hiiiiii

 

Well i want to ask a question .. is the question has arg(mod(z)) ??/

 

If it is so then according to me the answer shu dbe zero ..

 

Tell me .. i am waitng for ur reply ..

 

cheers

 

 

hiiiiiii

 

There have beeen correct answers to this query already .. however lemme make things a bit more clear and neat ...

 

well ... now we know x² + x + 1 divides g(x²) and x²h(x²)

 

Also x² + x + 1 is (x-)(x-²)

 

So this means that  and ² must make g(x²) and x²h(x²) zero ...

 

Now, g(²) = g(⁴) = 0                ..... (1)

 

and,  ²h(²) = ^4.h(⁴) = 0     ...... (2)

 

Now ⁴ =    

 

And thus from (1) we get g() = 0

 

And from (2) we get .h() = 0 or since  is not zero

 

                      we have h() = 0

 

So, the correct answer is (a)

 

cheers

 

 

And so we get

hiiiiii

 

Your queries have been answered on the post

 

Just check the link .. okay ..

 

cheers

 

hiiiiiii

 

Well dude there is no need to take tutions if u decide to work hard and regularly on ur own ..

The tutions are sugggested so that a regular routine of problem solving is maintained .. and for no other reason ..

 

Also we ppl here at goIIT are to help u ppl only .. well u just need to post a query in case of any doubt and there are so many ppl to help u .. rite ..

 

Moreover there is study material also available on the site ... so u need not worry at all .. if u have some problems avoiding tutions then u can easily manage without them also .. okay .. moreover lemme tell u a expert of ours made it into JEE without any coaching ...so just relax ..

 

I hope this helps ..

 

cheers

 

hiiiiii

 

Is ur question .. to solve the equation 2.x + 2.[x] = 1

 

well if yes . thn ...

 

let us try ...

 

we take few cases

 

case 1:  x > 1/2

 

so 2[x] >= 0

 

also 2x > 1

 

and thus 2.x + 2.[x] = 1 is not possible ..

 

case 2:  0<=x<1/2

 

[x] = 0 in this case

 

and 2x<1

 

thus 2.x + 2.[x] = 1 is again not possible ..

 

case 3:  x < 0

[x] <= -1

 

and 2x < 0

so, again 2.x + 2.[x] = 1 is not possible ..

 

case 4: x = 1/2

 

Now [x] = 0

and 2x = 1

 

thus, 2x + 2[x] = 2.(1/2) + 0  =1

 

and thus the only solution is x = 1/2 ..

 

I hope this helps u ..

 

cheers

 

 

 

hiiiiii

 

well .. hyperbolic functions are sometimes useful in solving questions....and thus it is better if u know about them ..... ok ...

 

However the imaginary and real part of trignometric function are not part of JEE syllabus .. and trust me will not be of any use in JEE at any stage ...

 

so u need not take tension  ... just read the hyperbolic functions ..

 

cheers

 

hiiiii

 

The discussions followed ur question have answered all ur queries ...

 

rite ? 

 

Even then if there are some specific queries u shud post as a separate query as suggested by the moderator .. rite ..

 

 

I hope ur questions are clear ..

 

cheers

 

hiiiiiii

 

Well indeed FIITJee is one of the premium institutes in the country .. and it is good .. no doubts on that ...

 

 

Well but a word of caution here .. do not get carried away with just the fact that u join a good institute .at the end wat will matter is ur hard work and u shud be ready to work hard ...

 

The teachers can guide u , but the ultimate thing is in ur hand and u shud work properly to make it into JEE ..

 

OKies .. cheers

 

hiiiiii

 

well ... dude this is a difficult one . well, the preparation for IIT shud be systematic and the hard work is the key of success. The focus of the preparation shud be understanding of the concepts and not just the cramming of things ....

 

So, my dear frnd ... u shud ideally have a plan of action in mind , u shud be ready to work hard and u shud be regular at work .. and results will follow ...

 

I hope this helps u ..

 

cheeers

 

hiiiii

 

Well not at all ...

 

It is for those who can work hard .. take me for example .. well i was a normal student .. not very smart .. but thn i worked hard .. n see the results ..

 

I was jee 89 .. the way to crack jee is to work harad and systematically ..

 

By systematically i mean that u shud try to be regular in ur work .... n sucess will follow ..

 

i Hope this reply help s

 

cheers

 

 

hiiiiiii

 

well KAB has told the rite approach .. lemme complete the question ..

 

 ₀ⁿ [x]dx=₀ ¹ (0)dx+₁² (1)dx+........+  _n-1^[n  (n-1)dx

 

I hope u are clear how this can be broken .. rite ..

 

Now,    ₀ⁿ [x]dx=₀ ¹ (0)dx+₁² (1)dx+........+  _n-1^[n  (n-1)dx

     

                        = 0 + 1.1 + 2.1 + 3.1 + .... + (n-1).1

 

                        = 1 + 2 + 3 + ..... + (n-1)

 

                       = n(n-1)/2

 

I hope the answer is clear ..

 

cheers

 

hiiiii

 

we at goIIT answer only one query at one time ..

 

I am also not sure about the (2) question .. i donno wat it means ..

 

well .. for the fisrt one ..

 

 

sin3x + cos4x - 4sin7x = cos10x + sin17x

 

or, sin3x - sin17x - 4sin7x = cos10x - cos4x

 

or, -2sin7x.cos10x - 4sin7x = -2sin3x.sin7x

 

or,  sin7x( cos10x - sin7x + 2) = 0   

 

so, either sin7x = 0 or cos10x - sin7x + 2 = 0

 

if sin7x = 0 .. this means x = 0 , /7 , 2/7 , 3/7 , 4/7 , 5/7 , 6/7 ,  

 

else if cos10x - sin7x + 2 = 0

 

this clearly means that sin7x = 1 and cos10x = -1 which is not possible ..

 

Thus the solution is x = 0 , /7 , 2/7 , 3/7 , 4/7 , 5/7 , 6/7 ,

 

cheers

 

hiiiii

 

Well the answer is (b)

 

In a equilateral triangle with unit area ..

 

the area of the triangle DEF is 1/4 ..

 

so the area in general shud be close to 1/4 only ... it cannot be half in any case ..

 

cheers

 

hiiiii

 

When I first read this question i cud not understand ... but after some time I realise wat u mean ...

 

Well it is just impossible .. never get confused ..

 

The example given by cab shud answer ur doubt i guess ...

 

ok ... cheers

 

 

hiiiiii

 

no .. the rule of thumb is that a function has to be continous if we wanna check its differentiablity ... just remember this ..

 

do not get confused at any stage and just think that if function is not continous then it cannot be differentiable .. rite ..

 

Now about this function .. here the function is not defined at sqrt(2) .. then how can u say about it differentiablity .. i do not know how u have calculated the left and right hand derivative .. ??

 

okay ... i hope u get my point ..

 

 

cheers