[vineet.nitd]

Originally Posted by warrior

take a= 72/b and sub in second eq y = b^(a^1/2). Take log on both sides and diff

we get b= e^2 ( b= 7.46 and a = 9.64) and b= 1

obviously we have min at b= 1 and max at b= 7.46

so any of 6 and 8 can give max

b=6 a= 12 a^1/2 = 3.46 y= 492
b=8 a=9 a^1/2 = 3 y= 512

so max at b= 8

I have the same solution here .... ...

But ur graph is not correct ....there will be step changes since b can take only whole number values ...and hence the graph cant be a continuous one . as for intermidiate values y is not defined ...

[vineet.nitd]

Originally Posted by varun nakra1

If it is the latter(posting good quesns) then lemme start with two of mine

2)Two points are taken at random on the given straight line AB of length 'a'. Prove that the probability of their distance exceeding a given length 'c' (<a) is equal to (1-c/a)^2.

Lets take the line segment and divide it into 3 parts of length x,c and y

now , for point not to be contained in the c lengthened region probability = (x+y)/a and

this elongation of length or in other words shift of the extreme end points can happen on either side of the length C region with equal probabilty ... hence

net probability =( (x+y)/a) ^2

but x+c+y = a =>x+y = a-c

hence P = ((a-c)/a)^2 = (1-c/a)^2

[sakshi 00786]

Hi all ,
its nice to have such innovative thread!
I've a problem...,to star with I have a question.,
a team of minners planned to mine1800 tons of ore during a certain number of days.
due to technical difficulties in one third of the planned number of days, the team was able to achieve an output of 20 tonsof ore less than the planned output.To make up for this, the team overachieved for the rest of the days by 20 tons.Theend result was that the team completed the task one day ahead of the time. How many tons of ore did the teaminitially plan to ore per day?
well a long question but is said to analyse and interpret each line......not able to do it!
just try to solve it anyways!!!

[warrior]

The formula for the method of integration by parts is given by


or to put it in other way


The ILATE rule

Sometimes it is more convenient to express this formula using differentials:


for choosing which of two functions is to be u and which is to be dv is to choose u by whichever function comes first in this list:
I: inverse trigonometric functions: arctan x , arcsec x, etc.
L: the logarithmic function: ln x
A: algebraic functions: x^2,3x^50, etc.
T: trigonometric functions: sin x, tan x, etc.
E: exponential functions: e^x, 13^x, etc.

Then make dv the other function. You can remember the list by the mnemonic ILATE. The reason for this is that functions longer down in the list have easier antiderivatives than the functions above them.
To demonstrate this rule, consider the integral


Following the ILATE rule, u = x and dv = cos x dx , hence du = dx and v = sin x , which makes the integral become
which equals
In general, one tries to choose u and dv such that du is simpler than u and dv is easy to integrate. If instead cos x was chosen as u and x as dv, we would have the integral
which, after recursive application of the integration by parts formula, would clearly result in an infinite recursion and lead nowhere.
Although a useful rule of thumb, there are exceptions to the ILATE rule. For example, to integrate
we would set
This results in





Now apply the above concept and solve these

1. .

2. .

For list of formulas on integration see teh attachment ( I have posted the same on CAT and Discussion forum previous )

Ya that's my 1900 :g m:

[warrior]

The foci of an ellipse are at (9, 20) and (49, 55), and it touches the x-axis. What is the length of its major axis?

[prade]

x (lnx)^3 - 3(x.(lnx)^2) -6xlnx-6x


sin3xcos5x dx
= sin3x sin5x/5 - (3cos3xsin5x dx)
cos3xsin5xdx = cos3x (-cos5x/5) + (sin3x.cos5x)dx/5
I = sin3x sin5x/5 -cos3x cos5x/5) + I/5
I = 5(sin3x sin5x/5 -cos3x cos5x/5)/4

[vineet.nitd]

Originally Posted by warrior

The foci of an ellipse are at (9, 20) and (49, 55), and it touches the x-axis. What is the length of its major axis?

IS the ans 85 ??

[vineet.nitd]

These Are this year's RMO's problems conducted few weeks before ..Got them from my sis ..There are no solution for these available till date any where ..We will solve them here ....


1. Let ABC be an acute-angled triangle and let D,E,F be the feet of perpendiculars from A,B,C respectively to BC,CA,AB. Let the perpendiculars from F to CB,CA,AD,BE meet them in P,Q,M,N respectively.Prove that P,Q,M,N are collinear.


2. Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.


3. If a,b,c are +ve real numbers ,prove that
(a^2 +1/b+c) + (b^2+1/c+a) + (c^2+1/a+b) >=3

4. A 6*6 square is dissected into 9 rectangles by lines parallel to its sides such that all these rectangles have only integer sides .Prove that there are always two congruent triangles.

5. Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD;AB=3CD and the area of the quadrilateral is 4. If a circle is drawn touching all the sides of the quadrilateral,find its radius.

6 Prove that there are infinitely many +ve integers n such that n(n+1) can be expressed as the sum of two +ve squares in at least two different ways.(Here a^2+b^2 and b^2+a^2 are considered as the same representation.)

7. Let X be the set of all +ve integers greater than or equal to 8 and let f:X ®X be a function such that f(x+y)=f(xy) for all x>=4,y>=4.
If f(8 ) =9,determine f(9).

[Aarav]

Originally Posted by vineet.nitd

These Are this year's RMO's problems conducted few weeks before ..Got them from my sis ..There are no solution for these available till date any where ..We will solve them here ....


1. Let ABC be an acute-angled triangle and let D,E,F be the feet of perpendiculars from A,B,C respectively to BC,CA,AB. Let the perpendiculars from F to CB,CA,AD,BE meet them in P,Q,M,N respectively.Prove that P,Q,M,N are collinear.


2. Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.


3. If a,b,c are +ve real numbers ,prove that
(a^2 +1/b+c) + (b^2+1/c+a) + (c^2+1/a+b) >=3

4. A 6*6 square is dissected into 9 rectangles by lines parallel to its sides such that all these rectangles have only integer sides .Prove that there are always two congruent triangles.

5. Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD;AB=3CD and the area of the quadrilateral is 4. If a circle is drawn touching all the sides of the quadrilateral,find its radius.

6 Prove that there are infinitely many +ve integers n such that n(n+1) can be expressed as the sum of two +ve squares in at least two different ways.(Here a^2+b^2 and b^2+a^2 are considered as the same representation.)

7. Let X be the set of all +ve integers greater than or equal to 8 and let f:X ®X be a function such that f(x+y)=f(xy) for all x>=4,y>=4.
If f(8 ) =9,determine f(9).

Here is my attempt to non-geometry problems.

2) a=23, b=5
3) (a^2 + a + 1/a) and 2 similar terms, in any case a >=1 or a <1 (a^2 + a + 1/a) > 1. Why is the equality given?
6) Took ages to do this. Put n = 4*k^4; one is obvious and another is (4*k^3)^2 + (2*k^2*(2*k^2-1))^2. I'm sure a better solution exists.
7) f(9) = f(20) = f(64) = f(16) = f(8 )

Couldn't solve (5) and got tired to even attempt (1) and (4). Will visit here again and try solving them :-)

(5) Radius is 1/CD and just above 1. Obviously, I'm not following the right path, just not striking right now and eventually had to do with co-ordinate geometry and messing with eq. of degree 8

Vineet, can you look at (3) again. LHS > 6 as (a+1/a) >= 2 for positive reals. Why 3 on RHS and that too with equality?

[prade]

2. Find the least possible value of a+b,where a,b are +ve integers such that 11 divides a+13b and 13 divides a+11b.