11no one interested??
A Few reasoning and assertion questions -:
f(x) = lim cos(n!2∩x) n is an integer .
n--->∞
g(x) = lim cosn(2∩x)
n--->∞
h(x) = lim {cos(n.x)+2}-n
Range of f = R1 , Domain Of f is D1
Range of g = R2 , Domain Of g is D2
Range of h = R3 , Domain Of h is D3
Which Of the following is not the whole real line ?
1) D1UD2 2) R1cUR3 3) D3 4) R2cUR3
Which is true about R3 ?
1) it has 1 point 2) It has infinitely many points
3) It has 2 points 3) It is an empty set
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UP 0 DOWN 0 0 4
4 Answers
1f(x)= 1 ; when x is rational
≠1 ; when x is irrational
proof
If x is rational, it can be written as p/q
since n→∞, n! contains q as a factor, so f(x) becomes, cos(kp2∩)
where k is n!/q, cos of an even integral multiple of ∩ is 1
If x is irrational, it cannot expressed as p/q, so, it is not an integral multiple of ∩, or even ∩/2, so it has some value between (-1,1)
Since between every two rationals there lie infinite irrationals and vice versa, it is discontinuous at infinite points.
Therefore, limit does not exist!!
[339]
1g(x)= 1 ; when x is integral
= 0 ; when x is not integral
proof
If x is integral,
cos(2∩x)=1
and cosn(2∩x) =1n =1
If x is non integral,
-1<cos(2∩x)<1
So, cosn(2∩x)= (fraction)∞ =0
Therefore g(x) is continuous at all non integral values of x and is equal to zero
and limit is not defined at integral values of x
[339]
1Aditya, what does n tend to for the question in h(x) ??
That will solve the problem.
Clearly, D1UD2 is (R-Z), which is not the whole real line,so it is one correct option.
But others need to be verified once h(x) is found