3/4x + 3/4 ............. m(1) = 3/4
3x + 4y - 15 = 0 => y = -3/4x + 15/4............m(2) = -3/4

tan x = gradient
When solving for x with these gradients, we work out that the solution is the same angle, but in different quadrants

Since the triangle has one side as y = 0, therefore the triangle is isosceles

x-intercepts of two lines
(1) x = -1
(2) x = 5
Two vertices of triangle (-1,0) (5,0)
Since we know the triangle is isosceles, we can work out that the x co-ordinate of the centre of the circle is x = 2

The centre of the circle lies at the intersection of the angle bisectors
m(1) = 3/4
tan x = 3/4
x = 36.969897 degrees
To find angle bisector we halve this angle x/2 = 18.43494882 degrees

(I don't know how this turns out so nicely) tan (x/2) = 1/3

Angle bisector: y = 1/3(x - 1)
y = 1/3x - 1/3
sub x = 2
y = 1/3

Centre of circle is (2 , 1/3)

Radius is perpendicular distance to x-axis = 1/3
Equation (x - 2)² + (y - 1/3)² = 1/9

cutemouse · Jan 14, 2010

xV1P3R said:
Centre of circle is (2 , 1/3)

Radius is perpendicular distance to x-axis = 1/3
Equation (x - 2)² + (y - 1/3)² = 1/9

Wrong.

xV1P3R · Jan 14, 2010

ahh I see where I went wrong, near the end

Angle bisector: y = 1/3(x + 1)
y = 1/3x + 1/3
sub x = 2
y = 1

Centre of circle is (2 , 1)

Radius is perpendicular distance to x-axis = 1
Equation (x - 2)² + (y - 1)² = 1
Hope I didn't do something hopelessly wrong

shaon0 · Jan 14, 2010

xV1P3R said:
ahh I see where I went wrong, near the end

Angle bisector: y = 1/3(x + 1)
y = 1/3x + 1/3
sub x = 2
y = 1

Centre of circle is (2 , 1)

Radius is perpendicular distance to x-axis = 1
Equation (x - 2)² + (y - 1)² = 1
Hope I didn't do something hopelessly wrong

dw about it. Close enough. Can you post another question? Preferably more difficult than previous as questions should get sequentially harder

xV1P3R · Jan 14, 2010

Don't know if this can be classed as 3 unit but:

Question: Given quadrilateral ABCD such that Angle ABC + Angle ADC = 180 degrees
Using contradiction or otherwise, prove that A, B, C and D are concyclic points (Given you know nothing about cyclic quadrilaterals)

cutemouse · Jan 14, 2010

You want a hard one? haha

(i) Show that 1 is a root of $h^3-9h^2+8=0$ and find the other roots.

(ii) A hemi-spherical bowl has a radius of 3m. Oil is poured into the container at a constant rate of $\frac{\pi}{3}m^3/min$ . When the depth is h metres, the volume of the oil is $V=\frac{\pi}{3}(9h^2-h^3)m^3.$

(A) How deep is the oil after 8 minutes?

(B) At what rate is h increasing at this time?

shaon0 · Jan 14, 2010

xV1P3R said:
Don't know if this can be classed as 3 unit but:

Question: Given quadrilateral ABCD such that Angle ABC + Angle ADC = 180 degrees
Using contradiction or otherwise, prove that A, B, C and D are concyclic points (Given you know nothing about cyclic quadrilaterals)

Not sure is this a very good proof by contradiction.

Assume, ABC+ADC=180 deg but A,B,C,D aren't concyclic.
This must mean either A,B,C,D lie outside the circle.
Assume, A lies inside the circle.
Since, D's inside the circle. AB or CD intersect the circle.
Now, suppose AD intersects the circle at E.
This means AED is an inscribed angle that intersects arc AC but it also intersects AB. Thus, BCA=BEA ie. BDA=BEA.
Additionally, AED is an ecterior angle of triangle AED ie AED is interior angle. Thus, BDA>BEA ie contradiciton. So, D's not inside
If D's outside. BEA's an inscribed angle intersecting BA ie BCA=BEA. Thus, BDA=BEA. As same as before BDA>BEA. So, contradiction ie D isn't exterior to circle.

