a= (2011^2 + 2012)^(1/2)
if f(x)=[x],[x] denotes greatest integer fn.
then value of f(a) is?

Question

a= (2011^2 + 2012)^(1/2) 
if f(x)=[x],[x] denotes greatest integer fn.
then value of f(a) is?

anonymous · Answer

a= $$\sqrt{2011^{2}+2012}$$

ganeshie8 · Answer

$\large    \sqrt{2011^2 +  2011 + 1}$

ganeshie8 · Answer

complete the square,
and mess wid the property  :  $(a+1)^2 - a^2 =  2a + 1$

ganeshie8 · Answer

$\large    \sqrt{(2011 + \frac{1}{2})^2 + \frac{3}{4}}$

ganeshie8 · Answer

$\large   2011 \lt     \sqrt{(2011 + \frac{1}{2})^2 + \frac{3}{4}} \lt    2012$

anonymous · Answer

what is the value of a @ganeshie8

ganeshie8 · Answer

^^look at above
it lies between 2011 and 2012

ganeshie8 · Answer

you dont need the exact value to evaluate greatest integer function

anonymous · Answer

not that..
the a in ur eq @ganeshie8

ganeshie8 · Answer

$ a = \sqrt{2011^{2}+2012} = \sqrt{2011^2 +  2011 + 1} \  =   \sqrt{2011^2 +  2\times \frac{1}{2} \times 2011 + \frac{1}{4} + \frac{3}{4}} \ =   \sqrt{(2011 + \frac{1}{2})^2 + \frac{3}{4}}$

anonymous · Answer

@ganeshie8  how can you assure it is leass than 2012 ?

ganeshie8 · Answer

use below :

$(a+1)^2 - a^2 =  2a + 1 $

ganeshie8 · Answer

the difference between any two consecutive squares is $2a + 1$

ganeshie8 · Answer

since 3/4 is no way near to 2011, 
you can be confident that  it will not go near 2012

ganeshie8 · Answer

$(2011+1)^2 - 2011^2 =  2*2011 + 1 $

ganeshie8 · Answer

to jump from *square* of 2011 to 2012, u need an extra of  2*2011 + 1
but ur 3/4 is very less - so it will not result in approaching 2012

anonymous · Answer

|dw:1396475742448:dw|  @ganeshie8   it is plus not minus