Check the graph below, i think that will give you all the answer.

Draw the line y=-x

and on that line draw sin2x graph, this is what your graph would exactly be.

As you know, mode always give a positive value. so, |2x-5| at min can be zero.

So, max value possible is 2.

and ofcourse, the |2x-5| can result into infinity. So, -infinity is lower limit.

Range: (-infinity, 2]

it's easy to analyze.

Whenever, you have variables in exponent, simply take log to ease your headache.

so, taking log:

log y = x*log3.

now, diff

(1/y) dy/dx = log3.

now, is it clear from where log3 came into picture :)

ok, apply product rule on right side, you'll still get the same ans.

you have x*ln3.

diff it as per product rule:  x*(d(ln3)/dx) + ln3 (dx/dx)

As, ln3 is a constant. So, it's diff. will be zero.

So, we are left with ln3 only.

Hope it's clear now.

Feel free to ask/nudge.

@Filokak: it's not f(-x) = f(x)

it's actually, f(g(-x)) = f(g(x)). So, you can't say, it's an even function.

Both are different things. Here, Sin and Cos can be treated as a different funtion.

Ok, try this way:

put x=x+pi

f(-Sinx) = f(-Cosx).

Also, we have, f(-Sinx) = f(Sinx) = f(Cosx)

=> f(Cosx) = f(-Cosx).

So, it can be said as constant function b'cos to satisfy the same values for Cos and Sin, there are some particular values are possible only.

It's worth thinking a bit,

See, in n-digit number each place has 3 options only,i.e, 2,5 and 7.

so, for a 3 digit number u'll have how much possible numbers: 3*3*3 = 3^3

for an n-digit number, we'll have 3^n numbers possible.

so, all what we need is : 3^n >= 900.

i hope now you can easily find n. :)

put x= -x,

f(sin(-x)) = f(cos(-x)).

so, f(-Sinx)) = f(cosx)

we get, f(-Sinx)) = f(Sinx)

It implies, it can be true only when f(x) is constant, beacuse it's giving f(-1/2) =f(1/2) and f(-1) =f(1)

and so on...

Possible for constant function only. => f(x) =c.

f(x) = 1- f(1/x)

f(x) = 2 f(1+2x)

so, putting x=1,  f(1) = 1-f(1)

=>f(1) = 1/2.

and, f(1) = 2 f(3)

=>f(3) =1/4.

putting, x=1/2.

f(1/2) = 1-f(2)  and f(1/2) = 2 f(2)

=> f(2) = 1/3.

hence, we can say, f(x) = 1/(1+x)

please read the last post first.
I have told that graph of cosx approaches 1 for both +0 and -0.
But it's greatest integer is always 0.
Because it's never exactly 1.
That's what i have written.
Hope it's clear now. :)

please read the last post first.
I have told that graph of cosx approaches 1 for both +0 and -0.
But it's greatest integer is always 0.
Because it's never exactly 1.
That's what i have written.
Hope it's clear now. :)

dude!! what i am trying to make u understand that,

cos(-x) = Cos x.

Do u remember that?

First think of it.

So, cos x where x is approaching from negative side results in cos x

see, you can see it from graph of cosx, what value it attains from +0 as well as -0.

u'll find it's just near to 1 and not 1.

clear now?

look, it's easy :

first when x tends from positive side to zero.

greatest integer of cosx will result in zero =>sin0 =0 absolutely,

now for negative side,

as cos is positive in first and 4th quad.

and it's value is always less than 1,

so, it again results in zero when greatest integer is applied to it.

So, everytime your numerator is ablsolutely, Sin0, means 0.

Hence, the ans.

Hope you got it now. :)

it can be solved as:

((x-1)*(x^2+x+1))/x >=0

now, x^2+x+1 is always greater than 0.

so, we are just to check (x-1)/x >=0

=> x not equal to 0.

and (x-1) >=0

hance, ans is x>=1

let y= i^i

logy = i log(i)

i=e^(i*pie/2)

so, logy= i*i*pie/2 loge

=> log y = -pie/2.

now, you have to just take antilog to get y.

Note: base is "e" here.

Already told you.

I have posted the complete solution.

Refer to the post :http://www.goiit.com/posts/list/algebra-domain-of-2-x-2-y-2-1019436.htm

2^y = 2 - 2^x
taking log:
ylog2 = log (2-2^x)

y = log (2-2^x)  with base 2.

Now, properties for log t  are:

1)t>0

2) base should be positive and not equal to 1. (which is 2 in our case)

So, 2-2^x >0

=> 2 > 2^x

=> x<1. (domain)

Even profs do not agree with decisions..

There are a lots of problems coming on.

See, How they are sorting it out ..

Do read the story :

http://timesofindia.indiatimes.com/NEWS/India/IIT-issues-clarification-on-mistakes-in-JEE-papers-/articleshow/5884640.cms

These are the clarification given by IIT.

Look at this problem in this way:

(a^n)/n! = (a/1)(a/2)........(a/a)......(a/k)....

these all the terms are in multiplication.

So, in denominator, it'll start from 1 and in b/w it'll be the time when denominator will be equal to a and then terms in deno starts increasing.

The number in denominator will approach to infinity.

and (a/n) where n is close to infinity is Zero.

So, the whole term multiplied by 0 becomes 0.

Hence, the ans is ZERO.