Thus, D's on the circle ie A,B,C,D are concyclic

That was epic, really hard to type up and think about. I've done another question like this form 1993 HSc or something. Really hard contradiciton question.

xV1P3R · Jan 14, 2010

Now, suppose AD intersects the circle at E.
This means AED is an inscribed angle that intersects arc AC but it also intersects AB. Thus, BCA=BEA ie. BDA=BEA.

I sorta got lost here thinking AED was a straight line

Anyway you could assume that 3 of the points, say A, B and C were concyclic, because any three points are. Then prove that D lies on this circle

shaon0 · Jan 14, 2010

xV1P3R said:
I sorta got lost here thinking AED was a straight line

Anyway you could assume that 3 of the points, say A, B and C were concyclic, because any three points are. Then prove that D lies on this circle

AED's an angle. I didn't write Angle AED as it's too long. Yeah, thats what i did. I proved for D outside or inside the circle and lead to contradictions.

addikaye03 · Jan 15, 2010

Someone post a Q. . .

cutemouse · Jan 15, 2010

Solve for x:

$\frac{|2x-1|}{x} \ge x+1$

shaon0 · Jan 15, 2010

jm01 said:
Solve for x:

$\frac{|2x-1|}{x} \ge x+1$

(2x-1)/x >= x+1
x(2x-1)>= x^2(x+1)
2x^2-x>=x^3+x^2
x^3-x^2+x=<0
x(x^2-x+1)=<0
x((x-1/2)^2+3/4)=<0
x<0 since x E R
OR
-2x^2+x>=x^3+x^2
x^3+3x^2-x=<0
x(x^2+3x-1)=<0

0< x =< 0.5*(sqrt(13)-3) or x =< -0.5 (3+sqrt(13))

cutemouse · Jan 15, 2010

shaon0 said:
(2x-1)/x >= x+1
x(2x-1)>= x^2(x+1)
2x^2-x>=x^3+x^2
x^3-x^2+x=<0
x(x^2-x+1)=<0
x((x-1/2)^2+3/4)=<0
x<0 since x E R
OR
-2x^2+x>=x^3+x^2
x^3+3x^2-x=<0
x(x^2+3x-1)=<0

0< x =< 0.5*(sqrt(13)-3) or x =< -0.5 (3+sqrt(13))

NB: x<0 is not a solution.

NEW QUESTION:

If $sinx - 7cosx=-5 , \ 0 \le x \le 2 \pi$ , find x correct to 2 dp.

Shadowdude · Jan 15, 2010

Finally, a question I can answer!

$sinx-7cosx=-5,\; 0\leq x\leq 2\pi \\ R=\sqrt{a^{2}+b^{2}}\\ =\sqrt{1^{2}+(-7)^{2}}\\ =\sqrt{1+49}\\ =\sqrt{50}=5\sqrt{2}\\ \\ tan\alpha =\frac{b}{a}=7\\ \alpha=81.86989765\\ \\ sinx-7cosx=5\sqrt{2}sin(x-\alpha)=-5\\ 5\sqrt{2}sin(x-\alpha)=-5\\ \sqrt{2}sin(x-\alpha)=-1\\ sin(x-\alpha)=-\frac{1}{\sqrt{2}}\\ x-\alpha=-\frac{\pi}{4}=-45\\ x-81.86989765=-45\\ x=36.86989765\\ \\ x=36.87\; (rounding\,to\,2\,decimal\,places)$

I hope that's right. Looks right.

Anyway, my question I shall add to the fray is:

$Solve\,the\,equation\,2x^{3}-7x^{2}-12{x}+45=0\,given\,that\,two\,of\,its\,roots\,are\,equal.$

Yeah, I'm very unoriginal...

cutemouse · Jan 15, 2010

Shadowdude said:
Finally, a question I can answer!

