Preliminary mathematics marathon (3 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 (3 Viewers)

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