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#1 2009-03-12 18:32:28

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,305

Level of Difficulty - III Questions.

1. If the sides of a triangle are in Arithmetic Progression, then find the value of


in terms of
.

2. A circle is inscribed in an equilateral triangle of side a. Find the area of the square inscribed in this circle.

3. Find the general value of θ satisfying the equation
tan[sup]2[/sup]θ + sec2θ = 1.

4. Solve the equation
sin x - 3sin 2x + sin 3x = cos x - 3cos 2x + cos 3x.


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#2 2009-03-15 00:43:26

ganesh
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Registered: 2005-06-28
Posts: 21,305

Re: Level of Difficulty - III Questions.

5. Find the area bounded by the curve x[sup]2[/sup]=4y and the
straight line x = 4y - 2.

6. If a, b, c are the three sides of a triangle and C = 60°, prove that

7. If a>0, b>0, and c>0, prove that

.

8.Find the equations of straight lines passing through (-2, -7) and having an interept of length 3 between the striaght lines
4x+3y = 12 and 4x+3y = 3.


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#3 2009-03-15 04:42:32

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Re: Level of Difficulty - III Questions.

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#4 2009-03-15 06:13:59

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,305

Re: Level of Difficulty - III Questions.

Answer to 7:-
Correct, JaneFairfax!
grtjb6.jpg


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#5 2009-03-16 03:06:18

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,305

Re: Level of Difficulty - III Questions.

9. For what values of m does the system of equations
3x + my = m and
2x - 5y = 20
have solutions satisfying the condtions x>0, y>0?

10. Let -1 ≤  p ≤  1. Show that the equation 4x[sup]3[/sup] - 3x - p = 0 has a unique root in the interval [½, 1] and identify it.


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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#6 2009-04-13 06:25:37

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 105,262

Re: Level of Difficulty - III Questions.

Hi Ganesh;
For #9

Solve for x and y in terms of m to understand how x and y behave when m varies:

After a lot of algebraic thrashing and some trial and error:

These values of m satisfy the constraints x>0 and y>0

Last edited by bobbym (2009-04-13 06:38:13)


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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#7 2009-04-20 15:35:26

bitus
Member
Registered: 2008-10-12
Posts: 20

Re: Level of Difficulty - III Questions.

#1: tan(A/2) +tan(C/2) = (2/3)cot(B/2)

#4: x=pi/8 or 3pi/8

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#8 2009-04-20 16:56:34

bitus
Member
Registered: 2008-10-12
Posts: 20

Re: Level of Difficulty - III Questions.

#6 : cosC=1/2⇒b² -(b/2)²=c² -(a-b/2)² ⇒ab=a²+b²-c² ⇒3ab=(a+b)² -c²=(a+b+c)(a+b-c)=(a+b+c)(a+b+c-2c)=
(a+b+c)²-2c(a+b+c)⇒3ab+2c(a+b+c)=(a+b+c)²⇒3ab+3c(a+b+c)=(a+b+c)²+c(a+b+c)⇒
3(a+c)(b+c)=(a+b+c)(b+c+a+c)⇒[(b+c)+(a+c)]/(a+c)(b+c)=3/(a+b+c)⇒1/(a+c)+1/(b+c)=
3/(a+b+c).

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#9 2009-04-20 22:44:35

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 105,262

Re: Level of Difficulty - III Questions.

Hi bitus,

For #4 I think you mean


doesn't work when I plug it in.


In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
No great discovery was ever made without a bold guess.

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#10 2010-05-16 19:26:06

BO
Member
Registered: 2010-05-16
Posts: 9

Re: Level of Difficulty - III Questions.

1.from sine rule we get sinA+sinC=2sinB, or cos([A-C]/2)=2cos(B/2)
tan(A/2)+tan(c/2)=cos(B/2)/[cos (A/2) cos (C/2)]=2cos(B/2)/[sin(B/2)+2cos(B/2)]=2/[1+cot (B/2)]

2.the radius of the circle=a*tan30=a/3
Then the diameter of the square=2a/3
area=(2a/3 sin45)^2=a^2/9

Last edited by BO (2010-05-16 19:32:48)

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#11 2010-05-16 19:40:23

BO
Member
Registered: 2010-05-16
Posts: 9

Re: Level of Difficulty - III Questions.

3. tan θ=1/2

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#12 2010-05-21 03:03:22

irspow
Member
Registered: 2005-11-24
Posts: 457

Re: Level of Difficulty - III Questions.

Last edited by irspow (2010-05-21 03:04:34)

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#13 2010-05-22 00:34:14

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,305

Re: Level of Difficulty - III Questions.

Well done,
irspow!!!


It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi. 

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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