Preliminary mathematics marathon (2 Viewers)

ohexploitable · Apr 12, 2010

The previous preliminary maths marathon was just for "basic algebra and arithmetic", this one is for more stuff.

Previous one: http://community.boredofstudies.org/12/mathematics/239269/basic-algebra-arithmetic-marathon.html

For those who don't know how to play, you just answer the questions and post your own ones. Simple.

To start off:

$\text{Use }t=tan\frac{x}{2}\text{ to find all the solutions to the equation }\frac{1-sinx}{cosx}=\frac{1}{2}$

UnrealAnchovies · Apr 12, 2010

Still working on the answer. LOL. Trig ain't my strong point.

Here's a question whilst I work on your one:

i) Find the coordinates of the CENTRE (C) and the RADIUS of the circle x^2+y^2-2x+4y-4=0.

Find the length of the perpendicular from C (Centre) on to the straight line 5x-12y+10=0 and hence prove this line is a tangent to the circle.

ohexploitable · Apr 12, 2010

After completing the square, you get (x-1)²+(y+2)²=9.

So the centre is C(1,-2) and radius is r=3.

For the second part, you use the perpendicular distance formula to find that it's 3, the same as the radius, so it only touches the circle one, therefore it's a tangent to the circle.

UnrealAnchovies · Apr 12, 2010

Considering I really need practice at trig equations, especially the t formulas (seeing as my teacher has yet to teach them), I'll post up another:

The point A(a,b) divides the interval from B(3,-2) to C(0,4) in the ratio M:1 internally.
i) Write down the coordinates of the point A in terms of M.
ii) If A lies on the line y=2x+3, determine the value of M and hence find the coordinates of A.

ohexploitable · Apr 12, 2010

$A\left(\frac{2}{m+1},\,4-\frac{6}{m+1}\right)$

For the second part, just substitute point A and solve for m, I got m=11.

UnrealAnchovies · Apr 12, 2010

Yeah, that's the right answer.

ohexploitable · Apr 12, 2010

New Question:

Find the normal to the curve y=x²-5x+4 at the point (2,-2).

UnrealAnchovies · Apr 12, 2010

I'm pretty sure I'm wrong, but I got y= -x

smileygirl · Apr 12, 2010

UnrealAnchovies said:
I'm pretty sure I'm wrong, but I got y= -x

I think that's the equation of the tangent, I got y=x-4...

UnrealAnchovies · Apr 12, 2010

smileygirl said:
I think that's the equation of the tangent, I got y=x-4...

I see. Anyways,

new question:

The normal and the tangent at P(2,-2) on the curve y = x^3-x^2-6 cuts the x-axis in N and T respectively. Find the area of triangle PNT.

adomad · Apr 12, 2010

for the Lolz

ohexploitable said:
$\text{Use }t=tan\frac{x}{2}\text{ to find all the solutions to the equation }\frac{1-sinx}{cosx}=\frac{1}{2}$

$LHS= \frac{1-\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}}$
= $\frac{\frac{1+t^2-2t}{1+t^2}}{\frac{(1-t)(1+t)}{1+t^2}}$
= $\frac{1+t^2-2t}{(1-t)(1+t)}$
= $\frac{(t-1)^2}{-(t-1)(1+t)}$
= $\frac{(t-1)}{-(1+t)}$
= $\frac{(t-1)}{-(1+t)} =\frac{1}{2}$
$\therefore -1-t=2t-2$
$3t=1$
$\therefore tan(\frac{x}{2})=\frac{1}{3}$
$\therefore x= 2tan^{-1}(\frac{1}{3}) +2n\pi$
$\therefore x= 36^\circ 52' $ $for 0<x<2\pi$
$$test x=180^\circ ... $ $ test is false \therefore x=180^\circ $ $is not a solution$

ohexploitable · Apr 12, 2010

smileygirl said:
I think that's the equation of the tangent, I got y=x-4...

Yar, it's this.

hscishard · Apr 12, 2010

decent questions.
Bout time, the other one was stupid

hscishard · Apr 12, 2010

unrealanchovies said:
i see. Anyways,

new question:

the normal and the tangent at p(2,-2) on the curve y = x^3-x^2-6 cuts the x-axis in n and t respectively. Find the area of triangle pnt.

65/4?

hscishard · Apr 12, 2010

New question:

A cubic curve has a minimum point at point (-2,-15), a maximum point at (1,12)
and it cuts the y axis at 5. What is the equation of the cubic?

A good question

hscishard · Apr 12, 2010

UnrealAnchovies said:
Still working on the answer. LOL. Trig ain't my strong point.

Here's a question whilst I work on your one:

i) Find the coordinates of the CENTRE (C) and the RADIUS of the circle x^2+y^2-2x+4y-4=0.

Find the length of the perpendicular from C (Centre) on to the straight line 5x-12y+10=0 and hence prove this line is a tangent to the circle.

5(1) -12(-2)+10
--------------------
root 5
=39/root5

Does no equal to 3, hence not a tangent.
WTF. What did I do wrong?

OOOO used coodinates with the sqaure bit.

39/ root 25 +144
= 39/root 169
=39/13
=3.

damn, been a long time since i've used that forumla

smileygirl · Apr 13, 2010

hscishard said:
5(1) -12(-2)+10
--------------------
root 5
=39/root5

Does no equal to 3, hence not a tangent.
WTF. What did I do wrong?

OOOO used coodinates with the sqaure bit.

39/ root 25 +144
= 39/root 169
=39/13
=3. damn, been a long time since i've used that forumla

I'd never actually seen that formula before till now. But I get it

AAEldar · Apr 13, 2010

Post some questions that I can answer!

Haven't done any of this stuff yet >.< Haha.

hscishard · Apr 13, 2010

AAEldar said:
Post some questions that I can answer!

Haven't done any of this stuff yet >.< Haha.

Lol yea, quite a few people are quite ahead.
Think they studied this last year:\ so...

hscishard · Apr 13, 2010

ohexploitable said:
$A\left(\frac{2}{m+1},\,4-\frac{6}{m+1}\right)$

For the second part, just substitute point A and solve for m, I got m=11.

3/m+1****

Preliminary mathematics marathon (2 Viewers)

hey

Member

hey

Member

hey

Member

hey

Member

Member

Member

HSC!!

hey

Active Member

Active Member

Active Member

Active Member

Member

Premium Member

Active Member

Active Member

Users Who Are Viewing This Thread (Users: 0, Guests: 2)