Try to analyze it's graph.

Check the graph on this link:

http://library.thinkquest.org/20991/alg2/graphs.html

now, u'll get to know that [x] is 1 for (1,2) and 2 for (2,2.5).

so, u'll get :

1.dx for (1,2) = 1

and 2.dx for (2,2.5) =1.

So, ans is 2.

Do ask again if any doubt is there.

see, x=[x] +{x}.

now, use this:

(x+1) -[x+1] + 2x = 4[x+1] - 6

3x +7=5[x+1].

3x +7=5[x] + 5.

3x +2 = 5[x].

Now, draw the graphs of y=3x+2 and y=5[x].

the intersection points will give you the desired solutions. You can try hit n trial also.

like 8/3 and 1 are two solutions. I haven't checked for more.

but for clarity try to draw graphs.

If still any doubt exists, do tell us.

see, it's simple to analyze.
since, x can't be negative it can't go to 2nd and 3rd quad
because sinx will become negative and log of negative numbers is not defined.

now, as Sinx cannot be greater than 1.
and log(m) is always negative for 0<m<1.
at m=1, log becomes zero.
For more clarification you can ask again and read the properties of log for better understanding.

@sdfgh: logm/logn is not equal to logm - logn

log (m/n) =log m - log n.

yes, differentiation is possible, logm with base n can be written as logm/logn.

Now, you can easily differentiate it by division rule.

If still any problem, feel free to ask.

the value of sinx varies regularly.
So, graph needs to be analyzwd in intervals.
first thing is that:  log(x) is defined for x>0.
So, sinx>0.
first take interval : 0<x<=pi/2.
sinx is an increasing function for this period and logx is an increasing function.
So, log(sinx) will also increase.
log1 = 0 so, log (sin(pi2))=0.
log(m) is negative when 0<m<1.
So, break it into the intervals and draw the graph.
Note: keep in mind the domain of sinx.

let  the end points of chord be (x1,y1) and (x2,y2).
mid point: (x,y)= ((x1+x2)/2),(y1+y2)/2)

centre= (3/2,-1) and radius is 1/2.
Since, the angle subended at the centre is of 90.
So, distance between the end-points of chord will be sqrt(2). (check diagramatically).
so, (x1-x2)^2 + (y1-y2)^2 = 1/2.
put the value of x1= 2x-x2 and similarly for y1.
so, we get (x-x2)^2 + (y-y2)^2 = 1/8.

now, join centre and mid point, and see carefully that angles in each half are two angles 45 and 1 angle is of 90.
apply, pythogoras theorem in half section.
((x-3)^2 + (y+2)^2)  + ((x-x2)^2 + (y-y2)^2) =1/4.

solve both equations:
u'll get x^2 -6x +y^2 +4x +13 =1/8.

If still you have any doubt do tell me...

if that is the problem then approach it in this way.
40C4 will form one plate and persons can be arranged in 3! ways in one plate.
Since, we get 10 plates, so these plates can also be arranged in 10! ways.
I am not able to get you exactly. So, if you want to store them in circular manner then arrangement will be 9!.

So, previous ans needed to be multiplied by 10!.
Do clarify your question if still there's any doubt.

Sir, she can ofcourse opt for ISRO.

After engg, there's an exam for ISRO and she need to give that exam.

You can find all the updates on this site : http://www.isro.gov.in/

Hope this is helpful

look, it's very simple.

131 : 1+3+1=5 and 5!=120.

323: 3+2+3=8 and 8!=40320.

now, i hope you know what should be 727?

7+2+7=16 and your ans is 16!.

I hope you got it now..

One triangle can be formed by using 3 points.

So, total possible triangles can be 12C3 but 7 points are collinear and if

we choose 3 points of these 7 points then triangle can't be formed.

so, these possibilities are 7C3.

hence, possible triangles are : 12C3-7C3

Pick any four persons out of 40 for 1 plate.

so, 40C4 and their arrangement in a circle is (4-1)!=3!.

so, now you are left with 36. Pick 4 out of them and their arrangement will be 3! again.

So, answer is  ((3!)^10)(40C4*36C4.........4C4).

google keeps on providing some of the very cool stuffs.

Have a look at it:

google gravity .............

http://mrdoob.com/projects/chromeexperiments/google_gravity/

:)

 Student inventor tackles challenge of hydrogen storage news

04 March 2010

Determined to play a key role in solving global dependency on fossil fuels, Tehran-born Javad Rafiee, a doctoral student in the department of mechanical, aerospace, and nuclear engineering at Rensselaer Polytechnic Institute, has developed a new method for storing hydrogen at room temperature. (See: Enabling greener transportation with graphene)

Lemelson-MIT Rensselaer Student Prize-Winner Javad Rafiee.
Rafiee has created a novel form of engineered graphene that exhibits hydrogen storing capacity far exceeding any other known material.

Well, you posted the qun under wrong category, buddy!!

Anyways, Aeroscope engg. is in very limited colleges in India in fact in a very few IITs.

Punjab Engg. College (P.E.C.) also offers aeronautical engg.

And what do you want to know about IITs?? (be specific)

Regarding scholarship, it's offered to top 2% students of a branch (may vary as per diff. IIT)

and also, other schols are offered like merit-cum-means and SC/ST etc.

ya!! i totally agree with the last answer.

As much no. of hydrogen atoms are attached with carbon atoms, more is hyperconjugation.

It shows +I effect.

Since, it doesn't have any lone pair, so no resonance effect.

Refer to the post:

http://www.goiit.com/posts/list/trignometry-i-m-having-a-prblm-to-solve-ques-of-half-angle-help-1009723.htm

I hope it helped.

hey!! try to use : Cos(2x) formulas that make it easy to get u half angle.

i.e., cos²x -1.

regarding  Tanx use : tan(2x).

that'll make it easy.If still nay problem let us know.

Since, the equation has imaginary roots.
So, on solving we get roots as omega(w) and w^2.
we know that w^3=1.
So, ans comes out to be ((w^3 +1)/(w^3))^3.
hence, ans is 8.

look, it's easy try to approach the problem in this way:
assume tanx=y.
dy=(Secx)^2.

now u have to integrate:
log(y)/(1+y^2) dy.

Later substitute y.
I think now u'll be able to solve it.

Excellent play by Indian Hockey team.
4-1 win over Pak on the day of holi ..
pleasant to watch....