$sinx-7cosx=-5,\; 0\leq x\leq 2\pi \\ R=\sqrt{a^{2}+b^{2}}\\ =\sqrt{1^{2}+(-7)^{2}}\\ =\sqrt{1+49}\\ =\sqrt{50}=5\sqrt{2}\\ \\ tan\alpha =\frac{b}{a}=7\\ \alpha=81.86989765\\ \\ sinx-7cosx=5\sqrt{2}sin(x-\alpha)=-5\\ 5\sqrt{2}sin(x-\alpha)=-5\\ \sqrt{2}sin(x-\alpha)=-1\\ sin(x-\alpha)=-\frac{1}{\sqrt{2}}\\ x-\alpha=-\frac{\pi}{4}=-45\\ x-81.86989765=-45\\ x=36.86989765\\ \\ x=36.87\; (rounding\,to\,2\,decimal\,places)$

I hope that's right. Looks right.

1) There are 2 solutions
2) Your answer should be in radians.

cutemouse · Jan 15, 2010

$Let P(x)=2x^{3}-7x^{2}-12{x}+45 \\ \therefore P'(x)=6x^2-14x-12 \\ \text{Now P(x)=0 has a root of mult. 2 and P'(x)=0 has a root of mult. 1} \\ \therefore \ If \ P'(x)=0 \ \therefore 2(3x^2-7x-6)=0 \\ \therefore 2(x-3)(3x+2)=0 \\ \therefore \ x=3 \ or \ \frac{-2}{3} \\ Now \ P(3)=0 \ and \ P(\frac{-2}{3}) \neq 0 \\ \\ \therefore \ P(x)=(x-3)^2(2x+5) \\ \therefore \ \text{Roots of } P(x)=0 \ are \ x=3,3,-2\frac{1}{2}.$

Q: Find the derivative of $e^{ln2x}}$

kaz1 · Jan 15, 2010

jm01 said:
$let p(x)=2x^{3}-7x^{2}-12{x}+45 \\ \therefore p'(x)=6x^2-14x-12 \\ \text{now p(x)=0 has a root of mult. 2 and p'(x)=0 has a root of mult. 1} \\ \therefore \ if \ p'(x)=0 \ \therefore 2(3x^2-7x-6)=0 \\ \therefore 2(x-3)(3x+2)=0 \\ \therefore \ x=3 \ or \ \frac{-2}{3} \\ now \ p(3)=0 \ and \ p(\frac{-2}{3}) \neq 0 \\ \\ \therefore \ p(x)=(x-3)^2(2x+5) \\ \therefore \ \text{roots of } p(x)=0 \ are \ x=3,3,-2\frac{1}{2}.$

q: Find the derivative of $e^{ln2x}}$

2

addikaye03 · Jan 15, 2010

jm01 said:
$Let P(x)=2x^{3}-7x^{2}-12{x}+45 \\ \therefore P'(x)=6x^2-14x-12 \\ \text{Now P(x)=0 has a root of mult. 2 and P'(x)=0 has a root of mult. 1} \\ \therefore \ If \ P'(x)=0 \ \therefore 2(3x^2-7x-6)=0 \\ \therefore 2(x-3)(3x+2)=0 \\ \therefore \ x=3 \ or \ \frac{-2}{3} \\ Now \ P(3)=0 \ and \ P(\frac{-2}{3}) \neq 0 \\ \\ \therefore \ P(x)=(x-3)^2(2x+5) \\ \therefore \ \text{Roots of } P(x)=0 \ are \ x=3,3,-2\frac{1}{2}.$

Q: Find the derivative of $e^{ln2x}}$

Although i would have used the same method. That's not a 3U method. Let roots be: a, a, b, c.

Answer: e^(ln2x)=2x therefore d/dx(2x)=2

Question: (easy one) prove nCk+nC(k-1)=(n+1)Ck

3 Unit Revising Marathon HSC '10 (1 Viewer)

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