24here is the link for 1st two
http://www.targetiit.com/iit-jee-forum/posts/area-262.html
find the Domain and Range(of course give explanation)these are also last yrs question
1)f(x)=limn->∞cos(n! 2Πx)
2)g(x)=limn->∞cosn( 2Πx)
3)h(x)=limn->∞{cos(nx) + 2}1/n
-
UP 0 DOWN 0 0 7
7 Answers
24
3yeah it is a weird ,i remember that this type of qn was created by nishant sir.
1061)f(x)=limn->∞cos(n! 2Πx)
2)g(x)=limn->∞cosn( 2Πx)
3)h(x)=limn->∞{cos(nx) + 2}1/n
(1)
CASE I: x is rational
Let x=p/q where p,q is an integer
as n→∞, so n! will cancel out any possible value of q
So the expression inside cos will be cos(2mÎ ) where m is an integer
So, its value will be 1
CASE II: x is irrational
Then for no value of n! the irrational value of x can be canceled. So,
here cos() can take any value from -1 to 1 depending upon the value of x
So, domain of x = R
and range = (-1,1]
(2)
x can take any value. And the range will be [-1,1]
(3) x can take any value and range will be 1